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+% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %
+%                                                                         %
+% Project Gutenberg's Leçons de Géométrie Supérieure, by Ernest Vessiot   %
+%                                                                         %
+% This eBook is for the use of anyone anywhere at no cost and with        %
+% almost no restrictions whatsoever.  You may copy it, give it away or    %
+% re-use it under the terms of the Project Gutenberg License included     %
+% with this eBook or online at www.gutenberg.org                          %
+%                                                                         %
+%                                                                         %
+% Title: Leçons de Géométrie Supérieure                                   %
+%        Professées en 1905-1906                                          %
+%                                                                         %
+% Author: Ernest Vessiot                                                  %
+%                                                                         %
+% Editor: Anzemberger                                                     %
+%                                                                         %
+% Release Date: January 24, 2011 [EBook #35052]                           %
+% Most recently updated: June 11, 2021                         %
+%                                                                         %
+% Language: French                                                        %
+%                                                                         %
+% Character set encoding: UTF-8                                           %
+%                                                                         %
+% *** START OF THIS PROJECT GUTENBERG EBOOK LEÇONS DE GÉOMÉTRIE SUPÉRIEURE ***
+%                                                                         %
+% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %
+
+\def\ebook{35052}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%                                                                  %%
+%% Packages and substitutions:                                      %%
+%%                                                                  %%
+%% book:     Required.                                              %%
+%% inputenc: Standard DP encoding. Required.                        %%
+%% babel:    French language features. Required.                    %%
+%%                                                                  %%
+%% calc:     Infix arithmetic. Required.                            %%
+%%                                                                  %%
+%% ifthen:   Logical conditionals. Required.                        %%
+%%                                                                  %%
+%% amsmath:  AMS mathematics enhancements. Required.                %%
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+%%                                                                  %%
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+%% indentfirst: Required.                                           %%
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+%% geometry: Enhanced page layout package. Required.                %%
+%% hyperref: Hypertext embellishments for pdf output. Required.     %%
+%%                                                                  %%
+%%                                                                  %%
+%% Producer's Comments:                                             %%
+%%                                                                  %%
+%%   Minor spelling and punctuation corrections are marked with     %%
+%%   \DPtypo{original}{corrected}. Errata listed in the original    %%
+%%   typed manuscript are applied with \Err{}{}. Punctuation added  %%
+%%   for uniformity is marked with \Add{}. Spelling modernizations  %%
+%%   are marked with \DPchg{}{}. Other changes are [** TN: noted]   %%
+%%   in this file.                                                  %%
+%%                                                                  %%
+%%   The original typed manuscript contained an unusually large     %%
+%%   number of abbreviations, errors, and inconsistencies. To the   %%
+%%   extent feasible, these have been regularized. Particularly,    %%
+%%                                                                  %%
+%%   1. In Chapter 3, there were two sections "3". Section numbers  %%
+%%      3--7 were incremented to 4--8.                              %%
+%%                                                                  %%
+%%   2. The original used numerals for both cardinals and ordinals. %%
+%%      The \Card{} and \Ord{}{} macros convert these to words.     %%
+%%                                                                  %%
+%%   3. Exercises were moved to the end of the respective chapters. %%
+%%                                                                  %%
+%% PDF pages: 244                                                   %%
+%% PDF page size: A4 (210 × 297 mm)                                 %%
+%% PDF document info: filled in                                     %%
+%% 50 PDF diagrams.                                                 %%
+%%                                                                  %%
+%% Summary of log file:                                             %%
+%% * Six harmless overfull hboxes.                                  %%
+%% * One underfull vbox, sixteen underfull hboxes.                  %%
+%%                                                                  %%
+%%                                                                  %%
+%% Compile History:                                                 %%
+%%                                                                  %%
+%% January, 2011: adhere (Andrew D. Hwang)                          %%
+%%              texlive2007, GNU/Linux                              %%
+%%                                                                  %%
+%% Command block:                                                   %%
+%%                                                                  %%
+%%     pdflatex x3                                                  %%
+%%                                                                  %%
+%%                                                                  %%
+%% January 2011: pglatex.                                           %%
+%%   Compile this project with:                                     %%
+%%   pdflatex 35052-t.tex ..... THREE times                         %%
+%%                                                                  %%
+%%   pdfTeXk, Version 3.141592-1.40.3 (Web2C 7.5.6)                 %%
+%%                                                                  %%
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+\usepackage{alltt}[1997/06/16]   %% boilerplate, credits, license
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+\usepackage{array}[2005/08/23]   %% extended array/tabular features
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+\usepackage{indentfirst}[1995/11/23]
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+\usepackage{graphicx}[1999/02/16]%% For a diagram,
+\usepackage{wrapfig}[2003/01/31] %% wrapping text around it,
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+% for running heads; no package date available
+\usepackage{fancyhdr}
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+  pdfkeywords={Andrew D. Hwang, Laura Wisewell,
+               Project Gutenberg Online Distributed Proofreading Team,
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+  pdfpagelabels=true,
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+
+% Set title, author here to avoid numerous hyperref warnings from accents
+\hypersetup{pdftitle={The Project Gutenberg eBook \#\ebook:%
+    L'\texorpdfstring{Leçons de Géométrie Supérieure}{Lecons de Geometrie Superieure}},
+  pdfauthor={Ernest Vessiot}}
+
+
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+\newcommand{\SetHead}[1]{%
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+        \protect\large TABLE DES MATIÈRES%
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+
+\newcommand{\ExSection}[1]{%
+  \SetHead{EXERCICES}
+  \section*{\large\Heading\MakeUppercase{Exercices.}}
+  \pdfbookmark[1]{Exercices.}{Exercices.#1}
+}
+
+\newcommand{\Chapitre}[2]{%
+  \cleardoublepage
+  \thispagestyle{empty}
+  \SetPageNumbers
+  \phantomsection
+  \section*{\LARGE\Heading\MakeUppercase{Chapitre #1}.}
+  \subsection*{\normalsize\Heading\MakeUppercase{#2}}
+  \ifthenelse{\equal{#1}{Premier}}{%
+    \addcontentsline{toc}{chapter}{Chapitre~I. #2}%
+    \SetHead{CHAPITRE~I.}%
+  }{%
+    \addcontentsline{toc}{chapter}{Chapitre~#1. #2}
+    \SetHead{CHAPITRE~#1.}%
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+}
+
+\newcommand{\SubChap}[1]{%
+  \phantomsection
+  \subsection*{\normalsize\Heading\MakeUppercase{#1}}
+  \addtocontents{toc}{%
+    \protect\subsection*{\protect\centering\protect\normalsize\protect#1}%
+  }
+}
+
+%\Section[ToC entry]{Centered Heading.}{Number.}{Run-in heading}
+\newcommand{\Section}[4][]{%
+  \medskip\par%
+  \MarginNote{#2}%
+  \phantomsection%
+  % If there's a section number, add a ToC entry
+  \ifthenelse{\not\equal{#3}{}}{%
+    \ifthenelse{\not\equal{#1}{}}{%
+      \addcontentsline{toc}{section}{#3 #1}%
+    }{%
+      \addcontentsline{toc}{section}{#3 #2}%
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+  }{}%
+  % Use #3 and/or #4 as run-in heading
+  \ifthenelse{\not\equal{#4}{}}{%
+    \ifthenelse{\not\equal{#3}{}}{%
+      #3 \RunInHeadFont{#4}%
+    }{%
+      \RunInHeadFont{#4}%
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+    #3%
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+  \quad\ignorespaces
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+\newcommand{\Paragraph}[1]{%
+  \medskip\par \RunInHeadFont{#1}\quad\ignorespaces
+}
+
+\newcommand{\ParItem}[2][]{%
+  \medskip\par%
+  \ifthenelse{\not\equal{#1}{}}{\MarginNote{#1}}{}#2 \ignorespaces
+}
+
+% Illustrations
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+  \includegraphics[width=#1]{./images/#2.pdf}
+}
+
+\newcommand{\Illustration}[2][2in]{%
+  \begin{wrapfigure}{O}{#1+0.125in}
+    \Input[#1]{#2}
+  \end{wrapfigure}%
+  \ignorespaces
+}
+\newcommand{\Figure}[2][2in]{%
+  \begin{figure}[hbt]
+    \centering\Input[#1]{#2}
+  \end{figure}%
+  \ignorespaces
+}
+
+\newcommand{\Figures}[3][2in]{%
+  \begin{figure}[hbt]
+    \centering\Input[#1]{#2}\hfil\Input[#1]{#3}
+  \end{figure}%
+  \ignorespaces
+}
+
+
+\newenvironment{Exercises}{%
+  \begin{list}{}{%
+      \setlength{\leftmargin}{\parindent}%
+      \setlength{\labelwidth}{\parindent}%
+      \setlength{\listparindent}{\parindent}%
+      \small%
+    }%
+  }{%
+  \end{list}%
+  \tb
+  \normalsize%
+}
+
+
+% Change from the book's list of errata
+\newcommand{\Err}[2]{#2}
+
+% Changes and notes made for stylistic or notational consistency
+\newcommand{\DPtypo}[2]{#2} % presumed error
+\newcommand{\DPchg}[2]{#2}  % modernization of spelling
+\newcommand{\DPnote}[1]{}
+\newcommand{\Add}[1]{\DPtypo{}{#1}}
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+
+\newcommand{\tb}{%
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+\newcommand{\Tag}[1]{\tag*{\ensuremath{#1}}}
+\newcommand{\Eq}[1]{\ensuremath{#1}}
+
+\DeclareMathOperator{\arc}{arc}
+%[** Original uses Cos and cos indiscriminately. Macros match original]
+\DeclareMathOperator{\Cos}{cos}
+
+\DeclareMathOperator{\arctg}{arctg}
+\DeclareMathOperator{\cotg}{cotg}
+\DeclareMathOperator{\tg}{tg}
+
+%[** TN: Matrix fraction]
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+  \ifthenelse{\equal{#2}{1}}{%
+    \ifthenelse{\equal{#1}{f}}{une}{un}%
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+}
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+    \ifthenelse{\equal{#3}{mes}}{deuxièmes}{deuxième}%
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+      \ifthenelse{\equal{#2}{4}}{quatrième}{% else #2 = 1
+        \ifthenelse{\equal{#3}{e}}{%
+          \ifthenelse{\equal{#1}{f}}{première}{premier}%
+        }{% Not \Ord{1}{e}
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+
+\begin{document}
+
+\pagestyle{empty}
+\pagenumbering{alph}
+
+%%%% PG BOILERPLATE %%%%
+\phantomsection
+\pdfbookmark[0]{PG Boilerplate.}{Boilerplate}
+
+\begin{center}
+\begin{minipage}{\textwidth}
+\small
+\begin{PGtext}
+Project Gutenberg's Leçons de Géométrie Supérieure, by Ernest Vessiot
+
+This eBook is for the use of anyone anywhere at no cost and with
+almost no restrictions whatsoever.  You may copy it, give it away or
+re-use it under the terms of the Project Gutenberg License included
+with this eBook or online at www.gutenberg.org
+
+
+Title: Leçons de Géométrie Supérieure
+       Professées en 1905-1906
+
+Author: Ernest Vessiot
+
+Editor: Anzemberger
+
+Release Date: January 24, 2011 [EBook #35052]
+Most recently updated: June 11, 2021
+
+Language: French
+
+Character set encoding: UTF-8     
+
+*** START OF THIS PROJECT GUTENBERG EBOOK LEÇONS DE GÉOMÉTRIE SUPÉRIEURE ***
+\end{PGtext}
+\end{minipage}
+\end{center}
+
+\clearpage
+
+
+%%%% Credits and transcriber's note %%%%
+\begin{center}
+\begin{minipage}{\textwidth}
+\begin{PGtext}
+Produced by Andrew D. Hwang, Laura Wisewell, Pierre Lacaze
+and the Online Distributed Proofreading Team at
+http://www.pgdp.net (The original copy of this book was
+generously made available for scanning by the Department
+of Mathematics at the University of Glasgow.)
+\end{PGtext}
+\end{minipage}
+\end{center}
+\vfill
+
+\begin{minipage}{0.85\textwidth}
+\small
+\pdfbookmark[0]{Note sur la Transcription.}{Note sur la Transcription}
+\subsection*{\centering\normalfont\scshape%
+\normalsize\MakeLowercase{Notes sur la transcription}}%
+
+\raggedright
+  Ce livre a été réalisé à l'aide d'un manuscrit dactylographié, dont
+  les images ont été fournies par le Département des Mathématiques de
+  l'Université de Glasgow.
+  \bigskip
+
+  Des modifications mineures ont été apportées à la présentation,
+  l'orthographe, la ponctuation et aux notations mathématiques. Le
+  fichier \LaTeX\ source contient les notes de ces corrections.
+\end{minipage}
+
+%% -----File: 001.png---Folio xx-------
+\clearpage
+\frontmatter
+\setlength{\TmpLen}{18pt}%
+\begin{center}
+\small PUBLICATIONS DU LABORATOIRE DE MATHÉMATIQUES \\[\TmpLen]
+\textsf{\bfseries De l'Université de Lyon}
+
+\vfill
+% [** Decoration]
+
+\textbf{\LARGE LEÇONS} \\[2\TmpLen]
+\footnotesize DE \\[3\TmpLen]
+\textbf{\Huge GÉOMÉTRIE SUPÉRIEURE} \\[2\TmpLen]
+\textsf{\bfseries Professées en 1905--1906} \\[2\TmpLen]
+\Large PAR M. E. VESSIOT \\[2\TmpLen]
+\textsc{\small Rédigées par M. ANZEMBERGER}
+\vfill
+% [** Decoration]
+\vfill
+\setlength{\TmpLen}{9pt}
+\footnotesize IMPRIMERIES RÉUNIES \\[\TmpLen]
+\scriptsize ANCIENNES MAISONS \\[\TmpLen]
+\normalsize DELAROCHE ET SCHNEIDER \\[\TmpLen]
+\scriptsize\textsf{\bfseries 8, rue Rachais} \\[\TmpLen]
+\footnotesize BUREAUX $\bigl\{$%
+\settowidth{\TmpLen}{\scriptsize\textit{85, rue de la République}}%
+\parbox[l]{\TmpLen}{\scriptsize\itshape%
+    85, rue de la République \\
+    9, quai de l'Hôpital} \\[9pt]
+\textsf{\bfseries\footnotesize LYON}
+\end{center}
+%% -----File: 002.png---Folio xx-------
+\clearpage
+\pagestyle{fancy}
+\fancyhf{}
+\thispagestyle{empty}
+\SetPageNumbers
+\SetHead{TABLE DES MATIÉRES}
+\tableofcontents
+
+\iffalse
+TABLE  DES  MATIERES.
+
+Pages
+
+CHAPITRE I.--REVISION DES POINTS ESSENTIELS DE LA THEORIE DES
+COURBES GAUCHES ET DES SURFACES DEVELOPPABLES:
+
+I.--Courbes gauches:
+
+1. Trièdre de Serret-Frenet............................  1
+
+2. Formules de Serret-Frenet...........................  2
+
+3. Courbure et Torsion.................................  4
+
+4. Discussion. Centre de Courbure .....................  5
+
+5. Signe de la torsion. Forme de la courbe.............  6
+
+6. Mouvement du trièdre de Serret-Frenet...............  8
+
+7. Calcul de la courbure...............................  9
+
+8. Calcul de la Torsion................................ 10
+
+9. Sphère osculatrice.................................. 11
+
+II.--Surfaces développables:
+
+10. Propriétés générales............................... 12
+
+11. Réciproques........................................ 15
+
+12. Surface rectifiante. Surface polaire............... 16
+
+CHAPITRE II.--SURFACES.
+
+1. Courbes tracées sur une surface. Longueurs d'arc et
+angles................................................. 19
+
+2. Déformation et représentation conforme.............. 20
+
+3. Les directions conjuguées et la forme \Sigma ld^2x.. 24
+
+4. Formules fondamentales pour une courbe de la surface 27
+%% -----File: 003.png---Folio xx-------
+
+CHAPITRE III.--ETUDE DES ELEMENTS FONDAMENTAUX DES COURBES
+D'UNE SURFACE.
+
+1. Courbure normale................................... 33
+
+2. Variations de la courbure normale.................. 35
+
+3. Lignes minima...................................... 40
+
+3. Lignes asymptotiques............................... 43
+
+4. Surfaces minima.................................... 48
+
+5. Lignes de courbure................................. 50
+
+6. Courbure géodésique. Propriétés des géodésiques.... 52
+
+7. Torsion géodésique. Théorèmes de Joachimsthal...... 57
+
+CHAPITRE IV.--LES SIX INVARIANTS.--LA COURBURE TOTALE.
+
+1. Les six invariants................................. 61
+
+2. Les conditions d'intégrabilité..................... 66
+
+3. Courbure totale.................................... 69
+
+4. Coordonnées orthogonales et isothermes............. 71
+
+5. Relations entre la courbure totale et la courbure
+géodésique............................................ 74
+
+CHAPITRE V.--SURFACES REGLEES.
+
+1. Surfaces développables............................. 80
+
+2. Développées des courbes gauches.................... 84
+
+3. Lignas de courbure................................. 87
+
+4. Développement d'une surface développable sur un plan
+Réciproque............................................ 89
+
+5. Lignes géodésiques d'une surface développable...... 93
+
+6. Surfaces réglées gauches--trajectoires orthogonales
+des génératrices...................................... 97
+
+7. Cône directeur. Point central. Ligne de striction.. 98
+%% -----File: 004.png---Folio xx-------
+
+8. Variations du plan tangent le long d'une génératrice.. 101
+
+9. Elément linéaire...................................... 106
+
+10. La forme \Sigma ld^2x et les lignes asymptotiques.... 110
+
+11. Lignes de courbure................................... 118
+
+12. Centre de courbure géodésique........................ 118
+
+CHAPITRE VI.--CONGRUENCES DE DROITES[**.]
+
+1. Points et plans focaux................................ 121
+
+2. Développables de la congruence. Examen des divers
+cas possibles. Cas singuliers ........................... 129
+
+3. Sur le point de vue corrélatif. Congruences de
+Koenigs. Surfaces de Joachimsthal........................ 136
+
+4. Détermination des développables d'une congruence...... 145
+
+CHAPITRE VII.--CONGRUENCES DE NORMALES.
+
+1. Propriété caractéristique des congruences de normales. 150
+
+2. Relations entre une surface et sa développée. Surface
+canal. Cyclide de Dupin. Cas singulier................... 153
+
+3. Etude des surfaces enveloppes de sphères. Correspondance
+entre les droites et les sphères. Equation de
+la cyclide de Dupin. Surface canal isotrope.............. 158
+
+4. Lignes de courbure et lignes asymptotiques. Bandes
+asymptotiques et bandes de courbure...................... 164
+
+5. Lignes de courbure des enveloppes de sphères.......... 168
+
+6. Cas où l'une des nappes de la développée est une
+développable............................................. 171
+
+CHAPITRE VIII.--LES CONGRUENCES DE DROITES ET LES CORRESPONDANCES
+ENTRE DEUX SURFACES.
+
+1. Nouvelle représentation des congruences............... 181
+
+2. Emploi des coordonnées homogènes...................... 183
+%% -----File: 005.png---Folio xx-------
+
+3. Correspondance entre les points M, M_1 de deux
+surfaces, telle que les développables de la
+congruence des droites MM_1 coupent les deux
+surfaces suivant deux réseaux conjugués homologues.... 189
+
+4. Correspondance par plans tangents parallèles....... 197
+
+CHAPITRE IX.--COMPLEXES DE DROITES.
+
+1. Eléments fondamentaux d'un complexe de droites..... 201
+
+2. Surfaces du complexe............................... 205
+
+3. Complexes spéciaux. Surface des singularités. Surfaces
+et courbes des complexes spéciaux..........  211
+
+4. Surfaces normales aux droites du complexe.......... 218
+
+CHAPITRE X.--COMPLEXES LINEAIRES.
+
+1. Généralités sur les complexes algébriques............ 220
+
+2. Coordonnées homogènes................................ 221
+
+3. Complexe linéaire.................................... 226
+
+4. Faisceau de complexes linéaires...................... 226
+
+5. Complexes linéaires en involution.................... 228
+
+6. Droites conjuguées................................... 230
+
+7. Réseau de complexes linéaires........................ 235
+
+8. Courbes d'un complexe linéaire. Leurs propriétés..... 236
+
+9. Surfaces normales aux droites d'un complexe linéaire. 240
+
+10. Surfaces réglées d'un complexe linéaire............. 243
+
+CHAPITRE XI.--TRANSFORMATIONS DUALISTIQUES. TRANSFORMATION DE
+SOPHUS LIE.
+
+1. Eléments de contact et multiplicités................. 245
+
+2. Transformations de contact. Transformations dualistiques.... 249
+%% -----File: 006.png---Folio xx-------
+
+3. Transformation de Sophus Lie........................ 255
+
+4. Transformation des droites en sphères............... 260
+
+5. Transformation des lignes asymptotiques............. 263
+
+6. Transformation des lignes de courbure............... 265
+
+CHAPITRE XII.--SYSTEMES TRIPLES ORTHOGONAUX.
+
+1. Théorème de Dupin................................... 268
+
+2. Equation aux dérivées partielles de Darboux......... 269
+
+3. Systèmes triples orthogonaux contenant une surface.. 274
+
+4. Systèmes triples orthogonaux contenant une famille de
+plans.................................................. 275
+
+5. Systèmes triples orthogonaux contenant une famille de
+sphères ............................................... 275
+
+CHAPITRE [** VIII missing].--CONGRUENCES DE SPHERES ET SYSTEMES CYCLIQUES.
+
+1. Généralités......................................... 280
+
+2. Congruences spéciales............................... 283
+
+3. Théorème de Dupin................................... 285
+
+4. Congruence des droites D............................ 289
+
+5. Congruence des droites \Delta....................... 291
+
+6. Le système triple de Ribaucour...................... 293
+
+7. Congruences de cercles et systèmes cycliques.
+Transformation de contact de Ribaucour................. 294
+
+8. Surfaces de Weingarten.............................. 301
+
+EXERCICES.                                             307
+\fi
+%% -----File: 007.png---Folio xx-------
+
+
+\Preface
+
+Ces leçons ont été professées en 1905--1906, pour répondre
+au programme spécial d'Analyse Mathématique de l'Agrégation.
+Elles ont été autographiées à la demande de mes étudiants, et
+rédigées par l'un d'eux.
+
+Peut-être pourront-elles être utiles aux étudiants désireux
+de s'initier à la géométrie supérieure, et leur être
+une bonne préparation à l'étude des livres de M.~Darboux et
+des mémoires originaux.
+
+J'ai supposé connus seulement les principes les plus
+simples de la théorie du contact; j'ai repris les points essentiels
+de la théorie des courbes gauches et de la théorie
+des surfaces, en mettant en évidence le rôle essentiel des
+formules de Frenet et des deux formes quadratiques différentielles
+de Gauss.
+
+L'objet principal de mes leçons était l'étude des systèmes
+de droites, et leur application à la théorie des surfaces.
+Il était naturel d'y joindre l'étude des systèmes de
+sphères, que j'ai poussée jusqu'aux propriétés élémentaires,
+si attrayantes, des systèmes cycliques de Ribaucour. J'ai
+insisté sur la correspondance des droites et des sphères, je
+l'ai éclairée par l'emploi des notions d'éléments de contact
+et de multiplicités, qui est également utile dans la théorie
+des congruences de droites; j'ai montré comment elle se traduisait
+par la transformation de contact de Lie.
+%% -----File: 008.png---Folio xx-------
+
+J'ai cherché à développer les diverses questions par
+la voie la plus naturelle et la plus analytique; voulant
+montrer à mes élèves comment la recherche méthodique, la discussion
+approfondie des questions même les plus simples,
+l'étude attentive et l'interprétation des résultats conduisent
+aux \DPtypo{consequences}{conséquences} les plus intéressantes.
+
+\null\hfil\hfil
+\parbox[c]{2in}{\centering
+Le 1\textsuperscript{er} Juin 1906. \\
+\textsc{E.~Vessiot}.}\hfil
+%% -----File: 009.png---Folio 1-------
+
+\mainmatter
+
+\Chapitre{Premier}{Révision des points essentiels de la théorie des Courbes Gauches et des Surfaces \DPtypo{Developpables}{Développables}.}
+
+\SubChap{I. Courbes Gauches.}
+
+\Section{Trièdre de Serret-Frenet\Add{.}}
+{1.}{} Les coordonnées d'un point d'une courbe gauche peuvent
+s'exprimer en fonction d'un paramètre~$t$
+\[
+x = f(t)\Add{,}\qquad y = g(t)\Add{,}\qquad z = h(t)\Add{.}
+\]
+Nous considérerons dans une telle courbe la \emph{tangente}, qui a
+pour paramètres directeurs $\dfrac{dx}{dt}$, $\dfrac{dy}{dt}$, $\dfrac{dz}{dt}$ et le \emph{plan osculateur}
+qui contient la tangente $\left(\dfrac{dx}{dt}, \dfrac{dy}{dt}, \dfrac{dz}{dt}\right)$ et l'accélération
+$\left(\dfrac{d^2x}{dt^2}, \dfrac{d^2y}{dt^2}, \dfrac{d^2z}{dt^2}\right)$ et dont par suite les coefficients sont les
+déterminants du \Ord{2}{e} ordre déduits du tableau
+\[
+\begin{Vmatrix}
+\mfrac{dx}{dt}     & \mfrac{dy}{dt}     & \mfrac{dz}{dt} \\
+\mfrac{d^2x}{dt^2} & \mfrac{d^2y}{dt^2} & \mfrac{d^2z}{dt^2}
+\end{Vmatrix}
+\]
+
+\Paragraph{Remarque.} Si on change de paramètre, en posant $t = \phi(u)$\Add{,}
+l'accélération nouvelle $\left(\dfrac{d^2x}{du^2}, \dfrac{d^2y}{du^2}, \dfrac{d^2z}{du^2}\right)$ est toujours dans le
+plan osculateur.
+
+Considérons en un point~$M$ d'une courbe la tangente~$MT$,
+la normale située dans le plan osculateur, ou \emph{normale principale}~$MN$,
+et la normale~$MB$ perpendiculaire au plan osculateur,
+ou \emph{binormale}. Ces \Card{3} droites forment un trièdre trirectangle
+que nous \DPtypo{appelerons}{appellerons} \emph{trièdre de Serret ou de Frenet}. L'une de
+ses faces, celle déterminée par la tangente et la normale principale,
+est le plan osculateur; celle déterminée par la normale
+principale et la binormale est le plan normal; enfin celle
+déterminée par la tangente et la binormale s'appelle le \emph{plan
+%% -----File: 010.png---Folio 2-------
+rectifiant}.
+
+Prenons sur la courbe une origine des arcs quelconques,
+et un sens des arcs croissants également quelconque. La différentielle
+de l'arc~$s$ est donnée par la formule
+\[
+ds^2 = dx^2 + dy^2 + dz^2
+\]
+d'où
+\[
+\frac{ds}{dt}
+  = ± \sqrt{\left(\frac{dx}{dt}\right)^2
+          + \left(\frac{dy}{dt}\right)^2
+          + \left(\frac{dz}{dt}\right)^2}
+\]
+et
+\[
+\left(\frac{dx}{ds}\right)^2
+  + \left(\frac{dy}{ds}\right)^2
+  + \left(\frac{dz}{ds}\right)^2 = 1\Add{,}
+\]
+$\dfrac{dx}{ds}$, $\dfrac{dy}{ds}$, $\dfrac{dz}{ds}$ sont ainsi les cosinus directeurs d'une des directions
+de la tangente, celle qui correspond au sens des arcs
+croissants; soient $a\Add{,} b\Add{,} c$ ces cosinus directeurs, nous avons
+\[
+\Tag{(1)}
+a = \frac{dx}{ds}\Add{,}\qquad
+b = \frac{dy}{ds}\Add{,}\qquad
+c = \frac{dz}{ds}\Add{.}
+\]
+
+Nous prendrons sur la normale principale une direction
+positive arbitraire de cosinus directeurs $a'\Add{,} b'\Add{,} c'$ et sur
+la binormale une direction positive de cosinus directeurs $a''\Add{,} b''\Add{,} c''$
+telle que le trièdre constitué par ces \Card{3} directions
+ait même disposition que le trièdre de coordonnées. On a alors
+\[
+\begin{vmatrix}
+a   & b   & c \\
+a'  & b'  & c' \\
+a'' & b'' & c''
+\end{vmatrix}
+= 1
+\]
+et chaque élément de ce déterminant est égal à son coefficient.
+
+\Section{Formules de Serret-Frenet\Add{.}}
+{2.}{} Il existe entre ces cosinus directeurs et leurs différentielles
+des relations importantes. Nous avons en effet
+\[
+a^2 + b^2 + c^2 = 1
+\]
+d'où en dérivant par rapport à~$s$
+\[
+\sum a\, \frac{da}{ds} = 0\Add{.}
+\]
+Mais d'après les relations~\Eq{(1)} on a
+\[
+\frac{da}{ds} = \frac{d^2x}{ds^2}\Add{,}\qquad
+\frac{db}{ds} = \frac{d^2y}{ds^2}\Add{,}\qquad
+\frac{dc}{ds} = \frac{d^2z}{ds^2}\Add{,}
+\]
+et la relation précédente s'écrit:
+%% -----File: 011.png---Folio 3-------
+\[
+\sum a\, \frac{d^2x}{\DPtypo{ds}{ds^2}} = 0\Add{.}
+\]
+
+La direction de coefficients directeurs
+\[
+\frac{d^2x}{ds^2},\quad
+\frac{d^2y}{ds^2},\quad
+\frac{d^2z}{ds^2}\quad \text{ou}\quad
+\frac{da}{ds},\quad
+\frac{db}{ds},\quad
+\frac{dc}{ds}
+\]
+est donc perpendiculaire à la tangente; d'autre part elle est
+dans le plan osculateur, c'est donc la normale principale, et
+on a des relations de la forme:
+\[
+\Tag{(2)}
+\frac{da}{ds} = \frac{1}{R}\, a'\Add{,}\qquad
+\frac{db}{ds} = \frac{1}{R}\, b'\Add{,}\qquad
+\frac{dc}{ds} = \frac{1}{R}\, c'\Add{.}
+\]
+On en déduit, pour le facteur $\dfrac{1}{R}$,
+\[
+\Tag{(3)}
+\frac{1}{R} = \sum a'\, \frac{da}{ds}\Add{.}
+\]
+De ces relations\DPnote{(2)} on tire, en multipliant par $a''\Add{,} b''\Add{,} c''$ et
+ajoutant
+\[
+\sum a''\, \frac{da}{ds} = 0\Add{.}
+\]
+D'autre part on a
+\[
+\sum aa'' = 0
+\]
+d'où en dérivant
+\[
+\sum a\, \frac{da''}{ds} + \sum a''\, \frac{da}{ds} = 0
+\]
+et par suite
+\[
+\sum a\, \frac{da''}{ds} = 0\Add{.}
+\]
+On a d'ailleurs
+\[
+\sum a''{}^2 = 1
+\]
+d'où
+\[
+\sum a''\, \frac{da''}{ds} = 0
+\]
+et les deux relations précédentes montrent que la direction
+$\dfrac{da''}{ds}, \dfrac{db''}{ds}, \dfrac{dc''}{ds}$ est perpendiculaire à la tangente et à la binormale.
+C'est donc encore la normale principale, et on a des
+relations de la forme\DPtypo{;}{:}
+\[
+\Tag{(4)}
+\frac{da''}{ds} = \frac{1}{T}\, a'\Add{,}\qquad
+\frac{db''}{ds} = \frac{1}{T}\, b'\Add{,}\qquad
+\frac{dc''}{ds} = \frac{1}{T}\, c'\Add{.}
+\]
+On en déduit, pour le facteur $\dfrac{1}{T}$,
+\[
+\Tag{(5)}
+\frac{1}{T} = \sum a'\, \frac{da''}{ds}\Add{.}
+\]
+Enfin de la relation
+\[
+\sum a'a'' = 0
+\]
+on tire
+\[
+\sum a'\, \frac{da''}{ds} + \sum a''\, \frac{da'}{ds} = 0\Add{,}
+\]
+ou
+\[
+\sum a''\, \frac{da'}{ds} = -\sum a'\, \frac{da''}{ds} = -\frac{1}{T}\Add{.}
+\]
+%% -----File: 012.png---Folio 4-------
+De la relation
+\[
+\sum a'a = 0
+\]
+on tire de même
+\[
+\sum a\, \frac{da''}{ds} = -\sum a'\, \frac{da}{ds} = -\frac{1}{R}\Add{,}
+\]
+et enfin de
+\[
+\sum a'{}^2 = 0
+\]
+on tire
+\[
+\sum a'\, \frac{da'}{ds} = 0\Add{.}
+\]
+On a ainsi \Card{3} équations en $\dfrac{da'}{ds}, \dfrac{db'}{ds}, \dfrac{dc'}{ds}$,
+\begin{align*}
+\sum a\, \frac{da'}{ds} &= -\frac{1}{R}\Add{,} \\
+\sum a'\, \frac{da'}{ds} &= 0\Add{,} \\
+\sum a''\, \frac{da'}{ds} &= -\frac{1}{T}\Add{,}
+\end{align*}
+et l'on en tire
+\[
+\Tag{(6)}
+\frac{da'}{ds} = -\frac{a}{R} - \frac{a''}{T},\qquad
+\frac{db'}{ds} = -\frac{b}{R} - \frac{b''}{T},\qquad
+\frac{dc'}{ds} = -\frac{c}{R} - \frac{c''}{T}.
+\]
+
+Les \Card{3} groupes de relations \Eq{(2)}\Add{,}~\Eq{(4)}\Add{,}~\Eq{(6)} constituent \emph{les
+formules de Serret ou de Frenet}.
+
+\Section{Courbure et \DPtypo{Torsion}{torsion}\Add{.}}
+{3.}{Interprétation de~$R$.} Considérons le point~$t$ de coordonnées
+$a\Add{,} b\Add{,} c$. Les formules~\Eq{(2)} expriment une propriété de la
+courbe lieu de ces points; cette courbe est tracée sur une
+sphère de rayon~$1$, on l'appelle \emph{indicatrice
+sphérique} de la courbe~$(C)$,
+et les formules~\Eq{(2)} montrent que \emph{la
+tangente en~$t$ à l'indicatrice sphérique
+est parallèle à la normale
+principale en~$M$ à la courbe~$C$}. Soit $u$
+l'arc de cette indicatrice compté à
+partir d'une origine arbitraire dans
+un sens également arbitraire, on aura
+\[
+\frac{da}{du} = ea',\qquad
+\frac{db}{du} = eb',\qquad
+\frac{dc}{du} = ec',\qquad (e = ±1)
+\]
+
+\Illustration[1.75in]{012a}
+\noindent d'où, en tenant compte des formules~\Eq{(2)}
+\[
+\frac{1}{R} = e\, \frac{du}{ds}\Add{.}
+\]
+%% -----File: 013.png---Folio 5-------
+
+Considérons alors les points $t$,~$t'$ correspondant aux
+points $M$,~$M'$; $\dfrac{du}{ds}$~est la limite du rapport $\dfrac{\arc tt'}{\arc MM'}$ quand $M'$~se
+rapproche indéfiniment de~$M$. L'arc~$tt'$ étant infiniment petit
+peut être remplacé par l'arc de grand cercle correspondant,
+qui n'est autre que la mesure de l'angle~$tOt'$ des \Card{2} tangentes
+infiniment voisines; c'est \emph{l'angle de contingence}; cette limite
+s'appelle la \emph{courbure} de la courbe au point~$C$; $R$~est le \emph{rayon
+de courbure}.
+
+\Paragraph{Interprétation de~$T$.} Pour interpréter~$T$, on \DPtypo{considèrera}{considérera}
+de même le lieu du point~$b$ de coordonnées $a\Add{,} b\Add{,} c$, ou \emph{deuxième
+indicatrice sphérique}. On pourra remarquer que d'après les
+formules \Eq{(2)}\Add{,}~\Eq{(4)}, \emph{les tangentes en $t$,~$b$ aux deux indicatrices
+sont parallèles à la normale principale en~$M$}. Si $v$~est l'arc
+de cette deuxième indicatrice sphérique, on trouvera comme
+précédemment que
+\[
+\frac{1}{T} = e'\, \frac{dv}{ds}\qquad (e' = ±1)
+\]
+et que $\dfrac{1}{T}$ est la limite du rapport de l'angle des plans
+osculateurs en $M$,~$M'$ à l'arc~$MM'$; c'est la \emph{torsion} en~$M$, et $T$~est
+le \emph{rayon de torsion}.
+
+\emph{Les deux indicatrices sont polaires réciproques sur la
+sphère.}
+
+
+\Section{Discussion. Centre de courbure.}
+{4.}{} Les cosinus directeurs que nous avons introduits dépendent
+de \Card{3} hypothèses arbitraires sur la disposition du
+trièdre de coordonnées, le sens des arcs croissants, et le sens
+positif choisi sur la normale principale. Si nous changeons
+ces hypothèses, et si nous désignons par $e_1, e_2, e_3$ des nombres
+égaux à~$±1$, $s$~sera remplacé par~$e_1s$, $a\Add{,} b\Add{,} c$ par $e_1a, e_1b, e_1c$;
+$a'\Add{,} b'\Add{,} c'$ par $e_2a, e_2b, e_2c$; et enfin, d'après les relations
+%% -----File: 014.png---Folio 6-------
+\[
+a'' = e_3 (bc' - cb'),\qquad
+b'' = e_3 (ca' - ac'),\qquad
+c'' = e_3 (ab' - ba'),
+\]
+$a''\Add{,} b''\Add{,} c''$ seront remplacés par $e_1e_2e_3a'', e_1e_2e_3b'', e_1e_2e_3c''$. Les
+formules~\Eq{(2)} donnent alors
+\[
+\frac{e_1\, da}{e_1\, ds} = \frac{1}{R}\, e_2a',\qquad \text{etc}\ldots,
+\]
+c'est à dire $R$~se change en~$e_2R$; et son signe ne dépend que
+de la direction positive choisie sur la normale principale.
+
+Donc le point~$C$ de la normale principale, tel que l'on
+%[** TN: MC has an overline accent; only instance, omitting]
+ait $MC = R$ ($R$~étant défini algébriquement comme précédemment),
+est un élément géométrique attaché à la courbe donnée. Ce
+point~$C$ s'appelle \emph{centre de courbure en~$M$}.
+
+Voyons maintenant~$T$. Les formules~\Eq{(4)} donnent
+\begin{align*}
+\frac{e_1e_2e_3\, da''}{\DPtypo{e}{e_1}\, ds}
+  &= \frac{1}{T}\, e_2a'\Add{,}\qquad \text{etc.} \\
+\intertext{ou}
+\frac{e_3\, da''}{ds} &= \frac{1}{T}\, a'\Add{,}\qquad \text{etc.}
+\end{align*}
+
+Donc $T$~se change en~$e_3T$; et le signe de~$T$ dépend uniquement
+de la disposition du trièdre de coordonnées. Il n'y a
+donc pas lieu de définir un centre de torsion.
+
+\Section{Signe de la torsion. Forme de la courbe.}
+{5.}{} Pour interpréter le signe de~$T$, nous allons étudier
+la rotation d'un plan passant par la tangente~$MT$ et par un
+point~$M'$ de la courbe infiniment voisin. Rapportons la courbe
+au trièdre de Serret, la tangente étant~$OX$, la normale principale~$OY$,
+la binormale~$OZ$. Alors $a = 1$, $a' = 0$, $a'' = 0$, $b = 0$,
+$b' = 1$, $b'' = 0$, $c = 0$, $c' = 0$, $c'' = 1$. Nous allons chercher
+les développements des coordonnées d'un point de la courbe
+infiniment voisin de~$M$ suivant les puissances croissantes de~$ds$,
+($ds$~étant l'arc de la courbe compté à partir du point~$O$).
+
+Nous avons
+\begin{align*}%[** TN: Added elided equations]
+X &= \frac{ds}{1}\, \frac{dx}{ds}
+   + \frac{ds^2}{2}\, \frac{d^2x}{ds^2}
+   + \frac{ds^3}{6}\, \frac{d^3x}{ds^3} + \dots\Add{,} \\
+Y &= \frac{ds}{1}\, \frac{dy}{ds}
+   + \frac{ds^2}{2}\, \frac{d^2y}{ds^2}
+   + \frac{ds^3}{6}\, \frac{d^3y}{ds^3} + \dots\Add{,} \\
+Z &= \frac{ds}{1}\, \frac{dz}{ds}
+   + \frac{ds^2}{2}\, \frac{d^2z}{ds^2}
+   + \frac{ds^3}{6}\, \frac{d^3z}{ds^3} + \dots\Add{.}
+\end{align*}
+%% -----File: 015.png---Folio 7-------
+
+Or:
+\begin{align*}
+\frac{dx}{ds} &= a = 1\Add{,} \\
+\frac{d^2x}{ds^2} &= \frac{da}{ds} = \frac{a'}{R} = 0\Add{,} \\
+\frac{d^3x}{ds^3} &= \frac{d^2a}{ds^2}
+  = \frac{1}{R}\, \frac{da'}{ds} + \frac{d\left(\dfrac{1}{R}\right)}{ds}\, a'
+  = \frac{1}{R} \left(-\frac{a}{R} - \frac{a''}{T}\right) - \frac{1}{R^2}\, a'\, \frac{dR}{ds}
+  = -\frac{1}{R^2}\Add{,}
+\end{align*}
+et de même pour les autres coordonnées. On trouve ainsi
+\[
+\Tag{(7)}%[** TN: Added brace]
+\left\{
+\begin{aligned}
+X &= ds &                    &-\frac{1}{6R^2}\, \Err{ds}{ds^3} + \dots\Add{,} \\
+Y &=    &\frac{1}{2R}\, ds^2 &-\frac{1}{6R^2}\, \frac{dR}{ds}\, ds^3 + \dots\Add{,} \\
+Z &=    &                    &-\frac{1}{6RT}\, ds^3 + \dots\Add{.}
+\end{aligned}
+\right.
+\]
+Tels sont les \DPtypo{developpements}{développements} des coordonnées, du point~$M'$ voisin
+de~$M$.
+
+Le plan que nous considérons passe par la tangente; le
+sens de sa rotation est donné par le signe de~$\dfrac{Z}{Y}$, coefficient
+angulaire de sa trace sur le plan des~$YZ$. Or,
+\[
+\frac{Z}{Y} = -\frac{ds}{3T}\, \bigl[1 + ds\, (\dots\dots)\bigr]\Add{.}
+\]
+
+\Illustration[3in]{015a}
+Ce coefficient angulaire est positif si $T\Err{\ }{<}0$, pour $s$~croissant,
+c'est à dire si le point se déplace dans la direction
+de la tangente; le plan va alors tourner dans le sens positif.
+Le point~$M'$ étant au-dessus du
+plan des~$\DPtypo{xy}{XY}$, l'arc~$MM'$ de la courbe
+est en avant du plan~$XZ$, si $T<0$; il
+est au contraire en arrière si $T>0$.
+
+Les formules~\Eq{(7)} permettent de
+représenter les projections de la
+courbe sur les \Card{3} faces du trièdre de
+Serret dans le voisinage du point~$M$.
+%% -----File: 016.png---Folio 8-------
+Nous supposerons pour faire ces projections $R > 0$ et $T < 0$.
+
+La considération des formules~\Eq{(7)} prises deux à deux
+montre que sur le plan rectifiant~$(XZ)$ la projection a au
+point~$m_1$ un point d'inflexion, la tangente inflexionnelle étant~$OX$.
+Sur le plan osculateur, la projection a au point~$m$ un point
+ordinaire, la tangente étant~$OX$; enfin sur le plan normal~$(Y\Add{,}Z)$
+la projection a en~$m_2$ un point de rebroussement, la tangente de
+rebroussement étant~$OY$.
+
+
+\Section{Mouvement du trièdre de Serret-Frenet.}
+{6.}{Remarque.} Considérons un point~$P$ invariablement lié
+au trièdre de Serret, et soient $X\Add{,} Y\Add{,} Z$ ses coordonnées constantes
+par rapport à ce trièdre; soient $\xi, \eta, \zeta$ les coordonnées de
+ce point par rapport à un système d'axes fixes. Lorsque le sommet
+du trièdre de Serret décrit la courbe donnée, les projections
+de la vitesse du point~$P$ sur les axes fixes sont, en remarquant
+que l'on a
+\begin{gather*}
+\begin{alignedat}{4}
+\xi   &= x &&+ aX &&+ a'Y &&+ a''Z, \\
+\eta  &= y &&+ bX &&+ b'Y &&+ b''Z, \\
+\zeta &= z &&+ cX &&+ c'Y &&+ c''Z\Add{,}
+\end{alignedat} \\[6pt]
+\begin{alignedat}{4}
+\frac{d\xi}{dt}
+  &= \frac{dx}{dt} &&+ X\frac{da}{dt} &&+ Y\frac{da'}{dt} &&+ Z\frac{da''}{dt},\\
+\frac{d\eta}{dt}
+  &= \frac{dy}{dt} &&+ X\frac{db}{dt} &&+ Y\frac{db'}{dt} &&+ Z\frac{db''}{dt},\\
+\frac{d\zeta}{dt}
+  &= \frac{dz}{dt} &&+ X\frac{dc}{dt} &&+ Y\frac{dc'}{dt} &&+ Z\frac{dc''}{dt}\Add{,}
+\end{alignedat}
+\end{gather*}
+ou encore
+\begin{align*}%[** TN: Added elided equations]
+\frac{d\xi}{dt}
+  &= \frac{ds}{dt}\, a + X\, \frac{a'}{R} + Y \left(-\frac{a}{R} - \frac{a''}{T}\right) + Z\, \frac{a'}{T}, \\
+\frac{d\eta}{dt}
+  &= \frac{ds}{dt}\, b + X\, \frac{b'}{R} + Y \left(-\frac{b}{R} - \frac{b''}{T}\right) + Z\, \frac{b'}{T}, \\
+\frac{d\zeta}{dt}
+  &= \frac{ds}{dt}\, c + X\, \frac{c'}{R} + Y \left(-\frac{c}{R} - \frac{c''}{T}\right) + Z\, \frac{c'}{T}\Add{.}
+\end{align*}
+
+Les projections de la vitesse sur les axes mobiles sont
+alors
+\begin{alignat*}{4}
+V_x &= a\, \frac{d\xi}{dt} &&+ b\, \frac{d\eta}{dt} &&+ c\, \frac{d\zeta}{dt}
+     &&= \frac{ds}{dt} \left(1 - \frac{Y}{R}\right)\Add{,} \\
+V_y &=  a'\, \frac{d\xi}{dt} &&+ b'\, \frac{d\eta}{dt} &&+ c'\, \frac{d\zeta}{dt}
+     &&= \frac{ds}{dt} \left(\frac{X}{R} + \frac{Z}{T}\right)\Add{,} \\
+V_z &=  a''\, \frac{d\xi}{dt} &&+ b''\, \frac{d\eta}{dt} &&+ c''\, \frac{d\zeta}{dt}
+     &&= -\frac{ds}{dt}\, \frac{Y}{T}\Add{,}
+\end{alignat*}
+$\dfrac{ds}{dt}$ est la vitesse du sommet du trièdre. Si nous ne considérons
+que la vitesse de rotation, nous savons que, si $p\Add{,} q\Add{,} r$ sont les
+composantes de la rotation instantanée sur les axes mobiles,
+%% -----File: 017.png---Folio 9-------
+on a
+\[
+V_x = qZ - rY\Add{,}\qquad
+V_y = rX - pZ\Add{,}\qquad
+V_z = pY - qX\Add{,}
+\]
+et nous trouvons ainsi, en identifiant avec les expressions
+précédentes (dans l'hypothèse $t = s$)
+\[
+p = -\frac{1}{T}\Add{,}\qquad
+q = 0\Add{,}\qquad
+r = \frac{1}{R}\Add{,}
+\]
+ce qui montre qu'\emph{à chaque instant, la rotation instantanée
+est dans le plan rectifiant et a pour composantes suivant la
+tangente et la binormale la torsion et la courbure}.
+
+Si l'on suppose le trièdre de Serret transporté à l'origine,
+il tourne autour de son sommet, l'axe instantané de rotation
+est dans le plan rectifiant, et le mouvement du trièdre
+est obtenu par le roulement de ce plan sur un certain cône.
+
+
+\Section[Calcul de la courbure.]{Calcul de~$R$.}
+{7.}{} Reprenons la formule~\Eq{(3)}
+\[
+\frac{1}{R} = \sum a'\, \frac{da}{ds}\Add{.}
+\]
+Nous avons
+\[
+a = \frac{dx}{ds},
+\]
+d'où
+\[
+\frac{da}{ds} = \frac{ds\, d^2x - dx\, d^2s}{\DPtypo{ds^2}{ds^3}}.
+\]
+Soit maintenant
+\[
+A = dy\, d^2z - dz\, d^2y\DPtypo{.}{,}\qquad
+B = dz\, d^2x - dx\, d^2z,\qquad
+C = dx\, d^2y - dy\, d^2x,
+\]
+et posons
+\[
+\sqrt{A^2 + B^2 + C^2} = D.
+\]
+$A\Add{,} B\Add{,} C$ sont les coefficients du plan osculateur, et par suite
+les cosinus directeurs de la binormale sont
+\[
+a'' = \frac{A}{D}\Add{,}\qquad
+b'' = \frac{B}{D}\Add{,}\qquad
+c'' = \frac{C}{D}\Add{,}
+\]
+et les cosinus directeurs de la normale principale, perpendiculaire
+aux deux droites précédentes, sont
+\begin{align*}% [** TN: Elided equations not added]
+a' &= \frac{B\, dz - C\, dy}{D\, ds}
+    = \frac{\DPtypo{dx^2}{dx^2}\, (dz^2 + dy^2) - dx\, (dz\, \DPtypo{dz^2}{d^2z} + dy\, \DPtypo{dy^2}{d^2y})}{D\, ds} \\
+   &= \frac{\DPtypo{dx^2}{d^2x}·ds^2 - dx·ds\, \DPtypo{ds^2}{d^2s}}{D\, ds}
+    = \frac{ds\, d^2x - dx\, d^2s}{D}\Add{,} \\[6pt]
+b' &= \dots \qquad  c' = \dots\Add{,}
+\end{align*}
+\iffalse%%%%[** TN: Code for elided equations]
+b' &= \frac{C\, dx - A\, dz}{D\, ds}
+    = \frac{dy^2\, (dx^2 + dz^2) - dy\, (dx\, d^2x + dz\, d^2z)}{D\, ds} \\
+   &= \frac{d^2y·ds^2 - dy·ds\, d^2s}{D\, ds}
+    = \frac{ds\, d^2y - dy\, d^2s}{D}\Add{,} \\[6pt]
+c' &= \frac{A\, dy - B\, dx}{D\, ds}
+    = \frac{dz^2\, (dy^2 + dx^2) - dz\, (dy\, d^2y + dx\, d^2x)}{D\, ds} \\
+   &= \frac{d^2z·ds^2 - dz·ds\, d^2s}{D\, ds}
+    = \frac{ds\, d^2z - dz\, d^2s}{D}\Add{,}
+\fi %%%% End of code for elided equations
+%% -----File: 018.png---Folio 10-------
+et alors
+\[
+\frac{1}{R} = \sum a'\, \frac{da}{ds}
+  = \sum \frac{B\, dz - C\, dy}{D\DPtypo{}{\,ds}}\,
+    \frac{ds\, d^2x - dx\, d^2s}{\DPtypo{ds^2}{ds^3}}
+\]
+ce qui peut s'écrire
+%[** TN: Original exponents unclear, but math checked by hand.]
+\begin{align*}
+\frac{1}{R}
+  &= \frac{1}{D\, ds^3}  \sum d^2x\, (B\, dz - C\, dy)
+   - \frac{d^2s}{D\, ds^4} \sum dx\, (B\, dz - C\, dy) \\
+\intertext{et se réduit à:}
+\frac{1}{R}
+  &= \frac{1}{D\, ds^3} \sum d^2x\, (B\, dz - C\, dy)
+   = \frac{1}{D\, ds^3}
+  \begin{vmatrix}
+  dx   &  dy  &  dz  \\
+  d^2x & d^2y & d^2z \\
+    A  &  B   &  C
+  \end{vmatrix} = \frac{D}{ds^3}\Add{,}
+\end{align*}
+d'où enfin:
+\[
+\frac{1}{R} = \frac{\sqrt{\sum (dy\, d^2z\DPtypo{_}{-} dz\, d^2y)^2}}
+                   {(dx^2 + dy^2 + dz^2)^{\tfrac{3}{2}}}.
+\]
+
+
+\Section[Calcul de la torsion]{Calcul de~$T$.}
+{8.}{} On aura de même
+\begin{align*}
+\frac{1}{T}
+  &= \sum a'\, \frac{da''}{ds}
+   = \sum \frac{B\, dz \DPtypo{_}{-} C\, dy}{D·ds}\,
+          \frac{D·dA - A\, dD}{D^2\, ds} \\
+\intertext{ce qui peut s'écrire}
+\frac{1}{T}
+  &= \frac{1}{D^2\, \DPtypo{ds}{ds^2}} \sum dA\, (B\, dz - C\, dy)
+   - \frac{dD}{\DPtypo{D^2}{D^3}\, ds^2} \sum A\, (B\, dz - C\, dy)
+\end{align*}
+et se réduit à
+\begin{align*}
+\frac{1}{T}
+  &= \frac{1}{D^2\, ds^2} \sum dA\, (B\, dz - C\, dy)
+   = \frac{1}{D^2\, ds} \sum (dy\, d^3z - dz\, d^3y) (ds\, d^2x - dx\, d^2s) \\
+\intertext{ou}
+\frac{1}{T}
+  &= \frac{1}{D^2} \sum d^2x (dy\, d^3z - dz\, d^3y)
+   - \frac{d^2s}{D^2\, ds} \sum dx\, (dy\, d^3z - dz\, d^3y); \\
+\intertext{la \Ord{2}{e} somme est nulle, et il reste}
+\frac{1}{T}
+  &= \frac{1}{\DPtypo{D}{D^2}} \sum d^2x\, (dy\, d^3z - dz\, d^3y)
+   = -\frac{1}{D^2}
+     \begin{vmatrix}
+     dx &  dy  &  dz \\
+     d^2x & d^2y & d^2z \\
+     d^3x & d^3y & d^3z
+  \end{vmatrix}
+\end{align*}
+avec
+\[
+D^2 = \sum (dy\, d^2z - dz\, d^2y)^2\Add{.}
+\]
+
+\Paragraph{Remarque.} Pour que la torsion d'une courbe soit constamment
+nulle, il faut et il suffit que l'on ait constamment
+\[
+\begin{vmatrix}
+  dx &  dy  &  dz \\
+  d^2x & d^2y & d^2z \\
+  d^3x & d^3y & d^3z
+\end{vmatrix} = 0,
+\]
+ce qui exige que $x, y, z$ soient liés par une relation linéaire,
+à coefficients constants, c'est-à-dire que la courbe soit plane.
+Ainsi \emph{les courbes à torsion constamment nulle sont des
+%% -----File: 019.png---Folio 11-------
+courbes planes.}
+
+\Section{Sphère osculatrice.}
+{9.}{} Cherchons les sphères qui ont en~$M$, avec la courbe
+considérée, un contact du second ordre. Le centre $(x_0\Add{,} y_0\Add{,} z_0)$ et
+le rayon~$R_0$ d'une telle sphère sont, d'après la théorie du
+contact, déterminés par les équations suivantes, que nous développons
+au moyen des formules de Serret-Frenet:
+\begin{align*}
+&\sum (x - x_0)^2 - R_0^2 = 0, \\
+&\frac{d}{ds} \left\{ \sum (x - x_0)^2 - R^2 \right\} = 0,\quad
+  \text{\DPchg{c.à.d.}{cést-à-dire}}\quad \sum a(x - x_0) = 0, \\
+&\frac{d^2}{ds^2} \left\{ \sum (x - x_0)^2 - R^2 \right\} = 0,\quad
+  \text{\DPchg{c.à.d.}{cést-à-dire}}\quad 1 + \frac{1}{R} \sum a' (x - x_0) = 0.
+\end{align*}
+
+Si on prend le trièdre de Serret-Frenet pour trièdre
+de coordonnées, comme on l'a fait plus haut, elles se réduisent
+à
+\[
+\sum X_0 - R_0^2 = 0,\qquad X_0 = 0,\qquad Y_0 = -R;
+\]
+et l'équation générale des sphères cherchées est, $Z_0$~restant
+arbitraire,
+\[
+X^2 + Y^2 + Z^2 - 2RY - 2Z_0Z = 0\Add{.}
+\]
+
+C'est un faisceau de sphères, dont fait partie le plan
+osculateur $Z = 0$. On vérifie ainsi la propriété de contact
+du plan osculateur.
+
+Le cercle commun à toutes ces sphères est, de plus,
+d'après la théorie du contact des courbes, celui qui a un
+contact du second ordre avec la courbe, \DPchg{c.à.d.}{cést-à-dire} le \emph{cercle
+osculateur}. Les équations sont
+\[
+Z = 0,\qquad X^2 + Y^2 - 2RY = 0,
+\]
+\DPchg{c.à.d.}{cést-à-dire} qu'il est dans le plan osculateur, a pour centre
+le centre de courbure~$C$ ($X = 0$, $Y = R$), et passe en~$M$. Le
+lieu des \Err{}{centres des }sphères considérées est l'axe du cercle osculateur.
+
+Parmi toutes ces sphères, il y en a une qui a un contact
+%% -----File: 020.png---Folio 12-------
+du troisième ordre avec la courbe. On l'obtient en introduisant
+la condition nouvelle:
+\begin{gather*}
+\frac{d^3}{ds^3} \left\{ \sum (x - x_0)^2 - R^2 \right\} = 0\Add{,} \\
+\intertext{\DPchg{c.à.d.}{cést-à-dire}}
+-\frac{1}{R^2}\, \frac{dR}{ds} \sum a'(x - x_0)
+  -\frac{1}{R} \left\{ \frac{1}{R} \sum a(x - x_0) + \frac{1}{T} \sum a''(x - x_0) \right\} = 0,
+\end{gather*}
+qui se réduit, avec les axes particuliers employés, à
+\[
+Z_0 = -T\, \frac{dR}{ds}.
+\]
+Le centre de cette \emph{sphère osculatrice} est donc défini par les
+formules:
+\[
+X_0 = 0,\qquad Y_0 = -R,\qquad Z_0 = -T\, \frac{dR}{ds}.
+\]
+Et son rayon est donné par
+\[
+R_0^2 = R^2 + T^2\, \frac{dR^2}{ds^2}.
+\]
+
+
+\SubChap{II. Surfaces développables.}
+
+\Section{Propriétés générales.}
+{10.}{} Une courbe gauche est le lieu de $\infty^{1}$~points; corrélativement
+nous \DPtypo{considèrerons}{considérerons} une surface développable, enveloppe
+de $\infty^{1}$~plans; la caractéristique de l'un de ces plans
+correspond corrélativement à la tangente en un point de la
+courbe, puisqu'elle est l'intersection de deux plans infiniment
+voisins.
+
+Soit
+\[
+\Tag{(1)}
+uX + vY + wZ + h = 0,
+\]
+l'équation générale des plans considérés, de sorte que $u, v, w, h$
+désignent des fonctions données d'un paramètre~$t$.
+
+Les caractéristiques ont, d'après la théorie des enveloppes,
+pour équations générales,
+\[
+\Tag{(2)}
+\left\{
+\begin{aligned}
+&uX + uY + wZ + h = 0\Add{,} \\
+&du·X + dv·Y + dw·Z + dh = 0.
+\end{aligned}
+\right.
+\]
+
+La surface développable, enveloppe des plans~\Eq{(1)}, est,
+%% -----File: 021.png---Folio 13-------
+d'après la théorie des enveloppes, le lieu des droites~\Eq{(2)},
+qui en sont, par conséquent, les génératrices rectilignes;
+et, toujours d'après la théorie des enveloppes, chacun des
+plans~\Eq{(1)} est tangent à la surface tout le long de la génératrice~\Eq{(2)}
+correspondant à la même valeur de~$t$.
+
+Considérons alors la courbe~$(C)$, lieu des points $(x,y,z)$
+définis par les équations:
+\[
+\Tag{(3)}
+\left\{
+\begin{aligned}
+&ux + vy + wz + h = 0\Add{,} \\
+&\DPtypo{x\, du}{u\, dx} + v\, dy + w\, dz + dh = 0\Add{,} \\
+&\DPtypo{x\, d^2u}{u\, d^2x} + v\, d^2y + w\, d^2z + d^2h = 0\Add{.}
+\end{aligned}
+\right.
+\]
+
+L'un quelconque de ses points~$M$ est sur la droite~\Eq{(2)},
+correspondant à la même valeur de~$t$, et, par conséquent, dans
+le plan~\Eq{(1)} correspondant. Cherchons la tangente à~$(C)$ en~$M$.
+Il faut différentier les équations~\Eq{(3)}; différentiant chacune
+des deux premières, en tenant compte de la suivante, nous
+trouvons
+\[
+\Tag{(4)}
+\left\{
+\begin{aligned}
+&u·dx +  v·dy +  w·dz = 0\Add{,} \\
+&du·dx + dv·dy + dw·dz = 0,
+\end{aligned}
+\right.
+\]
+ce qui exprime que la direction de la tangente est la même
+que celle de la droite~\Eq{(2)}. Donc les tangentes à~$(C)$ sont
+les génératrices de la développable.
+
+Cherchons encore le plan osculateur à~$(C)$ en~$M$. Il doit
+passer par la tangente, et être parallèle à la direction $(d^2x, d^2y, d^2z)$.
+Or\Add{,} si on différentie la première des équations~\Eq{(4)},
+en tenant compte de la seconde, on trouve
+\[
+u·d^2x + v·d^2y + w·d^2z = 0,
+\]
+ce qui montre que le plan~\Eq{(1)} satisfait à ces conditions.
+Donc le plan osculateur de~$(C)$ est le plan qui enveloppe la
+développable.
+
+$(C)$~s'appelle l'\emph{arête de rebroussement} de la développable
+%% -----File: 022.png---Folio 14-------
+
+Donc \emph{toute développable est l'enveloppe des plans osculateurs
+de son arête de rebroussement, et est engendrée par
+les tangentes à son arête de rebroussement}.%--
+
+\Paragraph{Remarques.} Nous avons fait implicitement diverses hypothèses.
+D'abord que le déterminant des équations~\Eq{(3)} n'est pas nul.
+S'il l'est, on a
+\[
+\begin{vmatrix}
+u  &  v  &  w  \\
+du &  dv &  dw \\
+d^2u & d^2v & d^2w
+\end{vmatrix} = 0\Add{,}
+\]
+ce qui exprime que $u, v, w$ sont liés par une relation linéaire
+homogène à coefficients constants; \DPchg{c.à.d.}{cést-à-dire} que les plans~\Eq{(1)}
+sont parallèles à une droite fixe. Dans ce cas, les droites~\Eq{(2)}
+sont parallèles à cette même direction, et la surface est
+un \emph{cylindre}. Dans ce cas figure, comme \emph{cas singulier}, celui où
+tous les plans~\Eq{(1)} passent par une droite fixe, qui est alors
+l'enveloppe.
+
+\DPchg{Ecartant}{Écartant} ce cas, nous avons admis qu'il y avait un lieu
+des points~$M$. Ceci suppose que $M$~n'est pas fixe. S'il en était
+ainsi les équations~\Eq{(3)} étant vérifiées par les coordonnées
+de ce point fixe, les plans~\Eq{(1)} passeraient par ce point fixe,
+ainsi que les droites~\Eq{(2)}. L'enveloppe serait un \emph{cône}.
+
+\DPchg{Ecartons}{Écartons} encore ce cas. Nous avons admis encore que les
+droites~\Eq{(2)} engendraient une surface. Mais cela n'est en défaut
+que si elles sont toutes confondues, ce qui est le cas
+singulier \DPtypo{déja}{déjà} examiné.
+
+Remarquons enfin que la courbe~$(C)$ est \DPtypo{forcèment}{forcément} gauche,
+car si elle était plane, son plan étant son plan osculateur
+unique, et nos raisonnements ne cessant pas de s'appliquer,
+tous les plans~\Eq{(1)} seraient confondus. Il n'y aurait donc
+pas $\infty^{1}$~plans~\Eq{(1)}.
+%% -----File: 023.png---Folio 15-------
+
+
+\Section{Réciproques.}
+{11.}{Réciproquement les plans osculateurs en tous les
+points d'une courbe gauche enveloppent une développable.} En
+effet, si nous reprenons les notations du §1, le plan osculateur
+en un point~$x\Add{,}y\Add{,}z$ d'une courbe a pour équation
+\[
+\sum a'' (X - x) = 0.
+\]
+Sa caractéristique est représentée par l'équation précédente
+et
+\[
+\sum  \frac{da''}{ds}\, (X - x) - \sum a''\, \frac{dx}{ds} = 0;
+\]
+mais on a
+\[
+\sum a''\, \frac{dx}{ds} = 0,\qquad \frac{da''}{ds} = \frac{1}{T}\, a';
+\]
+les équations de la caractéristique sont donc
+\[
+\sum a'(X - x) = 0,\qquad \sum a''(X - x) = 0.
+\]
+Et, si on prend comme trièdre de coordonnées le trièdre de
+Serret-Frenet, elles se réduisent à
+\[
+Y = 0,\qquad Z = 0.
+\]
+\emph{Donc la \DPtypo{caracteristique}{caractéristique} du plan osculateur en un point d'une
+courbe gauche est la tangente à cette courbe}, et l'enveloppe
+de ce plan est bien une surface développable. L'arête de rebroussement
+a pour équations
+\[
+\sum a''(X - x) = 0,\quad
+\sum a' (X - x) = 0,\quad
+\sum \frac{da'}{ds}\, (X - x) - \sum a'\, \frac{dx}{ds} = 0.
+\]
+
+Considérons la \DPtypo{3éme}{\Ord{3}{ème}}~équation; remarquons que l'on a
+\[
+\sum a'\, \frac{dx}{ds} = 0,\qquad
+\frac{da'}{ds} = -\left(\frac{a}{R} + \frac{a''}{T}\right);
+\]
+elle s'écrit alors
+\[
+\sum \left(\frac{a}{R} + \frac{a''}{T}\right) (X - x) = 0,
+\]
+ou encore, en tenant compte de la \Ord{1}{ère} équation
+\[
+\sum a(X - x) = 0.
+\]
+Nous avons ainsi \Card{3} équations linéaires et homogènes en $X - x$,
+$Y - y$, $Z - z$, dont le déterminant est~$1$; donc
+\[
+X - x = 0,\qquad
+Y - y = 0,\qquad
+Z - \DPtypo{Z}{z} = 0;
+\]
+\DPtypo{l'arète}{l'arête} de rebroussement est la courbe elle-même.
+%% -----File: 024.png---Folio 16-------
+
+\Paragraph{Remarque.} Le nom \DPtypo{d'arète}{d'arête} de rebroussement provient de ce fait
+que la \emph{section de la développable par le plan normal en~$M$ à
+l'arête de rebroussement présente au point~$M$ un point de rebroussement}.
+En effet, rapportons la courbe au trièdre de Serret
+relatif au point~$M$: les coordonnées d'un point de la courbe
+voisin du point~$M$ sont, d'après les formules établies au~\no5
+\begin{alignat*}{3}
+x &= ds &{}- \frac{1}{\Err{6R}{6R^2}}\, ds^3 &+ \dots, \\
+y &=    &  \frac{1}{2R}\, ds^2 &- \frac{1}{6R^2}\, \frac{dR}{ds}\, ds^3 + \dots, \\
+z &=    &                      &- \frac{1}{6RT}\, ds^3 + \dots\Add{.}
+\end{alignat*}
+Les coordonnées d'un point de la tangente au point~$x\Add{,}y\Add{,}z$ sont
+\begin{align*}
+X &= x + \lambda\, \frac{dx}{ds}
+   = \!\left(ds - \frac{1}{6R^2}\, ds^3 + \dots\right)
+       + \lambda\! \left(1 - \frac{1}{2R^2}\, ds^2 + \dots\right)\Add{\!,} \\
+Y &= y + \lambda\, \frac{dy}{ds}
+   = \!\left(\!\frac{1}{2R}\, ds^2 - \frac{1}{6R^2}\, \frac{dR}{ds}\, ds^3 + \dots\!\right)
+       + \lambda\! \left(\!\frac{1}{R}\, ds - \frac{1}{2R^2}\, \frac{dR}{ds}\, ds^2 + \dots\!\right)\Add{\!,} \\
+Z &= z + \lambda\, \frac{dz}{ds}
+   = \!\left(-\frac{1}{6RT}\, ds^3 + \dots\right)
+       + \lambda\! \left(-\frac{1}{2RT}\, ds^2 + \dots\right)\Add{\!.}
+\end{align*}
+Prenons l'intersection de cette tangente avec le plan normal
+$X = 0$, nous avons
+\[
+\lambda = -\frac{ds + \dots}{1 + \dots} = -ds + \dots
+\]
+et la courbe d'intersection \DPtypo{à}{a} pour équations
+%
+\begin{align*}
+&\smash{\raisebox{-0.25in}{\Input[1.5in]{024a}}}&&
+\begin{aligned}[b]
+Y &= -\frac{1}{2R}\, ds^2 + \dots\Add{,} \\
+Z &= \frac{1}{3RT}\, ds^3 + \dots\Add{.}
+\end{aligned}
+\end{align*}
+On voit qu'elle a au point~$M$ un point de rebroussement, la
+tangente de rebroussement étant la normale principale.%--
+
+\Section{Surface rectifiante. Surface polaire\Add{.}}
+{12.}{Remarques.} Cherchons les surfaces développables enveloppes
+des faces du trièdre de Serret dans une courbe \DPtypo{gauch}{gauche}~$(C)$.
+Nous venons de voir que \emph{le plan osculateur enveloppe la
+surface développable qui admet pour arête de rebroussement~$(C)$}\Add{.}
+%% -----File: 025.png---Folio 17-------
+
+Considérons maintenant le plan rectifiant
+\[
+\sum a'(X - x) = 0
+\]
+la caractéristique est représentée par l'équation précédente
+et par
+\[
+\frac{1}{R} \sum a(X - x) + \frac{1}{T} \sum a''(X - x) = 0\Add{.}
+\]
+Si on prend les axes de Serret ces équations deviennent
+\[
+Y=0,\qquad \frac{1}{R}\, X + \frac{1}{T}\, Z = 0,
+\]
+la caractéristique contient le point $Y = 0$, $X = -\dfrac{1}{T}$, $Z = \dfrac{1}{R}$,
+extrémité du \Err{secteur}{vecteur} qui représente la rotation instantanée
+du trièdre; \emph{c'est l'axe instantané de rotation du trièdre de
+Serret}. Son lieu s'appelle la \emph{surface rectifiante}. Elle contient
+la courbe~$(\DPtypo{c}{C})$.
+
+Considérons enfin le plan normal
+\[
+\sum a(X - x) = 0;
+\]
+la \Ord{2}{e} équation de la caractéristique est
+\[
+\sum \frac{da}{ds}\, (X - x) - \sum a\, \frac{dx}{ds} = 0,
+\]
+ou
+\[
+\frac{1}{R} \sum a'(X - x) - 1 = 0.
+\]
+Cette caractéristique s'appelle la \emph{droite polaire}, et son
+lieu s'appelle la \emph{surface polaire}.
+
+Prenant de nouveau les axes de Serret, les équations de la
+droite polaire deviennent
+\[
+X = 0,\qquad Y = R;
+\]
+Elle se confond donc avec \emph{l'axe du cercle osculateur}.
+
+Si nous cherchons le point d'intersection de la droite
+polaire avec l'arête de rebroussement de la surface polaire,
+nous avons les \Card{3} équations
+%% -----File: 026.png---Folio 18-------
+\[
+\sum a(X - x) = 0,\quad
+\sum a'(X - x) - R = 0,\quad
+\frac{1}{T} \sum a''(X - x) + \frac{dR}{ds} = 0\Add{,}
+\]
+qui deviennent, en prenant les axes de Serret,
+\[
+X = 0,\qquad Y = R,\qquad Z = -\frac{1}{T}\, \frac{dR}{ds}.
+\]
+
+Or\Add{,} ce sont les coordonnées du centre de la sphère osculatrice.
+(Voir \No9).
+
+Donc \emph{le point ou la droite polaire touche son enveloppe
+est le centre de la sphère osculatrice à la courbe~$(\DPtypo{c}{C})$.
+On peut dire encore que la courbe~$(\DPtypo{c}{C})$ est\DPtypo{}{ la} trajectoire orthogonale
+des plans osculateurs au lieu des centres de ses sphères
+osculatrices}.
+
+
+\ExSection{I}
+
+\begin{Exercises}
+\item[1.] Trouver l'axe instantané de rotation et de glissement
+du trièdre de Serret.
+
+\item[2.] Trouver les hélices circulaires osculatrices à une
+courbe gauche. Déterminer celle de ces hélices qui a même torsion
+que la courbe \DPtypo{donnee}{donnée}.
+
+\item[3.] Approfondir les relations entre une courbe et le lieu
+des centres de ses sphères osculatrices (courbure, torsion,
+\DPtypo{élement}{élément} d'arc).
+
+\item[4.] Chercher la condition nécessaire et suffisante pour
+qu'une courbe soit une courbe \DPtypo{spherique}{sphérique}.
+
+\item[5.] Déterminer toutes les courbes satisfaisant aux relations:
+\[
+\frac{dR}{ds} = F(R),\qquad T = G(R),
+\]
+où $F$~et~$G$ sont des fonctions données.
+
+\item[6.] Déterminer toutes les courbes à courbure constante.
+
+\item[7.] Déterminer toutes les courbes à torsion constante.
+\end{Exercises}
+
+%Voir les énoncés, page 18.
+%% -----File: 027.png---Folio 19-------
+
+
+\Chapitre{II}{Surfaces.}
+
+\Section[Courbes tracées sur une surface. Longeurs d'arc et angles.]
+{Le $ds^2$ de la surface, et les angles.}
+{1.}{Courbes tracées sur une surface. Longueurs d'arc et
+angles.} Les coordonnées d'un point d'une surface peuvent
+s'exprimer en fonction de deux paramètres arbitraires
+\[
+\Tag{(S)}
+x = f(u\Add{,} v),\qquad
+y = g(u\Add{,} v),\qquad
+z = h(u\Add{,} v);
+\]
+$u\Add{,} v$ sont les \emph{coordonnées curvilignes} d'un point de la surface~$(S)$.
+On définira une courbe~$(c)$ de la surface en établissant
+une relation entre $u, v$; ou, ce qui revient au même, en exprimant
+$u, v$ en fonction d'un même paramètre~$t$
+\[
+\Tag{(c)}
+u = \phi(t),\qquad
+v = \psi(t).
+\]
+La tangente à cette courbe a pour paramètres directeurs
+\[
+\Tag{(1)}
+dx = \frac{\dd x}{\dd u}\, du + \frac{\dd x}{\dd v}\, dv,\qquad
+dy = \frac{\dd y}{\dd u}\, du + \frac{\dd y}{\dd v}\, dv,\qquad
+dz = \frac{\dd z}{\dd u}\, du + \frac{\dd z}{\dd v}\, dv;
+\]
+la tangente est déterminée par les différentielles $du, dv$.
+
+L'élément d'arc a pour expression:
+\[
+\Tag{(2)}
+ds^2 = dx^2 + dy^2 + dz^2
+  = E\, du^2 + 2F\, du\, dv + G\, dv^2 = \Phi(du, dv)
+\]
+en posant
+\[
+E = \sum \left(\frac{\dd x}{\dd u}\right)^2,\qquad
+F = \sum \frac{\dd x}{\dd u}\, \frac{\dd x}{\Err{\dd v^2}{\dd v}},\qquad
+G = \sum \left(\frac{\dd x}{\dd v}\right)^2.
+\]
+
+Imaginons \Card{2} courbes passant par un même point $(u, v)$ de la
+surface; soient $du, dv$ les différentielles correspondant à l'une
+d'elles; $\delta u, \delta v$ celles correspondant à l'autre; $ds, \delta s$ les différentielles
+des arcs correspondants. Si $V$~est l'angle des deux
+courbes, nous avons
+\[
+\Cos V = \sum \frac{dx·\delta x}{ds·\delta s};
+\]
+or,
+\begin{align*}
+\sum dx·\delta x
+  &= \sum \left(\frac{\dd x}{\dd u}\, du       + \frac{\dd x}{\dd v}\,  dv\right)
+          \left(\frac{\dd x}{\dd u}\, \delta u + \frac{\dd x}{\dd \DPtypo{u}{v}}\, \delta v\right) \\
+  &=  E\, du\, \delta u + F (du\, \delta v + dv\, \delta u) + G\, dv\, \delta v;
+\end{align*}
+%% -----File: 028.png---Folio 20-------
+c'est une forme polaire de la forme quadratique $\Phi(du, dv)$
+et on a
+\[
+\Tag{(3)}
+\Cos V = \frac{1}{2}\, \frac{\delta u\, \dfrac{\dd\Phi(du, dv)}{\dd·du}
+                           + \delta v\, \dfrac{\dd\Phi(du, dv)}{\dd·dv}}
+   {\sqrt{\Phi(du, dv) · \Phi(\delta u, \delta v)}}
+\]
+
+Pour que les deux courbes soient orthogonales, il faut et
+il suffit que $\Cos V = 0$, ou
+\[
+\Tag{(4)}
+E\, du·\delta u + F(du·\delta v + dv·\delta u) + G·dv·\delta v = 0.
+\]
+En particulier, cherchons à quelles conditions les courbes coordonnées
+$u = \Cte$ et $v = \Cte$ forment un réseau orthogonal;
+alors $dv = 0$, $\DPtypo{du}{\delta u} =0$, la condition précédente se réduit à
+\[
+F\, du\, \delta v = 0,
+\]
+et comme $du\Add{,} \DPtypo{dv}{\delta v}$ ne sont pas constamment nuls, on a $F = 0$. Dans
+ce cas, le carré de l'élément d'arc prend la forme
+\[
+ds^2 = E\, du^2 + G\, dv^2.
+\]
+
+\Paragraph{Remarque.} Si on définit la surface par une équation de
+la forme
+\[
+Z = f(x\Add{,} y)
+\]
+en désignant comme d'habitude par $p, q$ les dérivées partielles
+de~$Z$ par rapport à $x, y$, on a
+\[
+ds^2 = dx^2 + dy^2 + (p\, dx + q\, dy)^2
+  = (1 + p^2)\, ds^2 + 2pq\, dx\, dy + (1 + q^2)\, dy^2\Add{,}
+\]
+\DPtypo{C.à.d.}{\DPchg{c.à.d.}{cést-à-dire}}
+\[
+E = 1 + p^2,\qquad F = pq,\qquad G = 1 + q^2.
+\]
+
+
+\Section{Déformation et représentation conforme.}
+{2.}{Surfaces applicables. Représentations conformes.}
+Considérons deux surfaces $(\Err{S'}{S})\Add{,} (S')$
+\begin{alignat*}{3}
+\Tag{(S)}
+x &= f(u, v),   &y &= g(u, v),   &z &= h(u\Add{,} v) \\
+\Tag{(S')}
+x &= F(u', v'),\qquad &y &= G(u', v'),\qquad &z &= H(u', v')
+\end{alignat*}
+on peut établir une correspondance point par point entre ces
+deux surfaces, et cela d'une infinité de manières. Il suffit
+de poser
+\[
+u' = \phi(u, v),\qquad
+v' = \psi(u\Add{,} v),
+\]
+les fonctions $\psi, \phi$ étant quelconques; à condition toutefois que
+%% -----File: 029.png---Folio 21-------
+les équations précédentes soient résolubles en $u, v$. Les équations
+de la surface~$(S')$ peuvent alors se mettre sous la forme
+\[
+\Tag{(S')}%[** TN: Duplicate label, re-expresses earlier equation (S')]
+x = \Err{F}{F_{1}}(u\Add{,} v),\qquad
+y = \Err{G}{G_{1}}(u\Add{,} v),\qquad
+z = \Err{H}{H_{1}}(u\Add{,} v),
+\]
+ce qui revient à dire que les points correspondants correspondent
+aux mêmes systèmes de valeurs des paramètres.
+
+Soient les éléments d'arcs sur ces \Card{2} surfaces
+\begin{alignat*}{3}
+ds^2   &= E\,   du^2 &&+ 2F\,   du·dv &&+ G\,   dv^2 \\
+ds_1^2 &= E_1\, du^2 &&+ 2F_1\, du·dv &&+ G_1\, dv^2
+\end{alignat*}
+Supposons ces éléments d'arc identiques, $E \equiv E_1$, $F \equiv F_1$, $G \equiv G_1$.
+Si alors $u, v$ sont exprimés en fonction du paramètre~$t$, les arcs
+des deux courbes correspondantes sur les deux surfaces compris
+entre \Card{2} points correspondants ont tous deux pour expression
+\[%[** TN: Not displayed in original]
+\int_{t_0}^{t_1} \sqrt{E\, du^2 + 2F\, du\, dv + G\, dv^2},
+\]
+$t_0, t_1$ étant les valeurs de~$t$ correspondant
+aux extrémités. Réciproquement, si les arcs homologues
+de deux courbes homologues sur les deux surfaces ont même
+longueur, les éléments d'arc sont identiques sur les deux surfaces.
+On dit que les deux surfaces sont \emph{applicables} l'une
+sur l'autre, ou résultent l'une de l'autre par \emph{déformation}.
+
+Dans cette correspondance, la fonction~$\Phi$ étant la même
+pour les \Card{2} surfaces, la formule~\Eq{(3)} montre que les angles se
+conservent. Mais la réciproque n'est pas vraie. L'expression
+de~$\Cos V$ est homogène et du \Ord{1}{er} degré en $E\Add{,} F\Add{,} G$; pour que les angles
+de deux courbes homologues soient égaux, il faut et il suffit
+que
+\[
+\Err{\frac{E_1}{E} = \frac{F_1}{F} = \frac{G_1}{G}}
+    {\frac{E}{E_1} = \frac{F}{F_1} = \frac{G}{G_1}} = \chi(u,v),
+\]
+ce rapport étant indépendant de $du, dv$. On dit dans ce cas qu'il
+y a \emph{représentation conforme} des deux surfaces l'une sur l'autre.
+
+
+\Section{Problème de la représentation conforme.}
+{}{\DPchg{Etant}{Étant} données deux surfaces, il est toujours possible d'établir
+entre elles une représentation conforme.} Ceci revient
+à dire que l'on peut exprimer $u_1, v_1$ en fonction de $u, v$ de
+%% -----File: 030.png---Folio 22-------
+telle sorte que l'on ait,
+\[
+E\, du^2 + 2F\, du·dv + G\, dv^2 = \chi(u, v)(E_1\, du^2 + 2F_1\, du·dv + G_1\, dv^2).
+\]
+Décomposons les deux $ds^2$ en facteurs du \Ord{1}{er} degré. Remarquons
+que $EG - F^2$ est la somme des carrés des déterminants déduits du
+tableau
+\[
+\begin{Vmatrix}
+\mfrac{\dd x}{\dd u} & \mfrac{\dd y}{\dd u} & \mfrac{\dd z}{\dd u} \\
+\mfrac{\dd x}{\dd v} & \mfrac{\dd y}{\dd v} & \mfrac{\dd z}{\dd v}
+\end{Vmatrix};
+\]
+$EG - F^2$ est positif pour toute surface réelle. Posons
+\[
+EG - F^2 = H^2;
+\]
+alors
+\[
+ds^2 = E\left(du + \frac{F + iH}{E}\, dv\right)
+        \left(du + \frac{F - iH}{E}\, dv\right);
+\]
+chacun des facteurs du \Ord{2}{e} membre admet un facteur intégrant.
+On a donc
+\begin{align*}
+du + \frac{F + iH}{E}\, dv &= M(u, v)\, d\alpha(u\Add{,} v)\Add{,} \\
+du + \frac{F - iH}{E}\, dv &= N(u, v)\, d\beta(u\Add{,} v)\Add{.}
+\end{align*}
+Les fonctions $\alpha\Add{,} \beta$ sont indépendantes; en effet $d\alpha$~et~$d\beta$ ne
+peuvent s'annuler en même temps si $H \neq 0$, ce que nous supposons.
+Nous pouvons donc prendre $\alpha, \beta$ comme coordonnées curvilignes
+sur la \Ord{1}{ère} surface, et nous avons
+\begin{align*}
+ds^2 &= P(u, v)\, d\alpha·d\beta
+  = \Theta(\alpha, \beta)\, d\alpha·d\beta\Add{.} \\
+\intertext{De même pour la \Ord{2}{e} surface, nous pourrons écrire}
+ds_1^2 &= P_1(u_1, v_1)\, d\alpha_1·d\beta_1
+  = \Theta_1(\alpha_1, \beta_1)\, d\alpha_1·d\beta_1.
+\end{align*}
+Nous aurons alors à satisfaire à l'équation
+\[
+\Theta(\alpha, \beta)\, d\alpha · d\beta
+  = \Omega(\alpha, \beta)\, \Theta_1(\alpha_1, \beta_1)\, d\alpha_1 · d\beta_1\Add{.}
+\]
+
+Remarquons que pour $d\alpha = 0$, on doit avoir $d\alpha_1 · d\beta_1 = 0$.
+Si nous prenons $d\alpha_1 = 0$,  $\alpha_1$~sera fonction de~$\alpha$ et de même $\beta_1$~sera
+fonction de~$\beta$
+\[
+\alpha_1(u_1, v_1) = \phi\bigl(\alpha(u, v)\bigr)\Add{,}\qquad
+\beta_1 (u_1, v_1) = \psi\bigl(\beta (u, v)\bigr)\Add{.}
+\]
+%% -----File: 031.png---Folio 23-------
+Au contraire en prenant $d\beta_1 = 0$, $\beta_1$~sera fonction de~$\alpha$ et de
+même~$\alpha_1$, de~$\beta$
+\[
+\beta_1 (u_1, v_1) = \phi\bigl(\alpha(u, v)\bigr)\qquad
+\alpha_1(u_1, v_1) = \psi\bigl(\beta (u, v)\DPtypo{}{\bigr)}\Add{.}
+\]
+On voit donc bien que l'on peut toujours établir une représentation
+conforme. Et nous avons de plus la solution \DPtypo{génerale}{générale}
+de ce problème.
+
+
+\Section{Condition pour que deux surfaces soient applicables.}
+{}{Deux surfaces données ne sont pas en général applicables
+l'une sur l'autre.}
+
+Autrement dit, étant données deux surfaces, il est impossible
+d'établir entre elles une correspondance telle que $ds^2 = ds_1^2$\Add{.}
+En effet, en reprenant le calcul précédent, il faudrait satisfaire
+à la relation
+\[
+\Theta(\alpha, \beta)\, d\alpha · d\beta = \Theta_1(\alpha_1, \beta_1)\, d\alpha_1 \Add{·} d\beta_1,
+\]
+il faudrait comme \DPtypo{précedemment}{précédemment}, prendre par exemple
+\[
+\alpha_1 = \phi(\alpha)\qquad \beta_1 = \psi(\beta);
+\]
+et la relation à satisfaire devient
+\[
+\Theta(\alpha, \beta)
+  = \Theta_1\bigl(\phi(\alpha), \psi(\beta)\bigr)\, \phi'(\alpha)\, \psi'(\beta);
+\]
+il est facile de voir que, les fonctions $\Theta, \Theta_1$\DPtypo{,}{} étant données\Add{,}
+il est impossible en général de trouver des fonctions $\phi, \psi$,
+satisfaisant à cette relation. Considérons en effet le cas
+particulier où la deuxième surface est le plan $z = 0$. Dans
+ce cas $ds_1^2 = dx^2 + dy^2 = d\alpha_1 · d\beta_1$ et on devrait avoir
+\[
+\Theta(\alpha, \beta) = \phi'(\alpha)\, \psi'(\beta)\Add{;}
+\]
+or\Add{,} la fonction $\Theta$ étant quelconque, n'est pas le produit d'une
+fonction de~$\alpha$ par une fonction de~$\beta$.
+
+Pour qu'il en soit ainsi, il faut et il suffit que l'on ait
+\[
+\log \Theta(\alpha, \beta) = \log \phi'(\alpha) + \log \psi'(\beta),
+\]
+ou
+\[
+\frac{\dd^2 \log \Theta(\alpha, \beta)}{\dd\alpha · \dd\beta} = 0.
+\]
+%% -----File: 032.png---Folio 24-------
+Nous venons ainsi de montrer qu'une surface n'est pas en
+général applicable sur un plan, et de trouver la condition
+pour qu'une surface soit applicable sur un plan.
+
+
+\Section{Les \DPtypo{Directions}{directions} conjuguées et la forme \texorpdfstring{$\sum l\, d^{2}x$}{}\Add{.}}
+{3.}{Développables circonscrites. Directions conjuguées\Add{.}}
+
+Corrélativement aux courbes tracées sur la surface,
+lieux de $\infty^{1}$~points de la surface, nous considérerons les
+développables circonscrites, enveloppes de $\infty^{1}$~plans tangents
+à la surface. Définissons le plan tangent en un point de
+la surface. Soient $l, m, n$ les coefficients directeurs de la
+normale, et supposons les coordonnées rectangulaires. Nous
+devons avoir pour toute courbe de la surface
+\[
+l\, dx + m\, dy + n\, dz = 0;
+\]
+en particulier, pour les courbes coordonnées, $u = \cte$ et $v = \cte$
+nous aurons
+\begin{align*}
+l\, \frac{\dd x}{\dd u} + m\, \frac{\dd y}{\dd u} + n\, \frac{\dd z}{\dd u} &= 0\Add{,} \\
+l\, \frac{\dd x}{\dd v} + m\, \frac{\dd y}{\dd v} + n\, \frac{\dd z}{\dd v} &= 0\Add{,}
+\end{align*}
+et ces relations montrent que $l, m, n$, sont proportionnels aux
+déterminants fonctionnels $A, B, C$,
+\[
+\Tag{(1)}
+A = \frac{\dd y}{\dd u}\, \frac{\dd z}{\dd v} - \frac{\dd z}{\dd u}\, \frac{\dd y}{\dd v}
+  = \frac{D(y, z)}{D(u, v)},\qquad
+B = \frac{D(z, x)}{D(u, v)},\qquad
+C = \frac{D(x, y)}{D(u, v)};
+\]
+nous avons vu d'ailleurs que $A^2 + B^2 + C^2 = H^2$;
+donc les \DPtypo{cosimus}{cosinus} directeurs de la normale sont
+\[
+\Tag{(2)}
+\lambda = \frac{A}{H},\qquad \mu = \frac{B}{H},\qquad \nu = \frac{C}{H}.
+\]
+
+Considérons une développable circonscrite; nous pourrons
+la définir en exprimant $u, v$ en fonction d'un paramètre~$t$,
+%% -----File: 033.png---Folio 25-------
+\[
+u = \phi(t),\qquad
+v = \psi(t);
+\]
+alors le point $(u\Add{,} v)$ décrit une courbe de la surface, soit~$(c)$,
+et les plans tangents à la surface aux divers points de~$(c)$
+enveloppent la développable considérée. Le plan tangent
+à la surface au point $(x\Add{,} y\Add{,} z)$ est, $X\Add{,} Y\Add{,} Z$ étant les coordonnées courantes,
+\[
+l·(X - x) + m·(Y - y) + n·(Z - z) = 0;
+\]
+la caractéristique est définie par l'équation précédente et
+par l'équation
+\[
+dl·(X - x) + dm·(Y - y) + dn·(Z - z) = 0
+\]
+obtenue en différentiant la précédente par rapport à~$t$, et
+remarquant que l'on~a
+\[
+l\, dx + m\, dy + n\, dz = 0\Add{.}
+\]
+
+Voyons quelle est la direction de cette caractéristique\Add{.}
+Soient $\delta x, \delta y, \delta z$ ses coefficients de direction. Elle est
+tangente à la surface, donc on peut poser
+\[
+\delta x = \frac{\dd x}{\dd u}\, \delta u + \frac{\dd x}{\dd v}\, \delta v,\quad
+\delta y = \frac{\dd y}{\dd u}\, \delta u + \frac{\dd y}{\dd v}\, \delta v,\quad
+\delta z = \frac{\dd z}{\dd u}\, \delta u + \frac{\dd z}{\dd v}\, \delta v;
+\]
+en remplaçant $X-x, Y-y, Z-z$ par les quantités proportionnelles
+$\delta x, \delta y, \delta z$, on obtient
+\[
+dl · \delta x + dm · \delta y + dn · \delta z = 0;
+\]
+or\Add{,} on a
+\[
+dl = \frac{\dd l}{\dd u}\, du + \frac{\dd l}{\dd v}\, dv,\quad
+dm = \frac{\dd m}{\dd u}\, du + \frac{\dd m}{\dd v}\, dv,\quad
+dn = \frac{\dd n}{\dd u}\, du + \frac{\dd n}{\dd v}\, dv;
+\]
+donc la relation
+\[
+\sum dl · \delta x = 0\DPtypo{.}{}
+\]
+s'écrit
+\[
+\sum \left(\frac{\dd l}{\dd u}\, du + \frac{\dd l}{\dd v}\, dv\right)
+     \left(\frac{\dd x}{\dd u}\, \delta u + \frac{\dd x}{\dd v}\, \delta v\right) = 0.
+\]
+%% -----File: 034.png---Folio 26-------
+Ordonnons par rapport à $du, dv$, $\delta u, \delta v$. Remarquons que l'on a
+\[
+\sum l\, \frac{\dd x}{\dd u} = 0;
+\]
+d'où en dérivant par rapport à~$u$
+\[
+\sum l\, \frac{\dd^2 x}{\dd u^2} + \sum \frac{\dd l}{\dd u}\, \frac{\dd x}{\dd u} = 0;
+\]
+de même, la relation
+\[
+\sum l\, \frac{\dd x}{\dd v} = 0
+\]
+donne
+\[
+\sum l\, \frac{\dd^2 x}{\dd v^2} + \sum \frac{\dd l}{\dd v}\, \frac{\dd x}{\dd v} = 0;
+\]
+et
+\[
+\sum l\, \frac{\dd^2 x}{\dd u\, \dd v} + \sum \frac{\dd l}{\dd u}\, \frac{\dd x}{\dd v} = 0;
+\]
+de sorte que la relation précédente s'écrit
+\[
+\Tag{(3)}
+\sum l\, \frac{\dd^2 x}{\dd u^2}\, du · \delta u +
+\sum l\, \frac{\dd^2 x}{\dd u\, \dd v}\, (du · \delta v + dv\Add{·} \delta u) +
+\DPtypo{}{\sum} l\, \frac{\dd^2 x}{\dd v^2}\, dv · \delta v = 0\Add{.}
+\]
+Telle est la relation qui existe entre les coefficients de direction
+de la caractéristique et de la tangente à la courbe de
+contact. Elle serait visiblement la même en coordonnées obliques,
+$l, m, n$ étant alors les coefficients de l'équation du plan tangent
+soit
+\[
+\Tag{(4)}
+E' = \sum l\, \frac{\dd^2 x}{\dd u^2},\qquad
+F' = \sum l\, \frac{\dd^2 x}{\dd u\, \dd v},\qquad
+G' = \sum l\, \frac{\dd^2 x}{\dd v^2},
+\]
+et
+\[
+\Tag{(5)}
+\Psi(du\Add{,} dv) = E'\, du^2 + 2F'\, du\, dv + G'\, dv^2.
+\]
+On a, en particulier, quand on prend $l = A$, $m = B$, $n = C$:
+\[
+E' = \begin{vmatrix}
+\mfrac{\dd^2 x}{\dd u^2} & \mfrac{\dd^2 y}{\dd u^2} & \mfrac{\dd^2 z}{\dd u^2} \\
+\mfrac{\dd x}{\dd u} & \mfrac{\dd y}{\dd u} & \mfrac{\dd z}{\dd u} \\
+\mfrac{\dd x}{\dd v} & \mfrac{\dd y}{\dd v} & \mfrac{\dd z}{\dd v}
+\end{vmatrix},\
+%
+F' = \begin{vmatrix}
+\mfrac{\dd^2 x}{\dd u\, \dd v} & \mfrac{\dd^2 y}{\dd u\, \dd v} & \mfrac{\dd^2 z}{\dd u\, \dd v} \\
+\mfrac{\dd x}{\dd u} & \mfrac{\dd y}{\dd u} & \mfrac{\dd z}{\dd u} \\
+\mfrac{\dd x}{\dd v} & \mfrac{\dd y}{\dd v} & \mfrac{\dd z}{\dd v}
+\end{vmatrix},\
+%
+G' = \begin{vmatrix}
+\mfrac{\dd^2 x}{\dd v^2} & \mfrac{\dd^2 y}{\dd v^2} & \mfrac{\dd^2 z}{\dd v^2} \\
+\mfrac{\dd x}{\dd u} & \mfrac{\dd y}{\dd u} & \mfrac{\dd z}{\dd u} \\
+\mfrac{\dd x}{\dd v} & \mfrac{\dd y}{\dd v} & \mfrac{\dd z}{\dd v}
+\end{vmatrix}\Add{.}
+\]
+La relation précédente s'écrira alors
+\[
+E' · du\, \delta u + F' · (du\, \delta v + dv\, \delta u) + G' · dv\, \delta v = 0,
+\]
+ou
+\[
+\Tag{(6)}
+\frac{\dd\Psi(du, dv)}{\dd\Add{·} du}\, \delta u + \frac{\dd\Psi(du, dv)}{\dd\Add{·} dv}\, \delta v = 0.
+\]
+%% -----File: 035.png---Folio 27-------
+Cette relation est symétrique par rapport à $d, \delta$; \emph{il y a donc
+réciprocité entre la direction de la tangente à la courbe de
+contact de la développable et la direction de la caractéristique
+du plan tangent à cette développable}. Ces deux directions
+sont dites \emph{directions conjuguées}.
+
+Cherchons en particulier la condition pour que les courbes
+$u = \cte$, $v = \cte$ forment un réseau conjugué. Alors, $dv = 0$\Add{,} $\delta u = 0$
+la condition est $F' = 0$.
+
+\Paragraph{Remarque.} On a
+\begin{gather*}
+dx = \frac{\dd x}{\dd u}\, du + \frac{\dd x}{\dd v}\, dv, \\
+d^2 x = \frac{\dd x}{\dd u}\, d^2 u
+      + \frac{\dd x}{\dd v}\, d^2 v
+      + \frac{\dd^2 x}{\DPtypo{\dd u}{\dd u^2}}\, du^2
+      + 2 \frac{\dd^2 x}{\dd u\, \dd v}\, du\, dv
+      + \frac{\dd^2 x}{\DPtypo{\dd v}{\dd v^2}}\, dv^2.
+\end{gather*}
+On en conclut, à cause de
+\[
+\sum l\, \frac{\dd x}{\dd u} = 0,\qquad
+\sum l\, \frac{\dd x}{\dd v} = 0,
+\]
+l'identité
+\[
+\sum l\, d^2x
+  = \left(\sum l\, \frac{\dd^2 x}{\dd u^2}\right) du^2
+  + 2 \left(\sum l\, \frac{\dd^2 x}{\dd u\, \dd v}\right) du\, dv
+  + \left(\sum l\, \frac{\dd^2 x}{\dd v^2}\right) dv^2\Add{,}
+\]
+c'est-à-dire
+\[
+\sum l\, d^2 x = \Psi(du, dv).
+\]
+
+
+\Section{Formules fondamentales pour une courbe de la surface.}
+{4.}{\DPchg{Eléments}{Éléments} fondamentaux d'une courbe de la surface.}
+
+Nous considérerons en un point de la courbe le trièdre
+de Serret, et un trièdre constitué par la tangente à la courbe,
+la normale~$MN$ à la surface, et la tangente~$MN'$ à la surface
+qui est normale à la courbe. Nous choisirons les directions
+positives de telle façon que le trièdre ainsi constitué
+ait même disposition que le trièdre de Serret, de sorte que
+%% -----File: 036.png---Folio 28-------
+si $l, m, n$ sont les cosinus directeurs de la normale à la surface,
+$a_1, b_1, c_1$ de la tangente à la surface normale
+à la courbe, on ait
+\[
+\begin{vmatrix}
+a   & b   & c   \\
+a_1 & b_1 & c_1 \\
+l   & m   & n
+\end{vmatrix} = 1.
+\]
+
+\Illustration[2.5in]{036a}
+Les \Card{2} trièdres considérés ont un axe
+commun et de même direction, qui est
+la tangente. Pour les définir l'un par rapport à l'autre, il
+suffira de se donner l'angle d'une des arêtes de l'un avec
+l'une des arêtes de l'autre. Nous nous donnerons l'angle
+dont il faut faire tourner la demi-normale principale~$MP$ pour
+l'amener à \DPtypo{coincider}{coïncider} avec la demi-normale à la surface~$MN$, le
+sens positif des relations étant défini par la direction positive
+$MT$ de l'axe de rotation. Cherchons les relations qui existent
+entre les cosinus directeurs des arêtes de ces trièdres.
+Quand on passe de l'un à l'autre, on fait en réalité une
+transformation de coordonnées autour de l'origine dans le
+plan normal. Considérons le point à l'unité de distance de~$M$ sur
+$MN(l, m, n)$. Rapporté au système~$PMB$ il a pour coordonnées $\cos\theta$
+et $\sin\theta$, donc
+\[
+\Tag{(1)}
+l = a' \cos \theta + a'' \sin \theta\Add{,}\qquad
+m = b' \cos \theta + b'' \sin \theta\DPtypo{;}{,}\qquad
+n = c' \cos \theta + c'' \sin \theta;
+\]
+de même le point à l'unité de distance sur $MN'(a_1, b_1, c_1)$ rapporté
+au système $PMB$ a pour coordonnées $\cos (\theta - \frac{\pi}{2}) = \sin \theta$
+et $\sin (\theta - \frac{\pi}{2}) = -\cos \theta$, donc
+%% -----File: 037.png---Folio 29-------
+\[
+\Tag{(1')}%[** TN: Renumbered duplicate equation (1)]
+a_1 = a' · \sin \theta - a'' · \cos \theta,\quad
+b_1 = b' · \sin \theta - b'' · \cos \theta,\quad
+c_1 = c' · \sin \theta - c'' · \cos \theta.
+\]
+On aura donc, en faisant la transformation de coordonnées
+inverse
+\[
+\Tag{(2)}
+\begin{aligned}
+a'  &= l \cos \theta + a_1 \sin \theta, &
+b'  &= m \cos \theta + b_1 \sin \theta, &
+c'  &= n \cos \theta + c_1 \sin \theta, \\
+a'' &= l \sin \theta - a_1 \cos \theta, &
+b'' &= m \sin \theta - b_1 \cos \theta, &
+c'' &= n \sin \theta - c_1 \cos \theta.
+\end{aligned}
+\]
+Différentions les formules~\Eq{(1)} par rapport à~$s$: il vient
+\begin{alignat*}{5}
+&\frac{dl}{ds}
+  = (-a'\Add{·}\sin \theta + a'' · \cos\theta)\, &&\frac{d\theta}{ds}
+  &&+ \cos \theta\, \frac{da'}{ds} &&+ \sin\theta\, \frac{da''}{ds},
+  \quad&&\text{et les analogues;} \\
+&\frac{da_1}{ds}
+  = (a'\Add{·}\cos\theta + a'' · \sin \theta )\, &&\frac{d\theta}{ds}
+  &&+ \sin \theta\, \frac{da'}{ds} &&- \cos\theta\, \frac{da''}{ds},
+  \quad&&\text{et les analogues;}
+\end{alignat*}
+d'où, en tenant compte des formules de Frenet et des relations \Eq{(1)}\Add{,}~\Eq{(2)}
+\begin{align*}
+\Tag{(3)}
+\frac{dl}{ds}
+  &= a_1 \left(\frac{1}{T} - \frac{d\theta}{ds}\right) - \frac{a \cos\theta}{R}
+    \qquad\text{et les analogues}; \\
+\Tag{(4)}
+\frac{da_1}{ds}
+  &= -l \left(\frac{1}{T} - \frac{d\theta}{ds}\right) \Err{}{-} \frac{a \sin\theta}{R}
+    \qquad\PadTxt{et les analogues}{(id.)}; \\
+\intertext{\DPtypo{Enfin}{enfin} nous avons}
+\Tag{(5)}
+\frac{da}{ds}
+  &= \frac{a'}{R} = l\, \frac{\cos \theta}{R} + a_1\, \frac{\sin \theta}{R}
+    \quad\qquad\PadTxt{et les analogues}{(id.)};
+\end{align*}
+\emph{les formules fondamentales \Eq{(3)}\Add{,} \Eq{(4)}\Add{,} \Eq{(5)} permettent de calculer
+$\theta, R\Add{,} T$, c'est-à-dire de déterminer le plan osculateur, la
+courbure et la torsion de la courbe considérée}.
+
+\MarginNote{Formule pour $\dfrac{\cos\theta}{R}$\Add{.}}
+En effet, les formules~\Eq{(5)} nous donnent d'abord
+\[
+\frac{\cos \theta}{R}
+  = \sum l\, \frac{da}{ds}
+  = \sum l\, \frac{d^2x}{ds^2}
+  = \frac{1}{H} \sum A\, \frac{d^2x}{ds^2},
+\]
+c'est-à-dire, d'après le calcul du paragraphe précédent, et
+%% -----File: 038.png---Folio 30-------
+et en posant:
+\begin{gather*}
+E' = \sum A\, \frac{\dd^2 x}{\dd u^2},\qquad
+F' = \sum A\, \frac{\dd^2 x}{\dd u\, \dd v},\qquad
+G' = \sum A\, \frac{\dd^2 x}{\dd v^2}, \\
+\frac{\cos\theta}{R}
+  = \frac{1}{H}\, \frac{E' · du^2 + 2F' · du\, dv + G'\Add{·} dv^2}{ds^2},
+\end{gather*}
+ou enfin
+\[
+\Tag{(6)}
+\frac{\cos \theta}{R} = \frac{1}{H} · \frac{\Psi(du, dv)}{\Phi(du, dv)}.
+\]
+
+\MarginNote{Formule pour $\dfrac{\sin\theta}{R}$\Add{.}}
+Les formules~\Eq{(5)} donnent encore
+\[
+\frac{\sin \theta}{R}
+  = \sum a_1\, \frac{da}{ds}
+  = \sum a_1\, \frac{d^2x}{ds^2}\Add{.}
+\]
+Remarquons que
+\[
+\sum a_1\, \frac{d^2x}{\Err{ds}{ds^2}}
+  = \frac{1}{ds^2}
+  \begin{vmatrix}
+    a     & b     & c     \\
+    d^2 x & d^2 y & d^2 z \\
+    l     & m     & n
+  \end{vmatrix}
+  = \frac{1}{ds^3}
+  \begin{vmatrix}
+    dx    & dy    & dz    \\
+    d^2 x & d^2 y & d^2 z \\
+    l     & m     & n
+  \end{vmatrix};
+\]
+pour calculer le déterminant, multiplions-le par
+\[
+\begin{vmatrix}
+\mfrac{\dd x}{\dd u} &  \mfrac{\dd y}{\dd u} &  \mfrac{\dd z}{\dd u} \\
+\mfrac{\dd x}{\dd v} &  \mfrac{\dd y}{\dd v} &  \mfrac{\dd z}{\dd v} \\
+l & m & n
+\end{vmatrix}
+= Al + Bm + Cn = \frac{A^2 + B^2 + C^2}{H} = H;
+\]
+le produit est
+\[
+\begin{vmatrix}
+\sum \mfrac{\dd x}{\dd u}\, dx   & \sum \mfrac{\dd x}{\dd v}\, dx   & \sum l\, dx   \\
+\sum \mfrac{\dd x}{\dd u}\, d^2x & \sum \mfrac{\dd x}{\dd v}\, d^2x & \sum l\, d^2x \\
+\sum l\, \mfrac{\dd x}{\dd u}    & \sum l\, \mfrac{\dd x}{\dd v}    & \sum l^2
+\end{vmatrix};
+\]
+or\Add{,} nous avons
+\begin{gather*}
+\sum \frac{\dd x}{\dd u} · dx  = \sum \frac{\dd x}{\dd u} · \left(\frac{\dd x}{\dd u}\, du + \frac{\dd x}{\dd v}\, dv\right) = E\, du + F\, dv, \\
+\sum \frac{\dd x}{\dd v} · dx  = \DPtypo{}{\sum} \frac{\dd x}{\dd v} · \left(\frac{\dd x}{\dd u}\, du + \frac{\dd x}{\dd v}\, dv\right)  = F\, du + G\, dv, \\
+\sum l\, dx = \sum l\, \frac{\dd x}{\dd u} = \sum l\, \frac{\dd x}{\dd v} = 0,
+\end{gather*}
+%% -----File: 039.png---Folio 31-------
+\begin{align*}
+\sum \frac{\dd x}{\dd u}\, d^2x
+  &= \sum \frac{\dd x}{\dd u} \left(\frac{\dd x}{\dd u}\, d^2u + \frac{\dd x}{\dd v}\, d^2v
+   + \frac{\dd^2 x}{\dd u^2}\, du^2 + 2\, \frac{\dd^2 x}{\dd u\, \dd v}\, du\, dv + \frac{\dd^2 x}{\dd v^2}\, dv^2\right) \\
+%
+  &= E\, d^2u + F\, d^2v + \frac{1}{2}\, \frac{\dd E}{\dd u}\, du^2 + \frac{\dd E}{\dd v}\, du·dv + \left(\frac{\dd F}{\dd v} - \frac{1}{2}\, \frac{\dd G}{\dd u}\right) dv^2, \\
+%
+\sum \frac{\dd x}{\dd v}\, d^2x
+  &= \sum \frac{\dd x}{\dd v} \left(\frac{\dd x}{\dd u}\, d^2u + \frac{\dd x}{\dd v}\, d^2v
+   + \frac{\dd^2 x}{\dd u^2}\, du^2 + 2\, \frac{\dd^2 x}{\dd u\, \dd v}\, du\, dv + \frac{\dd^2 x}{\dd v^2}\, dv^2\right) \\
+%
+  &= F\, d^2u + G\, d^2v + \left(\frac{\dd F}{\dd u} - \frac{1}{2}\, \frac{\dd E}{\dd v}\right) du^2 + \frac{\dd G}{\dd u}\, du\, dv + \frac{1}{2}\, \frac{\dd G}{\dd v}\, dv^2.
+\end{align*}
+
+Le produit précédent s'écrit donc
+\[
+\begin{vmatrix}
+E\, du + F\, dv & F\, du + G\, dv  & 0 \\
+\left[
+\begin{aligned}%[** TN: Reformatted very wide entry]
+  &E\, d^2u + F\, d^2v \\
+  &+ \mfrac{1}{2}\, \mfrac{\dd E}{\dd u}\, du^2
+  + \mfrac{\dd E}{\dd v}\, du\, dv
+  + \left(\mfrac{\dd F}{\dd v} - \mfrac{1}{2}\, \mfrac{\dd G}{\dd u}\right) dv^2
+\end{aligned}
+\right]
+  & F\, d^2u + G\, d^2v + \dots  & \sum l\, d^2x \\
+0 & 0 & 1
+\end{vmatrix}
+\]
+ou
+\[
+-\begin{vmatrix}
+E\, d^2u + F\, d^2v
+  + \mfrac{1}{2}\, \mfrac{\dd E}{\dd u}\, du^2
+  + \mfrac{\dd E}{\dd v}\, du\, dv
+  + \left(\mfrac{\dd F}{\dd v} - \mfrac{1}{2}\, \mfrac{\dd G}{\dd u}\right) dv^2
+  & E\, du + F\, dv \\
+F\, d^2u + G\, d^2v
+  + \left(\mfrac{\dd F}{\dd u} - \mfrac{1}{2}\, \mfrac{\dd E}{\dd v}\right) du^2
+  + \mfrac{\dd G}{\dd u}\, du\, dv
+  + \mfrac{1}{2}\, \mfrac{\dd G}{\dd v}\, dv^2
+  & F\, du + G\, dv
+\end{vmatrix}.
+\]
+Ce déterminant peut se décomposer en deux, dont le \Ord{1}{er} est
+\[
+-\begin{vmatrix}
+E\, d^2u + F\, d^2v & E\, du + F\, dv \\
+F\, d^2u + G\, d^2v & F\, du + G\, dv
+\end{vmatrix}
+  = H^2\, (du·d^2v - dv·d^2u),
+\]
+et on a finalement
+\begin{multline*}
+\Tag{(7)}
+\frac{\sin \theta}{R} = \frac{1}{H\, ds^3}
+\left[\MStrut
+\Err{\Omega^2}{-H^2}\, (du\Add{·}d^2v - dv·d^2u)\right. \\
+- \left.\begin{vmatrix}
+  \mfrac{1}{2}\, \mfrac{\dd E}{\dd u}\, du^2
+               + \mfrac{\dd E}{\dd v}\, du\, dv
+               + \left(\mfrac{\dd F}{\dd v}
+                 - \mfrac{1}{2}\, \mfrac{\dd G}{\dd u}\right) dv^2
+  & E\, du + F\, dv \\
+%
+  \left(\mfrac{\dd F}{\dd u} - \mfrac{1}{2}\, \mfrac{\dd E}{\dd v}\right) du^2
+  + \mfrac{\dd G}{\dd u}\, du\, dv
+  + \mfrac{1}{2}\, \mfrac{\dd G}{\dd v}\DPtypo{)}{}\, dv^2
+  & F\, du + G\, dv
+\end{vmatrix}\MStrut
+\right]\Add{.}
+\end{multline*}
+
+\MarginNote{Formule pour $\dfrac{1}{T} - \dfrac{d \theta}{ds}$\Add{.}}
+Enfin la formule~\Eq{(4)} nous donne
+\[
+\frac{1}{T} - \frac{d\theta}{ds} = \sum a_1\, \frac{dl}{ds}
+  = \frac{1}{ds}
+\begin{vmatrix}
+a  & b  & c  \\
+dl & dm & dn \\
+l  & m  & n
+\end{vmatrix}
+  = \frac{1}{ds^2}
+\begin{vmatrix}
+dx & dy & dz \\
+dl & dm & dn \\
+l  & m  & n
+\end{vmatrix};
+\]
+%% -----File: 040.png---Folio 32-------
+pour calculer le déterminant, nous le multiplierons encore
+par le même déterminant~$H$. Le produit sera,
+\[
+\begin{vmatrix}
+\sum \mfrac{\dd x}{\dd u}\, dx & \sum \mfrac{\dd x}{\dd v}\, dx            & \sum l\, dx \\
+\sum \mfrac{\dd x}{\dd u}\, dl & \sum \mfrac{\dd x}{\dd v}\, d\DPtypo{}{l} & \sum l\, dl \\
+\sum l\, \mfrac{\dd x}{\dd u}  & \sum l\, \mfrac{\dd x}{\dd v}             & \sum l^2
+\end{vmatrix}
+=
+\begin{vmatrix}
+E\, du + F\, dv                & F\, du + G\, dv                & 0 \\
+\sum \mfrac{\dd x}{\dd u}\, dl & \sum \mfrac{\dd x}{\dd v}\, dl & 0 \\
+0                              & 0                              & 1
+\end{vmatrix}.
+\]
+Nous avons d'ailleurs
+\[
+\sum l\, \frac{\dd x}{\dd u} = 0\Add{,}
+\]
+d'où en différentiant
+\[
+\sum \DPtypo{l}{dl}\, \frac{\dd x}{\dd u}
+  = -\sum \DPtypo{dl}{l}
+  \left(\frac{\dd^2 x}{\dd u^2}\, du + \frac{\dd^2 x}{\dd u\, \dd v}\, dv\right)
+  = -\frac{1}{H}\, (E'\, du + F'\, dv);
+\]
+de même
+\[
+\sum dl\, \frac{\dd x}{\dd v} = -\frac{1}{H}\, (F' · du + G' · dv)\Add{;}
+\]
+le produit est donc
+\[
+-\frac{1}{H} \begin{vmatrix}
+E\,  du + F\,  dv & F\,  du + G\,  dv \\
+E'\, du + F'\, dv & F'\, du + G'\, dv
+\end{vmatrix},
+\]
+et nous avons
+\[
+\Tag{(8)}
+\frac{1}{T} - \frac{d\theta}{ds} = \frac{1}{H^2\, ds^2}
+\begin{vmatrix}
+E'\, du + F'\, dv & E\, du + F\, dv \\
+F'\, du + G'\, dv & F\, du + G\, dv
+\end{vmatrix}.
+\]
+
+Les \Card{3} formules \Eq{(6)}\Add{,}~\Eq{(7)}\Add{,}~\Eq{(8)} permettent de calculer les
+\Card{3} éléments fondamentaux $\theta, R, T$.
+
+
+\ExSection{II}
+
+\begin{Exercises}
+\item[8.] On considère la surface~$S$ lieu des sections circulaires diamétrales
+d'une famille \DPchg{d'ellipsoides}{d'ellipsoïdes} homofocaux. Déterminer
+sur~$S$ les trajectoires orthogonales des sections circulaires
+qui l'engendrent.
+
+\item[9.] Déterminer toutes les représentations conformes d'une sphère
+sur un plan. Trouver celles qui donnent des systèmes connus
+de projections cartographiques.
+
+\item[10.] Sur une surface~$S$ on considère une courbe~$C$. Soit $M$ un de ses
+points, $MT$~la tangente à~$C$, $MN$~la normale à~$S$, et $MN'$~la normale
+à~$C$ qui \DPtypo{es}{et} tangente à~$(S)$. Montrer que les composantes
+de la rotation instantanée du trièdre~$\Tri{M}{T}{N'}{N}$ par rapport aux
+axes de ce trièdre sont les éléments fondamentaux $\dfrac{d\theta}{ds} - \dfrac{1}{T}$,
+$-\dfrac{\cos\theta}{R}$, $\dfrac{\sin\theta}{R}$.
+
+\item[11.]\phantomsection\label{exercice11}
+Si les courbes coordonnées de la surface~$S$, de l'exercice
+précédent, sont rectangulaires, soient $MU$~et~$MV$ leurs tangentes,
+et soit~$\phi$ l'angle~$(MU, MT)$. Déduire de la considération
+des mouvements des deux trièdres $\Tri{M}{T}{N'}{N}$~et~$\Tri{M}{U}{V}{N}$, lorsque $M$
+décrit~$C$, une formule de la forme
+\[
+\frac{\sin\theta}{R} - \frac{d\phi}{ds}
+  = r_{1}\frac{du}{ds} + r_{2}\frac{dv}{ds};
+\]
+et donner les expressions de $r_{1}$~et~$r_{2}$. Généraliser, en supposant
+les coordonnées $u$~et~$v$ quelconques.
+\end{Exercises}
+%% -----File: 041.png---Folio 33-------
+
+
+\Chapitre{III}{\DPchg{Etude}{Étude} des \DPtypo{Elements}{\DPchg{Eléments}{Éléments}} Fondamentaux des Courbes d'une
+Surface.}
+
+
+\Section{Courbure normale\Add{.}}
+{1.}{} Reprenons la \Ord{1}{ère} formule fondamentale
+\[
+\frac{\Cos \theta}{R}
+  = \frac{1}{H}\, \frac{E'\, du^2 + 2F'\, du\, dv + G'\, dv^2}
+                       {E\,  du^2 + 2F\,  du\, dv + G\,  dv^2},
+\]
+les différentielles secondes $d^2 u, d^2 v$ n'y figurent pas; $\dfrac{\Cos \theta}{R}$
+ne \DPtypo{depend}{dépend} que du rapport $\dfrac{dv}{du}$, c'est à dire de la direction
+de la tangente, \emph{$\dfrac{\Cos \theta}{R}$~est le même pour toutes les courbes de
+la surface tangentes à une même droite}. Considérons alors le
+\Figure{041a}
+centre de courbure~$C$ sur la normale principale~$MP$; si on prend
+pour axe polaire la normale~$MN$ à la surface,
+et pour pôle le point~$M$, $R\Add{,} \theta$ sont les
+coordonnées polaires du point~$C$. L'équation
+\[
+\frac{\Cos \theta}{R} = \cte,
+\]
+représente un cercle; le lieu du point~$C$
+est un cercle, ce qu'on peut encore voir comme il suit; considérons
+la droite polaire, elle est dans le plan normal à
+la courbe, donc elle rencontre la normale~$MN$ à la surface en
+un point~$K$, et nous avons
+\[
+R = M K \cos \theta,
+\]
+ou
+\[
+M K = \frac{R}{\Cos \theta}.
+\]
+%% -----File: 042.png---Folio 34-------
+$M K$~est constant, donc \emph{les droites polaires de toutes les
+courbes d'une surface passant par un même point~$M$ de cette
+surface et tangentes en ce point à une même droite rencontrent
+en un même point~$K$ la normale en~$M$ à la surface. Le lieu des
+centres de courbure de toutes ces courbes est le cercle de
+diamètre $M K$ \(cercle de Meusnier\)}. En particulier supposons
+$\theta = 0$, la normale principale se confond avec la normale à la
+surface, le plan osculateur passe par la normale, il est normal
+à la surface. Coupons la surface par ce plan, $K$~est le
+centre de courbure en~$M$ de la section, soit $R_n$~le rayon de
+courbure, nous avons
+\[
+\frac{\Cos \theta}{R} = \frac{1}{R_n},
+\]
+d'où
+\[
+R = R_n \Cos \theta.
+\]
+D'où le \emph{Théorème de Meusnier: Le centre de courbure en~$M$
+d'une courbe tracée sur une surface est la projection sur le
+plan osculateur en~$M$ à cette courbe du centre de courbure de
+la section normale tangente en~$M$ à la courbe}.
+
+Le Théorème est en \DPtypo{defaut}{défaut} si
+\[
+\Psi(du, dv) = E'\Add{·} du^2 + 2F'\, du · dv + G' · dv^2 = 0\Add{.}
+\]
+Alors $\dfrac{\Cos \theta}{R} = 0$, $R$~est en général infini. La formule devient
+complètement indéterminée si on a en même temps $\Cos \theta = 0$,
+alors la normale principale est perpendiculaire à la normale
+à la surface, le plan osculateur à la courbe est tangent à la
+surface. Les deux tangentes qui correspondent à ce cas d'exception
+s'appellent \emph{les deux directions asymptotiques} (correspondant
+au point~$M$ considéré).
+%% -----File: 043.png---Folio 35-------
+
+Le Théorème est également en défaut si
+\[
+\Phi(du, dv) = E\, du^2 + 2F · du · dv + G\, dv^2 = 0;
+\]
+alors $\dfrac{\cos \theta}{R}$ est infini, $R$~est nul en général. La direction de
+la tangente est telle que
+\[
+dx^2 + dy^2 + dz^2 = 0
+\]
+c'est une droite isotrope du plan tangent. Il y a donc \emph{deux
+directions isotropes correspondantes à chaque point~$M$ de la
+surface}.
+
+
+\Section{Variations de la courbure normale\Add{.}}
+{2.}{} Le Théorème de Meusnier nous montre que, pour étudier
+la courbure des diverses courbes passant par un point d'une
+surface, on peut se borner à considérer les sections normales
+passant par les différentes tangentes à la surface au point
+considéré.
+
+Nous avons
+\[
+\frac{1}{R_n} = \frac{1}{H}\,
+  \frac{E'\, du^2 + 2F'\, du · dv + G'\, dv^2}{E · du^2 + 2F\, du · dv + G · dv^2}\Add{.}
+\]
+Considérons dans la plan tangent en~$M$ les tangentes $M U, M V$
+aux courbes coordonnées $v = \Cte$ et $u = \Cte$ qui passent par~$M$,
+et considérons le trièdre constitué par $M U, M V$ et la normale
+$M N$ à la surface: les cosinus directeurs des axes sont
+\[
+\begin{array}{lr@{\,}lr@{\,}lr@{\,}l}
+MU: & \mfrac{dx}{ds}
+  = \mfrac{\dd x}{\dd u}\Add{·} \mfrac{du}{ds} = \mfrac{1}{\sqrt{E}} · \mfrac{\dd x}{\dd u} =& l', &
+\mfrac{1}{\sqrt{E}} · \mfrac{\dd y}{\dd u} =& m', &
+\mfrac{1}{\sqrt{E}}\Add{·} \mfrac{\dd z}{\dd u} =& n'\Add{,} \\
+%
+MV: & \mfrac{dx}{ds}
+  = \mfrac{\dd x}{\dd v} · \mfrac{dv}{ds} = \mfrac{1}{\sqrt{G}} · \mfrac{\dd x}{\dd v} = & l'', &
+\mfrac{1}{\sqrt{G}} · \mfrac{\dd y}{\dd v} =& m'', &
+\mfrac{1}{\sqrt{G}}\Add{·} \mfrac{\dd z}{\dd v} =& n''\Add{,} \\
+%
+MN: && \Err{1}{l}, && m, && n\Add{.}
+\end{array}
+\]
+%% -----File: 044.png---Folio 36-------
+
+Considérons alors une tangente $M T$ quelconque, \DPtypo{definie}{définie}
+par les valeurs $du, dv$ des différentielles des coordonnées.
+Les cosinus directeurs sont:
+\begin{alignat*}{4}
+\frac{dx}{ds}
+  &= \frac{\dd x}{\dd u}\, \frac{du}{ds} &&+ \frac{\dd x}{\dd v}\, \frac{dv}{ds}
+   &&= \sqrt{E}\, \frac{du}{ds} · l' &&+ \sqrt{G} · \frac{dv}{ds}\, l'', \\
+%
+\frac{dy}{ds}
+  &= \frac{\dd y}{\dd u}\, \frac{du}{ds} &&+ \frac{\dd y}{\dd v}\, \frac{dv}{ds}
+   &&= \sqrt{E}\, \frac{du}{ds} · m' &&+ \sqrt{G}\Add{·} \frac{dv}{ds}\, m'', \\
+%
+\frac{dz}{ds}
+  &= \frac{\dd z}{\dd u}\, \frac{du}{ds} &&+ \frac{\dd z}{\dd v}\, \frac{dv}{ds}
+   &&= \sqrt{E}\, \frac{du}{ds} · n' &&+ \sqrt{G}\Add{·} \frac{dv}{ds}\, n''\Add{.}
+\end{alignat*}
+
+Ces formules montrent que le segment directeur de $M T$
+est la somme géométrique de deux segments, de valeurs algébriques
+\[
+\lambda = \sqrt{E}\, \frac{du}{ds},\qquad \mu = \sqrt{G}\, \frac{dv}{ds},
+\]
+portés respectivement sur $M U$~et~$M V$. En d'autres termes:
+$\lambda, \mu$~sont les paramètres directeurs de~$M T$ dans le système
+de coordonnées $U\Add{,} M\Add{,} V$.
+
+La formule de $R_n$ devient, en y introduisant ces paramètres
+directeurs:
+\begin{align*}%[** TN: Not broken in original]
+\frac{1}{R_n}
+  &= \frac{1}{H}\left( E' \Bigl(\frac{du}{ds}\Bigr)^2 + 2 F'\, \frac{du}{ds} · \frac{dv}{ds} + G' \Bigl(\frac{dv}{ds}\Bigr)^2 \right) \\
+  &= \frac{1}{H} \left(\frac{E'}{E}\, \lambda^2 + \frac{2F'}{\sqrt{EG}}\, \lambda \mu + \frac{G'}{G}\, \mu^2 \right)\Add{.}
+\end{align*}
+Et si on considère le point~$P$ obtenu en portant sur~$M T$ un
+segment égal à $±\sqrt{R_n}$, le lieu de ce point~$P$, dont les coordonnées,
+dans le système $M\Add{,} U\Add{,} V$, sont:
+\[
+U = ±\lambda \sqrt{R_n},\qquad
+V = ±\mu \sqrt{R_n},
+\]
+aura pour équation
+\[
+\frac{E'}{E}\, U^2 + \frac{2F'}{\sqrt{EG}}\, UV + \frac{G'}{G}\, V^2 = H.
+\]
+
+C'est une conique à centre située dans le plan tangent,
+qu'on appelle \emph{indicatrice} de la surface au point~$M$. La conique
+\DPtypo{tracee}{tracée}, on a immédiatement le rayon de courbure d'une section
+%% -----File: 045.png---Folio 37-------
+normale quelconque, et on suit sans peine la variation du rayon
+de courbure, quand $M T$ varie.
+
+La nature de l'indicatrice dépend du signe de $\dfrac{E'G' - F'{}^2}{E\Del{·} G}$,
+ou puisque $E\Add{,} G$ sont positifs, de $E'G' - F'{}^2$. Dans le cas
+ou l'équation de la surface est
+\[
+Z = f(x, y)
+\]
+on a en prenant les notations habituelles
+\[
+ds^2 = (1 + p^2) · dx^2 + 2pq · dx · dy + (1 + q^2)\Add{·} dy^2
+\]
+d'où
+\[
+E = 1 + p^2\Add{,}\qquad F = p \Del{·} q\Add{,}\qquad G = 1 + q^2
+\]
+et
+\[
+H = \sqrt{E · G - F^2} = \sqrt{1 + p^2 + q^2}\Add{.}
+\]
+Maintenant
+\[
+A = -p,\qquad B = -q,\qquad C = 1,
+\]
+et
+\[
+\sum A · d^2x = -\sum dA · dx  = dp · dx + dq\Add{·} dy\Add{.}
+\]
+Mais
+\[
+dp = r\, dx + s\, dy,\qquad
+dq = s\, dx + t · dy,
+\]
+donc
+\[
+\sum A · d^2x = r · \Err{d^2x}{dx^2} + 2s · dx\, dy + t\Add{·} dy^2;
+\]
+donc
+\[
+E' = r,\qquad F' = s,\qquad G' = t,
+\]
+et
+\[
+E'G' - F'{}^2 = rt - s^2.
+\]
+
+\ParItem{\Primo} $E'G' - F'{}^2 > 0$, la conique est une \Err{e lipse}{ellipse}, tous les
+rayons de courbure sont de même signe, on dit que la surface
+est \emph{convexe} au point~$M$; elle est toute entière d'un même côté
+du plan tangent en~$M$ dans le voisinage du point~$M$.
+
+\ParItem{\Secundo} $E'G' - F'{}^2 < 0$, l'indicatrice est une hyperbole. La surface
+traverse au point~$M$ son plan tangent; elle est dite \emph{à
+courbures opposées}.
+
+\ParItem{\Tertio} $E'G' - F'{}^2 = 0$\Add{,} l'indicatrice est du genre parabole, et
+comme elle est à centre, elle se réduit à un système de deux
+droites parallèles. Le point~$M$ est dit \emph{point parabolique}.
+%% -----File: 046.png---Folio 38-------
+
+Considérons le cas particulier où $\dfrac{1}{R_n}$ est constant, quelle
+que soit la section que l'on \DPtypo{considèré}{considéré}. Il faut et il suffit
+pour cela que $\dfrac{1}{R_n}$ soit indépendant de~$\dfrac{du}{dv}$, donc que l'on ait
+\[
+\frac{E'}{E} = \frac{F'}{F} = \frac{G'}{G}\DPtypo{,}{.}
+\]
+%[** TN: Removed paragraph indentation/break.]
+Or\Add{,} l'angle de $M U, M V$ est donné par
+\[
+\cos \theta = \sum l' · l'' = \frac{F}{\sqrt{EG}};
+\]
+%[** TN: Removed paragraph indentation/break.]
+\DPtypo{Ces}{ces} conditions peuvent donc s'écrire\Del{:}
+\[
+\frac{E'}{E} =  \frac{\ \dfrac{F'}{\sqrt{EG}}\ }{\Cos \theta} = \frac{G'}{G},
+\]
+et expriment que l'indicatrice est un cercle.
+Le point~$M$ est dit alors un \emph{ombilic}.
+
+Cherchons les directions des axes de l'indicatrice. Ce
+sont des directions conjuguées par rapport aux directions asymptotiques
+de l'indicatrice, définies par
+\[
+\Psi(du, dv) = 0
+\]
+et par rapport aux directions isotropes du plan tangent, définies
+par
+\[
+\Phi(du, dv) = 0\Add{.}
+\]
+Elles sont donc définies par l'équation
+\[
+\frac{\ \dfrac{\dd \Psi}{\dd\, du}\ }{\dfrac{\dd \Phi}{\dd\, du}}
+  = \frac{\ \dfrac{\dd \Psi}{\dd\, dv}\ }{\dfrac{\dd \Phi}{\dd\, dv}}
+  = \frac{\Psi(du, dv)}{\Phi(du, dv)} = \frac{H}{R} = S,
+\]
+puisque $du, dv$ sont des coordonnées homogènes pour les directions
+$M T$ du plan tangent.
+
+Ce sont les \emph{directions principales}. Les rayons de courbure
+correspondants sont dits \emph{rayons de courbure principaux}.
+%% -----File: 047.png---Folio 39-------
+
+L'équation qui définit les directions principales est
+donc:
+\[
+\begin{vmatrix}
+E · du + F · dv & F · du + G · dv \\
+E' ·du + F'· dv & F'· du + G'· dv
+\end{vmatrix} = 0;
+\]
+le \Ord{1}{er} membre est un covariant simultané des formes $\Phi, \Psi$.
+
+L'équation aux rayons de courbure principaux s'obtiendra
+en éliminant $du, dv$ entre les équations
+\[
+\frac{\dd \Psi}{\dd\Add{·} du} = S\, \frac{\dd \Phi}{\dd\Add{·} du},\qquad
+\frac{\dd \Psi}{\dd\Add{·} dv} = S\, \frac{\dd \Phi}{\dd\Add{·} dv}\DPtypo{,}{.}
+\]
+Ce qui donne
+\[
+\begin{vmatrix}
+E' - S E & F' - S F \\
+F' - S F & G' - S G
+\end{vmatrix} = 0,
+\]
+ou
+\[
+S^2 (E · G - F^2) - S (E · G' + G · E' - 2FF') + E'G' - F'{}^2 = 0
+\]
+avec
+\[
+S = \frac{H}{R}.
+\]
+
+Supposons maintenant que les courbes coordonnées soient
+tangentes aux directions principales. Ces directions sont rectangulaires;
+donc les courbes coordonnées constituent un réseau
+orthogonal, et $F = 0$; alors l'indicatrice étant rapportée
+à ses axes on a
+\[
+F' = 0,\quad H = \sqrt{E · G},\quad\text{et}\quad
+\frac{1}{R_n} = \frac{\lambda^2 E'}{E \sqrt{EG}} + \frac{\mu^2 G'}{G \sqrt{EG}}\Add{.}
+\]
+Si nous supposons $\lambda = 1$, $\mu = 0$, nous avons un des rayons de
+courbure principaux~$R_1$
+\[
+\frac{1}{R_1} = \frac{E'}{E \sqrt{EG}}\Add{;}
+\]
+pour $\mu = 1$, $\lambda = 0$, nous avons l'autre rayon de courbure principal~$R_2$
+\[
+\frac{1}{R_2} = \frac{G'}{G \sqrt{EG}}\Add{,}
+\]
+%% -----File: 048.png---Folio 40-------
+et la formule devient
+\[
+\frac{1}{R_n} = \frac{\lambda^2}{R_1} + \frac{\mu^2}{R_2}\Add{;}
+\]
+mais ici, les coordonnées étant rectangulaires, si $\phi$ est
+l'angle de la tangente $M T$ avec l'une des directions principales,
+nous avons $\lambda = \cos \phi$, $\mu = \sin \phi$, et nous obtenons
+la \emph{formule d'Euler}
+\[
+\frac{1}{R_n} = \frac{\cos^2 \phi}{R_1} + \frac{\sin^2 \phi}{R_2}\Add{.}
+\]
+Considérons la tangente $M T'$ perpendiculaire à~$M T$, il faudra
+remplacer $\phi$ par $\phi + \frac{\pi}{2}$, et nous aurons
+\[
+\frac{1}{R'_n} = \frac{\sin^2 \phi}{R_1} + \frac{\cos^2 \phi}{R_2}
+\]
+d'où
+\[
+\frac{1}{R_n} + \frac{1}{R'_n} = \frac{1}{R_1} + \frac{1}{R_2}
+\]
+donc \emph{la somme des courbures de \Card{2} sections normales rectangulaires
+quelconques est constante et égale à la somme des courbures
+des sections normales principales}. La quantité constante
+$\dfrac{1}{2} \left(\dfrac{1}{R_1} + \dfrac{1}{R_2}\right)$ s'appelle \emph{courbure moyenne} de la surface au point
+considéré.
+
+
+\Section{Lignes minima.}
+{3.}{} En chaque point d'une surface, il y a \Card{3} couples de
+directions remarquables: les droites isotropes du plan tangent,
+définies par $\Phi(du, dv) = 0$; les directions asymptotiques
+de l'indicatrice $\Psi(du, dv) = 0$\DPtypo{;}{,} et les directions des sections
+principales.
+
+Considérons les directions isotropes, et cherchons s'il
+existe sur la surface des courbes tangentes en chacun de leurs
+points à une direction isotrope; ceci revient à intégrer l'équation
+%% -----File: 049.png---Folio 41-------
+\[
+\Phi(du, dv) = 0,
+\]
+et on obtient ainsi les \emph{courbes minima}. L'équation \DPtypo{précedente}{précédente}
+se décompose en \Card{2} équations de \Ord{1}{er} ordre, donc \emph{il y a deux
+familles de courbes minima et par tout point de la surface
+passe en général une courbe de chaque famille}. Ces courbes
+sont imaginaires; on a le long de chacune d'elles
+\[
+ds^2 = dx^2 + dy^2 + dz^2 = 0;
+\]
+c'est pourquoi on les appelle aussi lignes de longueur nulle.
+Si on les prend pour lignes coordonnées, l'équation $\Phi(du, dv) = 0$
+devant alors être vérifiée pour $du = 0$, $dv = 0$, on a
+\[
+E = 0\Add{,}\quad G = 0,\quad \text{et}\quad ds^2 = 2F\, du · dv.
+\]
+
+En général les deux systèmes de lignes minima sont distincts.
+Pour qu'ils soient confondus, il faut que
+\[
+EG - F^2 = H^2 = 0,
+\]
+dans ce cas, on a $A^2 + B^2 + C^2 = 0$, et les formules fondamentales
+ne s'appliquent plus. Pour étudier la nature d'une telle surface
+\DPtypo{Considérons}{considérons} le plan tangent:
+\[
+A(X - x) + B(Y - y) + C(Z - z) = 0;
+\]
+ce plan est alors tangent à un cône isotrope, c'est un \emph{plan isotrope}.
+\emph{Tous les plans tangents à la surface sont isotropes.}
+Cherchons l'équation générale des plans isotropes. Soit
+\[
+ax + by + cz + d = 0
+\]
+nous avons la condition
+\[
+a^2 + b^2 + c^2 = 0
+\]
+ou
+\begin{gather*}
+a^2 + b^2 = -c^2\Add{,} \\
+(a + ib) · (a - ib) = \Err{-c}{-c^2}\Add{.}
+\end{gather*}
+%% -----File: 050.png---Folio 42-------
+Posons
+\[
+a + ib = tc,\qquad
+a - ib = -\frac{1}{t}\, c,
+\]
+ou
+\[
+a + ib - tc = 0,\qquad ta - ibt + c = 0;
+\]
+de ces deux relations nous tirons
+\[
+\frac{a}{1 - t^2} = \frac{ib}{-(1 + t^2)} = \frac{c}{-2t},
+\]
+ou
+\[
+\frac{a}{1 - t^2} = \frac{b}{i(1 + t^2)} = \frac{c}{-2t};
+\]
+d'où l'équation générale des plans isotropes
+\[
+\Tag{(1)}
+(1 - t^2)x + i(1 + t^2)y - 2tz + 2w = 0.
+\]
+Un plan isotrope dépend de deux paramètres. La surface considérée
+est l'enveloppe de plans isotropes; si ces plans dépendent
+de deux paramètres, elle se réduit au cercle imaginaire
+à l'infini. Supposons alors que $w$~soit fonction de~$t$ par exemple;
+le plan tangent ne dépendant que d'un paramètre, la surface
+est développable, c'est une \emph{développable isotrope}. Cherchons
+son arête de rebroussement. Différentions l'équation~\Eq{(1)}
+\Card{2} fois par rapport à~$t$. Nous avons
+\begin{gather*}
+\Tag{(2)}
+-tx + ity - z + w' = 0 \\
+\Tag{(3)}
+-x + iy + w'' = 0
+\end{gather*}
+les équations \Eq{(1)}\Add{,}~\Eq{(2)}\Add{,}~\Eq{(3)} définissent l'arête de rebroussement;
+\Eq{(3)}~donne
+\[
+x - iy = w''\Add{,}
+\]
+\Eq{(2)}~s'écrit
+\[
+z = -t(x - iy) + w' = w' - tw''\Add{,}
+\]
+et~\Eq{(1)}
+\[
+x + iy = t^2(x - iy) + 2tz - 2w = t^2w'' + 2t(w' - tw'') - 2w
+\]
+d'où, pour les équations de l'arête de rebroussement:
+\[
+\Tag{(4)}
+x - iy = w'',\qquad
+x + iy = -2w + 2tw' - t^2w'',\qquad
+z = w' - tw''.
+\]
+Nous en tirons
+\[
+d(x - iy) = w'''\, dt,\qquad
+d(x + iy) = -t^2w'''\, dt,\qquad
+dz = -tw'''\, dt;
+\]
+%% -----File: 051.png---Folio 43-------
+d'où
+\[
+d(x - i y) · d(x + i y)
+  = - t^2 \DPtypo{w'''}{(w''')^2}\, dt^2
+  = \DPtypo{}{-}dz^2\Add{,}
+\]
+ou
+\begin{gather*}
+d(x - i y) · d(x + i y) + dz^2 = 0, \\
+dx^2 + dy^2 + dz^2 = 0;
+\end{gather*}
+c'est une courbe minima. \emph{L'arête de rebroussement d'une développable
+isotrope est une courbe minima.}
+
+Réciproquement, considérons une courbe minima. Nous avons
+la relation
+\[
+dx^2 + dy^2 + dz^2 = 0\Add{.}
+\]
+Différentions
+\[
+dx · d^2 x + dy · d^2 y + dz · d^2 z = 0\Add{,}
+\]
+mais l'identité de Lagrange nous donne
+\[
+\sum dx^2 \sum (d^2 x)^2 - \DPtypo{\sum (dx·d^2 x)}{\left(\sum dx·d^2 x\right)^2}
+  = \sum (dy·d^2 z - dz·d^2 y)^2 = 0\Add{,}
+\]
+ou\DPtypo{,}{} $A\Add{,} B\Add{,} C$ désignant les coefficients du plan osculateur
+\[
+A^2 + B^2 + C^2 = 0\Add{.}
+\]
+\emph{Le plan osculateur en un point d'une courbe minima est isotrope.
+Toute courbe minima peut être considérée comme l'arête
+de rebroussement d'une développable isotrope.}
+
+Il en résulte que cette arête de rebroussement est la
+courbe minima la plus générale, et que les coordonnées d'un
+point d'une courbe minima quelconque sont données par les
+formules~\Eq{(4)}, ou $w$~est une fonction arbitraire de~$t$.
+
+%[** TN: Renumber 3 -> 4]
+\Section{Lignes asymptotiques.}
+{4.}{} Si nous cherchons maintenant les courbes d'une surface
+tangentes en chacun de leurs points à une asymptote de
+l'indicatrice, nous sommes ramenés à intégrer l'équation
+\[
+\Psi(du , dv) = 0\Add{,}
+\]
+et nous obtenons les \emph{lignes asymptotiques}. Comme précédemment,
+%% -----File: 052.png---Folio 44-------
+nous voyons qu'\emph{il y a deux familles de lignes asymptotiques,
+et par tout point de la surface passe en général une asymptotique
+de chaque famille}.
+
+L'équation différentielle précédente s'écrit
+\begin{align*}
+\sum A\, d^2 x &= 0\Add{,} \\
+\intertext{on a d'ailleurs}
+\sum A\, dx &= 0;
+\end{align*}
+mais $A\Add{,} B\Add{,} C$ sont les coefficients du plan tangent à la surface;
+les équations précédentes montrent qu'il contient les directions
+$dx, dy, dz$ et $d^2 x, d^2 y, d^2 z$, donc \DPtypo{coincide}{coïncide} avec le plan osculateur
+à la courbe; donc \emph{les lignes asymptotiques sont telles
+que le plan osculateur en chacun de leurs points soit tangent
+à la surface}. En particulier, \emph{toute génératrice rectiligne
+d'une surface est une ligne asymptotique}, car le plan osculateur
+en un point d'une droite étant indéterminé, peut être
+considéré comme \DPtypo{coincidant}{coïncident} avec le plan tangent en ce point
+à la surface. \emph{Si donc une surface est réglée, un des systèmes
+de lignes asymptotiques est constitué par les génératrices
+rectilignes.}
+
+Si nous prenons les lignes asymptotiques pour courbes
+coordonnées, nous aurons
+\[
+E' = G' = 0
+\]
+et
+\[
+\Psi(du, dv) = 2F'\, du · dv.
+\]
+
+Les lignes asymptotiques sont réelles aux points où la
+surface est à courbures opposées, imaginaires aux points où
+elle est convexe. Elles sont en général distinctes, et distinctes
+aussi des lignes minima. Nous allons examiner les cas
+d'exception.
+
+\ParItem{\Primo.} \emph{Les lignes asymptotiques sont confondues.} Prenons
+%% -----File: 053.png---Folio 45-------
+l'équation de la surface sous la \DPtypo{formé}{forme}
+\[
+Z = f(x, y):
+\]
+nous avons $E'G' - F^2 = 0$, condition qui se réduit ici à
+\[
+rt - s^2 = 0;
+\]
+tous les points de la surface doivent être paraboliques. L'équation
+différentielle précédente peut s'écrire
+\[
+dp \DPtypo{,}{·} dx + dq · dy = 0.
+\]
+Elle montre que si l'une des différentielles $dp, dq$ est nulle,
+l'autre est aussi nulle, donc $p, q$ sont fonctions l'un de l'autre.
+Le plan tangent en un point s'écrit
+\[
+p(X - x) \DPtypo{,}{+} q(Y - y) - (Z - z) = 0 ,
+\]
+ou
+\[
+pX + qY - Z = px + qy - z.
+\]
+Mais
+\[
+d(px + qy - z) = x·dp + y·dq
+\]
+et nous voyons que si $dp = 0$, puisque $dq = 0$, on a en même
+temps $d(px + qy - z) = 0$, donc $px + qy - z$ est fonction
+de~$p$, de même que~$q$, et alors le plan tangent ne dépend que
+d'un seul paramètre, et la surface est développable. La \DPtypo{reciproque}{réciproque}
+est évidente, car si l'équation $pX + qY - Z = px + qy - z$
+ne dépend que d'un paramètre~$\theta$, $dp$~et~$dq$ sont proportionnels
+à~$d\theta$, et les deux formes linéaires $dp = r·dx + s·dy$\Add{,}
+$dq = s·dx + t·dy$ ne sont pas indépendantes. On a donc bien
+\[
+\begin{vmatrix}
+r & s \\
+s & t
+\end{vmatrix} = rt - s^2 = 0.
+\]
+
+Donc \emph{les surfaces à lignes asymptotiques doubles sont
+les surfaces développables, et les lignes asymptotiques doubles
+sont les génératrices rectilignes. Pour les développables
+isotropes, les lignes asymptotiques doubles sont confondues
+avec les lignes minima doubles, qui sont les génératrices
+%% -----File: 054.png---Folio 46-------
+rectilignes isotropes}.
+
+\Paragraph{Remarque.} Pour les surfaces développables, l'arête de
+rebroussement ayant son plan osculateur tangent à la surface
+doit être considérée comme une ligne asymptotique. Son équation
+est en effet une \DPtypo{integrale}{intégrale} singulière de l'\DPtypo{equation}{équation}
+différentielle des lignes asymptotiques.
+
+\ParItem{\Secundo.} \emph{Une famille de lignes asymptotiques est confondue
+avec une famille de lignes minima.} \DPchg{Ecartons}{Écartons} le cas des développables
+isotropes, qui vient d'être examiné. Prenons les
+lignes minima comme courbes coordonnées, $E = 0$, $G = 0$, et si
+nous supposons que la famille $v = \cte$ constitue une famille
+d'asymptotiques, $dv = 0$ doit être solution de $\Psi(du, dv) = 0$,
+donc \DPtypo{$E' = 0$}{}
+\[
+E' =
+\begin{vmatrix}
+\mfrac{\dd x}{\dd u} & \mfrac{\dd y}{\dd u} & \mfrac{\dd z}{\dd u} \\
+\mfrac{\dd x}{\dd v} & \mfrac{\dd y}{\dd v} & \mfrac{\dd z}{\dd v} \\
+\mfrac{\dd^2 x}{\dd u^2} & \mfrac{\dd^2 y}{\dd u^2} & \mfrac{\dd^2 z}{\dd u^2}
+\end{vmatrix} = 0\Add{.}
+\]
+Il existe donc entre les éléments des lignes de ce déterminant
+une même relation linéaire et homogène.
+On a
+\[
+\left\{%[** Moved brace to left-hand side]
+\begin{aligned}
+\frac{\dd^2 x}{\dd u^2} &= M\, \frac{\dd x}{\dd u} + N\, \frac{\dd x}{\dd v}\Add{,} \\
+\frac{\dd^2 y}{\dd u^2} &= M\, \frac{\dd y}{\dd u} + N\, \frac{\dd y}{\dd v}\Add{,} \\
+\frac{\dd^2 z}{\dd u^2} &= M\, \frac{\dd z}{\dd u} + N\, \frac{\dd z}{\dd v}\Add{.}
+\end{aligned}
+\right.
+\]
+Multiplions respectivement par $\dfrac{\dd x}{\dd u}, \dfrac{\dd y}{\dd u}, \dfrac{\dd z}{\dd u}$ et ajoutons. Le
+coefficient de~$M$ est $E = 0$, le \Ord{1}{e} membre est $\dfrac{1}{2}\, \dfrac{\dd E}{\dd u} = 0$, donc
+$NF = 0$, et comme $F \neq 0$, $N = 0$, et nous avons:
+%% -----File: 055.png---Folio 47-------
+\[
+\frac{\ \dfrac{\dd^2 x}{\dd u^2}\ }{\dfrac{\dd x}{\dd u}} =
+\frac{\ \dfrac{\dd^2 y}{\dd u^2}\ }{\dfrac{\dd y}{\dd u}} =
+\frac{\ \dfrac{\dd^2 z}{\dd u^2}\ }{\dfrac{\dd z}{\dd u}} = M\Add{,}
+\]
+les courbes $v = \cte$ sont des droites, et comme ce sont des
+lignes minima, ce sont des droites isotropes. Et réciproquement
+si les courbes $v = \cte$ sont des droites, on a
+\[
+\frac{\dd^2 x}{\dd u^2} = M\, \frac{\dd x}{\dd u},\qquad
+\frac{\dd^2 y}{\dd u^2} = M\, \frac{\dd y}{\dd u},\qquad
+\frac{\dd^2 z}{\dd u^2} = M\, \frac{\dd z}{\dd u};
+\]
+et par suite
+\[
+\sum A\, \frac{\dd^2 x}{\dd u^2} = M \sum A\, \frac{\dd x}{\dd u} = 0
+\]
+les courbes $v = \cte$ qui sont des droites minima sont des lignes
+asymptotiques. Donc \emph{les surfaces qui ont une famille
+d'asymptotiques confondue avec une famille de lignes minima
+sont des surfaces \DPtypo{réglees}{réglées} à génératrices isotropes, et ces
+génératrices sont les asymptotiques confondues avec les courbes
+minima}.
+
+\ParItem{\Tertio.} \emph{Les deux systèmes d'asymptotiques sont des courbes
+minima.} En prenant toujours les lignes minima comme courbes
+coordonnées, il faut que l'équation $\Psi(du, dv) = 0$ soit satisfaite
+pour $du = 0$, $dv = 0$, il faut donc que $E' = G' = 0$.
+Alors les formes quadratiques $\Phi$~et~$\Psi$ sont proportionnelles.
+Il en est de même avec un système de coordonnées quelconques
+et on~a
+\[
+\frac{E'}{E} = \frac{F'}{F} = \frac{G'}{G}\Add{.}
+\]
+L'indicatrice en un point quelconque est un cercle, \emph{tous les
+points de la surface sont des ombilics}. En reprenant le calcul
+comme précédemment, on verra que la surface admet deux
+systèmes de génératrices rectilignes isotropes. \emph{C'est une
+sphère.}
+%% -----File: 056.png---Folio 48-------
+
+%[** TN: Renumber 4 -> 5]
+\Section{Surfaces minima\Add{.}}
+{5.}{} Ce dernier cas nous a conduit à étudier la surface
+telle que l'indicatrice soit toujours un cercle. Examinons
+maintenant \emph{le cas où cette indicatrice est toujours une hyperbole
+équilatère}. Ceci revient à chercher les surfaces
+pour lesquelles les lignes asymptotiques sont orthogonales.
+Il faut pour cela que l'on ait
+\[
+EG' + GE' - 2FF' = 0,
+\]
+ou
+\[
+\frac{1}{R_1} + \frac{1}{R_2} = 0.
+\]
+Les rayons de courbure en chaque point sont égaux et de signes
+contraires; la surface est dite une \emph{surface minima}.
+
+Prenons pour coordonnées les lignes minima. Alors $E = 0$\Add{,}
+$G = 0$, et
+\[
+ds^2 = 2F · du · dv;
+\]
+la condition précédente donne $F' = 0$, et
+\[
+\Psi(du, dv) = E'\, du^2 + G'\, dv^2.
+\]
+Mais on a
+\[
+F' =
+\begin{vmatrix}
+\mfrac{\dd x}{\dd u} & \mfrac{\dd y}{\dd u} & \mfrac{\dd z}{\dd u} \\
+\mfrac{\dd x}{\dd v} & \mfrac{\dd y}{\dd v} & \mfrac{\dd z}{\dd v} \\
+\mfrac{\dd^2 x}{\dd u\, \dd v} & \mfrac{\dd^2 y}{\dd u\, \dd v} & \mfrac{\dd^2 z}{\dd u\, \dd v}
+\end{vmatrix} = 0.
+\]
+Il existe donc une même relation linéaire et homogène entre
+les éléments des lignes. On a
+\[
+\left\{%[** TN: Moved brace to left]
+\begin{aligned}
+\frac{\dd^2 x}{\dd u\, \dd v} = M\, \frac{\dd x}{\dd u} + N\, \frac{\dd x}{\dd v}\Add{,} \\
+\frac{\dd^2 y}{\dd u\, \dd v} = M\, \frac{\dd y}{\dd u} + N\, \frac{\dd y}{\dd v}\Add{,} \\
+\frac{\dd^2 z}{\dd u\, \dd v} = M\, \frac{\dd z}{\dd u} + N\, \frac{\dd z}{\dd v}\Add{.}
+\end{aligned}
+\right.
+\]
+Multiplions respectivement par $\dfrac{\dd x}{\dd u}$, $\dfrac{\dd y}{\dd u}$, $\dfrac{\dd z}{\dd u}$ et ajoutons. Le
+\Ord{1}{er} membre est $\dfrac{1}{2}\, \dfrac{\dd E}{\dd u} = 0$; le coefficient de~$M$ est $E = 0$;
+%% -----File: 057.png---Folio 49-------
+nous avons donc $N F = 0$, donc $N = 0$. De même en multipliant
+par $\dfrac{\dd x}{\dd v}$, $\dfrac{\dd y}{\dd v}$, $\dfrac{\dd z}{\dd v}$ et ajoutant, on trouvera $M = 0$; donc on a
+\[
+\frac{\dd^2 x}{\dd u\, \dd v} = 0,\qquad
+\frac{\dd^2 y}{\dd u\, \dd v} = 0,\qquad
+\frac{\dd^2 z}{\dd u\, \dd v} = 0.
+\]
+Ce qui donne
+\[
+x = f(u) + \phi(v),\qquad
+y = g(u) + \psi(v),\qquad
+z = h(u) + \chi(v);
+\]
+les surfaces représentées par des équations de cette forme
+sont dites \emph{surfaces de translation. Elles peuvent être engendrées
+de deux façons différentes par la translation d'une
+courbe de forme invariable dont un point décrit une autre
+courbe}. Considérons en effet sur la surface les \Card{4} points
+$M_0(u_0\Add{,} v_0)$, $M_1(u\Add{,} v_0)$, $M_2(u_0\Add{,} v)$, $M(u\Add{,} v)$. D'après les formules précédentes
+ces points sont les sommets d'un parallélogramme. Si,
+laissant $v_0$~fixe, on fait varier~$u$, le point~$M_1$ décrit une
+courbe~$\Gamma$ de la surface; de même si, laissant $u_0$~fixe, on fait
+varier~$v$, le point~$M_2$ décrit une autre courbe~$\Gamma'$ de la surface.
+On peut donc considérer la surface comme engendrée par la
+courbe~$\Gamma$ animée d'un mouvement de translation dans lequel le
+point~$M_2$ décrit la courbe~$\Gamma'$, ou par la courbe~$\Gamma'$ animée d'un
+mouvement de translation dans lequel le point~$M_1$ décrit la
+courbe~$\Gamma$.
+
+Pour les surfaces minima, les \Card{6} fonctions $f\Add{,} g\Add{,} h\Add{,} \phi\Add{,} \psi\Add{,} \chi$ ne
+sont pas quelconques. Elles doivent satisfaire aux relations
+\[
+E = f'{}^2 + g'{}^2 + h'{}^2 = 0,\qquad G = \phi'{}^2 + \psi'{}^2 + \chi'{}^2 = 0;
+\]
+il en résulte que la courbe
+\[
+x = f(u),\qquad y = g(u),\qquad  z = \Err{h()}{h(u)}
+\]
+est une courbe minima, et si nous nous reportons aux équations
+générales d'une courbe minima, nous voyons que nous pouvons
+%% -----File: 058.png---Folio 50-------
+écrire, $F$~étant une fonction quelconque de~$u$
+\begin{align*}%[** TN: Unaligned in original]
+f(u) - ig(u) &= F''(u), \\
+f(u) + ig(u) &= -2F(u) + 2uF'(u) - u^2F''(u), \\
+h(u) &= F'(u) - uF''(u).
+\end{align*}
+De même la courbe
+\[
+x = \phi(v),\qquad
+y = \psi(v),\qquad
+z = \chi(v)
+\]
+étant une courbe minima, on aura, $\Phi$~étant une fonction quelconque
+de~$v$,
+\begin{align*}%[** TN: Unaligned in original]
+\phi(v) - i \psi(v) &= \Phi''(v)\Add{,} \\
+\phi(v) + i \psi(v) &= -2 \Phi(v) + 2v \Phi'(v) - v^2 \Phi''(v)\Add{,} \\
+\chi(v) &= \Phi'(v) - v \Phi''(v);
+\end{align*}
+d'où les coordonnées d'un point de la surface minima la plus
+générale
+\begin{align*}
+x + iy &= - 2F(u) + 2u F'(u) - u^2 F''(u) - 2 \Phi(v) + 2v \Phi'(v) + v^2 \Phi''(v)\Add{,} \\
+x - iy &= F''(u) + \Phi''(v), \\
+z &= F'(u) - u F''(u) + \Phi'(v) - v \Phi''(v).
+\end{align*}
+
+Dans le cas où l'équation de la surface est mise sous la
+forme
+\[
+z = f(x, y),
+\]
+l'équation aux dérivées partielles des surfaces minima est
+\[
+(1 + p^2)·t + (1 + q^2)·r + 2pq\, s = 0.
+\]
+
+%[** TN: Renumber 5 -> 6]
+\Section{Lignes courbure.}
+{6.}{} Les \emph{lignes de courbure} sont les lignes tangentes en
+chacun de leurs points aux directions principales ou axes de
+l'indicatrice. Ce sont les intégrales de l'équation
+\[
+\frac{\dd\Phi}{\dd·du} · \frac{\dd\Psi}{\dd·dv} -
+\frac{\dd\Phi}{\dd·dv} · \frac{\dd\Psi}{\dd·du} = 0,
+\]
+les directions principales étant conjuguées par rapport aux
+directions isotropes et aux directions asymptotiques. Si ces
+\Card{2} couples constituent \Card{4} directions distinctes, les directions
+%% -----File: 059.png---Folio 51-------
+principales seront aussi distinctes et distinctes des précédentes.
+Il en \DPtypo{resulte}{résulte} qu'il n'y aura pas d'autres cas singuliers
+pour les lignes de courbure que ceux déjà rencontrés
+pour les lignes minima et les lignes asymptotiques.
+
+\ParItem{\Primo.} \emph{Surfaces réglées non développables à génératrices
+isotropes \(la sphère exceptée\).} Une famille de lignes minima
+est constituée par des lignes asymptotiques. Prenant les lignes
+minima comme coordonnées, nous avons
+\[
+\Phi = 2 F·du·dv;
+\]
+prenons les lignes $u = \cte$ confondues avec les asymptotiques,
+$du = 0$ doit annuler~$\Psi$; donc
+\[
+\Psi = E' · du^2 + 2F' · du·dv;
+\]
+l'équation différentielle des lignes de courbure est
+\[
+F·dv · F'·du - F·du (E'·du + F'·dv) = 0,
+\]
+ou
+\[
+E'\Del{·}F · du^2 = 0.
+\]
+\emph{Les lignes de courbure sont doubles, ce sont les génératrices
+rectilignes isotropes qui sont déjà lignes minima et asymptotiques.}
+
+\ParItem{\Secundo.} \emph{Sphère.} $\Phi, \Psi$ sont proportionnels, l'équation différentielle
+est identiquement vérifiée. \emph{Sur la sphère toutes
+les lignes sont lignes de courbure.}
+
+\ParItem{\Tertio.} \emph{Surfaces développables non isotropes.} Prenons les
+génératrices rectilignes comme courbes $u = \cte$, ce sont des
+lignes asymptotiques doubles, nous avons
+\begin{align*}
+ds^2 &= E·du^2 + 2F·du·dv + G·dv^2, \\
+\Psi &= E'·du^2;
+\end{align*}
+l'équation différentielle des lignes de courbure est
+\[
+(F·du + G·dv)\, E'·du = 0.
+\]
+%% -----File: 060.png---Folio 52-------
+\emph{Les lignes de courbure sont les génératrices rectilignes, qui
+sont déjà lignes asymptotiques, et leurs trajectoires orthogonales.}
+
+\ParItem{\Quarto.} \emph{Surfaces développables isotropes.} Prenant pour courbe
+$v = \cte$ les lignes minima doubles confondues avec les lignes
+asymptotiques doubles, nous avons
+\[
+\Phi = E·du^2\Add{,}\qquad
+\Psi = E'·du^2\Add{.}
+\]
+L'équation aux lignes de courbure est identiquement vérifiée.
+\emph{Sur les développables isotropes toutes les lignes sont lignes
+de courbure.}
+
+\Paragraph{Remarque.} Pour un plan, les courbes minima sont des
+droites; et toute ligne du plan est asymptotique et ligne de
+courbure.
+
+%[** TN: Renumber 6 -> 7]
+\Section{Courbure géodésique.}
+{7.}{} Examinons maintenant la \Ord{2}{e} formule fondamentale
+\begin{multline*}
+\frac{\sin \theta}{R}
+  = \frac{1}{H\, ds^2} \left[\MStrut
+    H^2\, (du\, d^2v - dv\, d^2u) \right. \\
+  - \left.\MStrut
+    \begin{vmatrix}
+      \mfrac{1}{2}\, \mfrac{\dd E}{\dd u}\, du^2
+      + \mfrac{\dd E}{\dd v}\, du\, dv
+      + \left(\mfrac{\dd F}{\dd v}
+        - \mfrac{1}{2}\, \mfrac{\dd G}{\dd u}\right) dv^2 & E\, du + F\, dv \\
+      %
+      \left(\mfrac{\dd F}{\dd u} - \mfrac{1}{2}\, \mfrac{\dd E}{\dd v}\right) du^2
+      + \mfrac{\dd G}{\dd u}\, du\, dv
+      + \mfrac{1}{2}\, \mfrac{\dd G}{\dd v}\, dv^2 & F\, du + G\, dv
+    \end{vmatrix}
+  \right]\Add{;}
+\end{multline*}
+\Illustration[1.625in]{060a}
+$\theta$~est l'angle de la normale principale avec la
+normale à la surface. Soit $C$~le centre de courbure.
+Considérons la droite polaire, qui rencontre
+le plan tangent en~$G$, nous avons
+\[
+\DPtypo{MO}{MC} = MG \cos \left(\theta - \tfrac{\pi}{2}\right) = MG \sin\theta.
+\]
+$MG$ est ce qu'on appelle le \emph{rayon de courbure
+géodésique~$R_g$}. On a
+\[
+R = R_g \sin\theta.
+\]
+%% -----File: 061.png---Folio 53-------
+
+Le point~$G$ est le \emph{centre de courbure géodésique. La projection
+du centre de courbure géodésique sur la normale principale
+est le centre de courbure}. L'inverse du rayon de courbure
+géodésique s'appelle \emph{courbure géodésique}. Son expression
+ne dépend que de $E, F, G$ et de leurs dérivées; \emph{la courbure géodésique
+se conserve quand on déforme la surface}.
+
+Cherchons s'il existe des courbes de la surface dont le
+rayon de courbure géodésique soit constamment infini. De telles
+courbes sont appelées lignes géodésiques. Alors $\dfrac{\sin \theta}{R}$ est
+constamment nul, et comme $R$~n'est pas constamment infini, $\sin \theta = 0$.
+\emph{Le plan osculateur est normal à la surface en chaque
+point de la courbe.} Les lignes géodésiques sont définies par
+une équation différentielle de la forme
+\[
+v'' = \Phi (u, v, v').
+\]
+De l'étude des équations de cette forme il résulte qu'\emph{il y a
+une ligne géodésique et une seule passant par chaque point de
+la surface et tangente en ce point à une direction donnée du
+plan tangent. Il y en a une et une seule joignant deux points
+donnés dans un domaine suffisamment petit.}
+
+Prenons pour \Err{}{lignes }coordonnées les lignes minima. Alors $E = G = 0$
+et $H^2 = - F^2$. L'équation différentielle des lignes géodésiques
+devient
+\[
+- F^2 (du·d^2 v - dv·d^2 u)
+  - \begin{vmatrix}
+  \mfrac{\dd F}{\dd v}\, dv^2 & F\, dv \\
+  \mfrac{\dd F}{\dd u}\, du^2 & F\, du
+  \end{vmatrix} = 0,
+\]
+%% -----File: 062.png---Folio 54-------
+ou
+\[
+du · d^2v - dv · d^2u
+  + \frac{\dd · \log F}{\dd v}\, du · dv^2
+  - \frac{\dd · \log F}{\dd u}\, du^2\Add{·} dv = 0
+\]
+on voit qu'elle est vérifiée pour $du = 0$, $dv = 0$. Ainsi \emph{les
+lignes minima sont des lignes géodésiques}.
+
+\Paragraph{Remarque.} Si le plan osculateur se confond avec le plan
+tangent, le centre de courbure se confond avec le centre de
+courbure géodésique; et si en particulier on considère un plan
+\emph{dans ce plan la courbure géodésique n'est autre que la courbure}.
+Il en résulte que \emph{les lignes géodésiques du plan sont les
+droites de ce plan}, ce qu'on vérifie facilement par le calcul.
+
+\Illustration{062a}
+\Paragraph{Définition directe de la courbure géodésique.} Considérons
+sur la surface une courbe~$(C)$ et une famille
+de courbes~$(K)$ orthogonales à~$(C)$. Sur chaque
+courbe~$(K)$ portons à partir du point~$M$ où elle
+rencontre la courbe~$(C)$ une longueur d'arc
+constante~$M N$. Pour chaque valeur de cette
+constante nous obtenons une courbe~$(C')$ lieu
+du point~$N$. Prenons comme courbes coordonnées
+les courbes $(C)\Add{,} (C')\Add{,} \dots\Add{,} (v = \cte)$, la courbe~$(C)$ étant $v = 0$,
+et les courbes~$(K)\Add{,} (u = \cte)$. Alors $v$~ne sera autre que la
+longueur d'arc~$MN$. Nous avons
+\[
+ds^2 = E · du^2 + 2F \Add{·} du · dv + G · dv^2\Add{.}
+\]
+La courbe $v = 0$ est orthogonale à toutes les courbes~$(K)$, donc
+on a, quel que soit~$u$
+\[
+F(u, \DPtypo{o}{0}) = 0;
+\]
+$v$~représentant l'arc~$M N$, on a $ds^2 = dv^2$, d'où $G = 1$, et alors
+\[
+ds^2 = E · du^2 + 2F \Add{·} du · dv + dv^2.
+\]
+Nous pouvons de même supposer que sur la courbe~$(C)$\Add{,} $u$~\DPtypo{represente}{représente}
+%% -----File: 063.png---Folio 55-------
+l'arc. Alors pour $v = 0$, on a $ds = du$, donc
+\[
+E(u, 0) = 1,
+\]
+et pour cette courbe~$(C)$ on a
+\[
+H^2 = E · G - F^2 = 1,
+\]
+d'où~$H = 1$. Nous avons alors
+\[
+\frac{\sin \theta}{ R} = - \frac{1}{ds^2}
+  \begin{vmatrix}
+    \mfrac{1}{2}\, \mfrac{\dd E}{\dd u}\, du^2  & E\, du \\
+    \left(\mfrac{\dd F}{\dd u} - \mfrac{1}{2}\, \mfrac{\dd E}{\dd v} \right) du^2 & F\, du
+  \end{vmatrix}
+  = - \frac{1}{2}\, \dfrac{\dd E}{\dd v}.
+\]
+Pour la courbe~$(C')$ nous aurons
+\[
+ds^2 = E\, du^2,
+\]
+d'où
+\[
+ds = \sqrt{E}\, du;
+\]
+prenons la dérivée logarithmique par rapport à $v$
+\[
+\frac{\dd · \log ds}{\dd v}
+  = \frac{\dd · \log \DPtypo{E}{\sqrt{E}}}{\dd v}
+%[** TN: Explicit \cdot; Latin-1 char converts to thinspace]
+  = \frac{1}{\sqrt{E}} \cdot \frac{1}{2}\, \frac{\dfrac{\dd E}{\dd v}}{\sqrt{E}}
+  = \frac{1}{2E} · \frac{\dd E}{\dd v}\Add{.}
+\]
+Si on considère la courbe~$(C)$, $E = 1$, et on a pour cette courbe
+\[
+\frac{\dd · \log ds}{\dd v}
+  = \frac{1}{2}\, \frac{\dd E}{\dd v},
+\]
+d'où
+\[
+\frac{1}{R_{g}}
+  = \frac{\sin \theta}{R}
+  = -\frac{\dd \log · ds}{\dd v}\Add{,}
+\]
+ce qui donne une \DPtypo{definition}{définition} de la courbure géodésique n'empruntant
+aucun élément extérieur à la surface.
+
+\MarginNote{Propriétés des
+géodésiques.}
+Supposons en particulier que toutes les courbes~$(K)$ soient
+des géodésiques. Avec les mêmes conventions que précédemment,
+$du = 0$~doit être une solution de \DPtypo{l'equation differentielle}{l'équation différentielle} des
+lignes géodésiques, ce qui donne
+\[
+\begin{vmatrix}
+  \mfrac{\dd F}{\dd v} & F \\
+  0 & 1
+\end{vmatrix}
+= \dfrac{\dd F}{\dd v} = 0;
+\]
+%% -----File: 064.png---Folio 56-------
+donc $F$~est une fonction de $u$~seulement, et comme $F = 0$ pour
+$v = 0$, $F$~est identiquement nul, et on a
+\[
+ds^2 = E\, du^2 + dv^2;
+\]
+et alors toutes les courbes~$(C)$ coupent orthogonalement les
+géodésiques~$(K)$. Donc \emph{si nous considérons une courbe~$(c)$, si
+nous menons en chaque point de~$(c)$ la géodésique qui lui est
+orthogonale, et si nous portons sur chacune de ces géodésiques
+un arc constant, le lieu des extrémités de ces arcs est une
+courbe~$(c')$ normale aux géodésiques}. Nous obtenons ainsi les
+\emph{courbes parallèles} sur une surface quelconque.
+
+\emph{Réciproquement, si nous considérons une famille de géodésiques
+et leurs trajectoires orthogonales, ces trajectoires
+déterminent sur les géodésiques des longueurs d'arc égales.}
+Toujours avec les mêmes hypothèses, les courbes $u = \cte$ et
+$v = \cte$ étant orthogonales, on a $F = 0$. Les $u = \cte$ étant des
+géodésiques, nous avons
+\[
+\begin{vmatrix}
+-\mfrac{1}{2}\, \mfrac{\dd G}{\dd u}\, dv^2 & 0 \\
+\phantom{-} \mfrac{1}{2}\, \mfrac{\dd G}{\dd v}\, dv^2 & G
+\end{vmatrix}
+  = -\frac{1}{2}\, G \frac{\dd G}{\dd u}\, dv^2 = 0.
+\]
+$G \neq 0$, sans quoi les courbes $u = \cte$ seraient des courbes minima,
+donc $\dfrac{\dd G}{\dd u} = 0$ et $G = \phi(v)$. Calculons alors l'arc d'une
+courbe~$(K)$ compris entre la courbe $v = v_{0}$ et la courbe $v = v_{1}$\Add{.}
+Nous avons
+\begin{align*}
+ds^2 &= G\, dv^2 = \phi(v)\, dv^2, \\
+\intertext{et}
+s &= \int_{v_{0}}^{v_{1}} \sqrt{\phi(v)} · dv;
+\end{align*}
+$s$~est indépendant de~$u$, l'arc est le même sur toutes les géodésiques.
+%% -----File: 065.png---Folio 57-------
+
+Si on prend encore pour~$v$ l'arc sur les courbes $u = \cte$
+on a
+\[
+ds^{2} = E\, du^{2} + dv^{2},
+\]
+et \emph{cette forme est caractéristique du système de coordonnées}
+employé, \emph{constitué par une famille de géodésiques et leurs
+trajectoires orthogonales}.
+
+Prenons alors sur la surface deux points voisins~$A\Add{,} B$.
+Il existe une ligne géodésique et une seule dans le domaine
+de ces deux points et joignant ces deux points. Considérons
+une famille de géodésiques voisines ne se coupant pas dans
+le domaine, et leurs trajectoires orthogonales. Prenons-les
+comme courbes coordonnées. Considérons une ligne quelconque
+de la surface allant de $A$ à~$B$, soit
+\[
+u = f(v)\Add{.}
+\]
+Si $A$ a pour coordonnées $u_{0}, v_{0}$ et~$B$, $u_{0}\Add{,} v_{1}$, la longueur de l'arc~$AB$
+de cette ligne est
+\[
+\int_{v_{0}}^{v_{1}} \sqrt{E\, du^{2} + dv^{2}}
+  = \int_{v_{0}}^{v_{1}} \sqrt{E\bigl(f(v), v\bigr)\, f'{}^{2}(v) + 1}\, dv.
+\]
+Cette intégrale est visiblement minima si $f'(v) = 0$, c'est-à-dire
+si la courbe joignant~$A\Add{,} B$ est la géodésique. Ainsi
+donc, \emph{dans un domaine suffisamment petit entourant deux points
+d'une surface, la géodésique est le plus court chemin entre
+ces deux points}.
+
+%[** TN: Renumber 7 -> 8]
+\Section{Torsion géodésique.}
+{8.}{} Voyons enfin la \Ord{3}{e} formule fondamentale
+\[
+\frac{1}{T} - \frac{d\theta}{ds}
+  = \frac{1}{H^{2}\, ds^{2}}
+\begin{vmatrix}
+\Err{E}{E'}\, du + F'\, dv & F'\, du + G'\, dv \\
+E\, du + F\, dv & F\, du + G\, dv
+\end{vmatrix}.
+\]
+Si $\theta$ est constant, et en particulier constamment nul, la formule
+%% -----File: 066.png---Folio 58-------
+précédente donne la torsion; elle donne donc en particulier
+la torsion d'une géodésique. L'expression précédente ne
+dépend que de~$\dfrac{du}{dv}$, c'est-à-dire de la direction de la tangente\Add{.}
+Considérons alors sur la surface une courbe~$(c)$ et un point~$M$.
+Il existe une géodésique tangente à~$(c)$ au point~$M$ et $\dfrac{1}{T} - \dfrac{d \theta}{ds}$
+est la torsion de cette géodésique. C'est pourquoi $\dfrac{1}{T} - \dfrac{d \theta}{ds}$ s'appelle
+\emph{torsion géodésique}. On voit ainsi que \emph{la torsion géodésique
+en un point d'une courbe est la torsion de la géodésique
+tangente en ce point à la courbe \DPtypo{donnee}{donnée}}. Posons
+\[
+\frac{1}{T_{g}} = \frac{1}{T} - \frac{d \theta}{ds}\Add{;}
+\]
+$T_{g}$~est le \emph{rayon de torsion géodésique}. Contrairement au rayon
+de courbure géodésique, il change dans la déformation des surfaces.
+
+La formule précédente montre que la torsion géodésique
+est nulle si la direction $du, dv$ est une direction principale;
+\emph{la torsion géodésique est nulle pour toute courbe tangente à
+une ligne de courbure}. Il en résulte que \emph{les lignes de courbure
+ont une torsion géodésique constamment nulle \(Théorème
+de Lancret\)}.
+
+$\dfrac{1}{T_{g}}$ est le quotient de deux trinômes du \Ord{2}{e} degré en $du,
+dv$, on peut donc étudier sa variation. Prenons pour courbes
+coordonnées les lignes de courbure, elles sont conjuguées et
+rectangulaires, donc $F - F' = 0$, et
+\[
+\frac{1}{T_{g}} = \frac{1}{H^{2}\, ds^{2}} (E'G - G'E)\, du\, dv
+  = \left( \frac{E'}{E} - \frac{G'}{G} \right) \frac{du}{ds}\, \frac{dv}{ds}\Add{.}
+\]
+Si nous revenons aux notations employées au §1 pour l'étude
+de la courbure normale, les coefficients de direction de la
+%% -----File: 067.png---Folio 59-------
+tangente dans le plan tangent sont
+\[
+\lambda = \sqrt{E}\, \frac{du}{ds}\Add{,} \qquad
+\mu     = \sqrt{G}\, \frac{dv}{ds}\Add{,}
+\]
+et alors
+\[
+\frac{1}{T_{g}}
+  = \frac{1}{\sqrt{EG}} \left(\frac{E'}{E} - \frac{G'}{G}\right) \lambda \mu;
+\]
+les rayons de courbure principaux sont
+\[
+\frac{1}{R_{1}} = \frac{1}{\sqrt{EG}} · \frac{E'}{E}\Add{,} \qquad
+\frac{1}{R_{2}} = \frac{1}{\sqrt{EG}} \Add{·} \frac{G'}{G}\Add{,}
+\]
+d'où
+\[
+\frac{1}{T_{g}} = \left(\frac{1}{R_{1}} - \frac{1}{R_{2}} \right) \lambda\mu\Add{,}
+\]
+d'où la \emph{formule d'Ossian Bonnet}, analogue à la formule d'Euler
+\[
+\frac{1}{T_{g}}
+  = \left(\frac{1}{R_{1}} - \frac{1}{R_{2}} \right) \sin\phi · \cos\phi\Add{.}
+\]
+
+\MarginNote{Théorèmes de
+Joachimsthal.}
+Considérons une courbe~$(c)$ intersection de deux surfaces;
+le plan normal à~$(c)$ en l'un de ses points~$M$ contient la normale
+principale à la courbe et les normales aux deux surfaces. Soit
+$V$~l'angle des normales $MN, MN'$; $\theta, \theta'$ leurs angles avec~$MP$.
+
+Nous avons
+\[
+V = \theta' - \theta;
+\]
+mais
+\[
+\frac{1}{T} - \frac{d \theta}{ds} = \frac{1}{T_{g}} \qquad
+\frac{1}{T} - \frac{d \theta'}{ds} = \frac{1}{T'_{g}}
+\]
+d'où en retranchant
+\[
+\frac{dV}{ds} = \frac{1}{T_{g}} - \frac{1}{T'_{g}}.
+\]
+Supposons alors que $(c)$ soit ligne de courbure des deux surfaces;
+$\dfrac{1}{T_{g}}$~et~$\dfrac{1}{T'_{g}}$ sont tous deux nuls, $\dfrac{dV}{ds} = 0$, $V$ est constant.
+D'où les \emph{Théorèmes de Joachimsthal: Si 2 surfaces se coupent
+suivant une ligne de courbure, leur angle est constant tout
+le long de cette ligne}, et la même formule montre immédiatement
+que réciproquement: \emph{si deux surfaces se coupent sous un
+angle constant, et si l'intersection est ligne de courbure
+pour l'une des surfaces, elle est aussi ligne de courbure pour
+l'autre.} Sur un plan ou sur une sphère, toutes les lignes sont
+%% -----File: 068.png---Folio 60-------
+lignes de courbure; donc \emph{si une ligne de courbure d'une surface
+est plane ou sphérique, le plan ou la sphère qui la contient
+coupe la surface sous un angle constant, et réciproquement, si
+un plan ou une sphère coupe une surface sous un angle constant
+l'intersection est une ligne de courbure de la surface}. Enfin
+si un cercle est ligne de courbure d'une surface, il y a une
+sphère passant par ce cercle qui est tangente à la surface en
+un point du cercle; et, par suite, en tous les points du cercle.
+Donc toute \emph{ligne de courbure circulaire est la courbe de
+contact d'une sphère inscrite ou circonscrite à la surface}.
+
+
+\ExSection{III}
+
+\begin{Exercises}
+\item[12.] On considère la surface
+\[
+x = \frac{c^{2} - b^{2}}{bc} · \frac{uv}{u + v},\quad
+y = \frac{\sqrt{c^{2} - b^{2}}}{b} · \frac{v \sqrt{b^{2} - u^{2}}}{u + v},\quad
+z = \frac{\sqrt{c^{2} - b^{2}}}{c} · \frac{u \sqrt{v^{2} - c^{2}}}{u + v};
+\]
+déterminer ses lignes de courbure, et calculer les rayons de
+courbure principaux.
+
+\item[13.] On considère la surface
+\begin{align*}
+x &= \frac{1}{2} \int (1 - u^{2})\, f(u)\, du
+   + \frac{1}{2} \int (1 - v^{2})\, \phi(v)\, dv, \\
+y &= \frac{1}{2} \int (1 + u^{2})\, f(u)\, du
+   - \frac{1}{2} \int (1 + v^{2})\, \phi(v)\, dv, \\
+z &= \int uf(u)\, du + \int v \phi(v)\, dv.
+\end{align*}
+Calculer les rayons de courbures principaux et les coordonnées
+des centres de courbures principaux. Former l'équation
+différentielle des lignes de courbure et des lignes asymptotiques.
+\DPchg{Etudier}{Étudier} les lignes de courbure en prenant
+\[
+f(u)    = \frac{2m^{2}}{(m^{2} + u^{2})^{2}},\qquad
+\phi(v) = \frac{2m^{2}}{(m^{2} + v^{2})^{2}},
+\]
+et en introduisant de nouvelles coordonnées par les formules
+\[
+u = m \tg \frac{\lambda + i \mu}{2},\qquad
+v = m \tg \frac{\lambda - i \mu}{2}.
+\]
+
+\item[14.] Soient, en coordonnées rectangulaires, les équations
+\begin{align*}
+x &= \frac{1}{2} e^{u} \cos(v - \alpha) + \frac{1}{2} e^{-u} \cos(v + \alpha),\\
+y &= \frac{1}{2} e^{u} \sin(v - \alpha) + \frac{1}{2} e^{-u} \sin(v + \alpha),\\
+z &= u \cos\alpha + v \sin\alpha.
+\end{align*}
+
+\Primo. Pour chaque valeur de $\alpha$, ces formules définissent une
+surface $S_{\alpha}$. Indiquer un mode de génération de cette surface.
+Que sent en particulier $S_{0}$ et $S_{\frac{\pi}{2}}$?
+
+\Secundo. On considère deux de ces surfaces $S_{\alpha}$~et~$S_{\beta}$, et on les fait
+correspondre point par point de manière que les plans tangents
+aux points correspondants soient parallèles. Démontrer que les
+tangentes à deux courbes correspondantes, menées en deux
+points homologues, font un angle constant.
+
+\Tertio. Chercher les lignes de courbure et les lignes asymptotiques
+de~$S_{\alpha}$ et trouver une propriété géométrique des courbes
+auxquelles elles correspondent sur~$S_{\beta}$, dans la transformation
+précédente. Qu'arrive-t-il pour $\alpha = \frac{\pi}{2}$?
+
+\item[15.] Chercher les surfaces dont les lignes de courbure d'un système
+sont les courbes de contact des cônes circonscrits ayant
+leurs sommets sur~$Oz$. Quelles sont les autres lignes de courbure?
+
+\item[16.] \DPchg{Etudier}{Étudier} les surfaces dont les lignes de courbure d'un système
+sont situées sur des sphères concentriques. Que peut-on dire
+des lignes de courbure de l'autre système?
+
+\item[17.] Si les courbes coordonnées $u = \const.$, $v = \const.$ sur une
+surface~$S$ sont les lignes asymptotiques de cette surface, et
+si $l, m, n$ sont les cosinus directeurs de la normale à~$S$, en un
+point quelconque de~$S$, montrer qu'il existe une fonction~$\theta$
+telle que l'on ait
+\begin{alignat*}{5}
+dx &= \theta \Biggl[
+  && m &&\left(\frac{\dd n}{\dd u}\, du - \frac{\dd n}{\dd v}\, dv\right)
+  &&-n &&\left(\frac{\dd m}{\dd u}\, du - \frac{\dd m}{\dd v}\, dv\right)
+  \Biggr], \\
+dy &= \theta \Biggl[
+  && n &&\left(\frac{\dd l}{\dd u}\, du - \frac{\dd l}{\dd v}\, dv\right)
+  &&-l &&\left(\frac{\dd n}{\dd u}\, du - \frac{\dd n}{\dd v}\, dv\right)
+  \Biggr], \\
+dz &= \theta \Biggl[
+  && l &&\left(\frac{\dd m}{\dd u}\, du - \frac{\dd m}{\dd v}\, dv\right)
+  &&-m &&\left(\frac{\dd l}{\dd u}\, du - \frac{\dd l}{\dd v}\, dv\right)
+  \Biggr].
+\end{alignat*}
+Calculer, en partant de ces formules, le~$ds^2$ de la surface,
+l'équation des lignes de courbure, l'équation aux rayons de
+courbure principaux. Calculer la torsion des lignes asymptotiques,
+et montrer qu'elle s'exprime au moyen des rayons de
+courbure principaux seulement.
+\end{Exercises}
+%% -----File: 069.png---Folio 61-------
+
+
+\Chapitre{IV}{Les Six Invariants --- La Courbure Totale.}
+
+
+\Section{Les six \DPtypo{Invariants}{invariants}.}
+{1.}{} Dans l'étude des courbes tracées sur une surface~$(S)$
+ne sont intervenus que les coefficients des deux formes quadratiques
+fondamentales:
+\begin{align*}
+\Phi (du, dv) &= ds^{2} = E\, du^{2} + 2F · du · dv + G\, dv^{2}, \\
+\Psi (du, dv) &= \sum A · d^{2}x = E'\, du^{2} + 2F'du · dv + G'\, dv^{2},
+\end{align*}
+et les différentielles de $u, v$, considérées comme les fonctions
+d'une variable indépendante~$t$ qui correspondent à chaque courbe
+particulière considérée.
+
+Si l'on déplace la surface~$(S)$ dans l'espace, sans la déformer,
+et sans changer les coordonnées superficielles $u, v$ employées,
+ces formes quadratiques demeureront les mêmes, de
+sorte que \emph{leurs six coefficients} $E, F, G$, $E', F', G'$ \emph{sont six invariants
+différentiels, pour le groupe des mouvements dans l'espace}\Add{.}
+
+Cela résulte, pour la forme $ds^{2} = \Phi (du, dv)$, de ce qu'elle
+représente le carré de la différentielle d'un arc qui reste
+le même dans les conditions énoncées.
+
+Dès lors $H = \sqrt{EG - F^{2}}$ est un invariant, et la formule
+\[
+\Psi(du, dv) = H · \Phi(du, dv) · \frac{\cos\theta}{R},
+\]
+dont tous les facteurs du second membre sont invariants, montre
+que $\Psi$~possède aussi la propriété d'invariance.
+
+Il n'y a du reste aucune difficulté à vérifier, par un
+calcul direct, l'invariance des six coefficients sur les formules
+%% -----File: 070.png---Folio 62-------
+qui les définissent:
+\begin{alignat*}{3}
+\Tag{(1)}
+\sum \left(\frac{\dd x}{\dd u} \right)^{2} &= E,\qquad
+&\sum \frac{\dd x}{\dd u}\, \frac{\dd x}{\dd v} &= F,\qquad
+&\sum \left(\frac{\dd x}{\dd v} \right)^{2} &= G, \\
+\Tag{(2)}
+\sum A\, \frac{\dd^{2} x}{\dd u^{2}} &= E',\qquad
+&\sum A\, \frac{\dd^{2} x}{\dd u\, \dd v} &= F',\qquad
+&\sum A\, \frac{\dd^{2} x}{\dd v^{2}} &= G',
+\end{alignat*}
+\DPtypo{où}{ou} l'on se rappelle que $A, B, C$ sont les trois déterminants
+fonctionnels
+\[
+A = \frac{D(y, z)}{D(u, v)}, \quad
+B = \frac{D(z, x)}{D(u, v)}, \quad
+C = \frac{D(x, y)}{D(u, v)}.
+\]
+Rappelons enfin l'identité
+\[
+H = \sqrt{A^{2} + B^{2} + C^{2}} = \sqrt{EG - F^{2}} \Add{.}
+\]
+
+\MarginNote{La forme de la
+surface définie
+par les six invariants.}
+Supposons maintenant que $E, F, G$, $E', F', G'$ aient été calculés,
+en fonction de $u, v$, pour une surface~$(S)$ particulière
+\[
+\Tag{(3)}
+x = f(u, v),\quad
+y = g(u, v),\quad
+z = h(u, v);
+\]
+et considérons les équations \Eq{(1)},~\Eq{(2)} comme un système d'équations
+aux dérivées partielles, où $x, y, z$ sont les fonctions inconnues,
+$u, v$ les variables indépendantes, et $E, F, G$, $E', F', G'$ des
+fonctions données. En vertu de l'invariance que nous venons
+d'expliquer, ce système \DPtypo{differentiel}{différentiel} admettra comme intégrales,
+non seulement les fonctions~\Eq{(3)}, qui \DPtypo{definissent}{définissent}~$(S)$,
+mais encore toutes les fonctions
+\[
+\Tag{(4)}
+\left\{%[** TN: Added brace]
+\begin{alignedat}{4}
+x &= x_{0} &&+ \alpha f &&+ \alpha' g &&+ \alpha'' h,\\
+y &= y_{0} &&+ \beta f &&+ \beta' g &&+ \beta'' h, \\
+z &= z_{0} &&+ \gamma f &&+ \gamma' g &&+ \gamma'' h,
+\end{alignedat}
+\right.
+\]
+qui \DPtypo{definissent}{définissent} les surfaces obtenues en déplaçant~$(S)$ de toutes
+les manières possibles, lorsqu'on donne à $x_{0}, y_{0}, z_{0}$, toutes
+les valeurs constantes possibles, et à $\alpha, \beta, \gamma$, $\alpha', \beta', \gamma'$,
+$\alpha'', \beta'', \gamma''$ toutes les valeurs constantes compatibles avec les
+conditions d'orthogonalité bien connues.
+%% -----File: 071.png---Folio 63-------
+
+Cela donne donc des intégrales dépendant de six constantes
+arbitraires. Nous prouverons que le système \Eq{(1)},~\Eq{(2)} n'en
+a pas d'autres; ce que l'on pourra exprimer en disant que \emph{la
+forme de la surface est entièrement définie par les six invariants
+$E, F, G$, $E', F', G'$}.
+
+On démontre dans la théorie des équations aux dérivées
+partielles que, dans tout système dont l'intégrale générale
+ne dépend que de constantes arbitraires, toutes les dérivées
+partielles d'un certain ordre peuvent s'exprimer en fonction
+des variables indépendantes et dépendantes et des dérivées
+d'ordre inférieur. Nous devons donc essayer de constater que
+cela a lieu pour notre système; et commencer par différentier
+les équations~\Eq{(1)}. Les résultats obtenus peuvent s'écrire:
+\[%[** TN: Rearranged in pairs, added brace]
+\Tag{(5)}
+\left\{
+\begin{aligned}
+&\sum \frac{\dd x}{\dd u} · \frac{\dd^{2}x}{\dd u^{2}}
+  = \frac{1}{2}\, \frac{\dd E}{\dd u},
+&&\sum \frac{\dd x}{\dd v}\Add{·} \frac{\dd^{2}x}{\dd u^{2}}
+  = \frac{\dd F}{\dd u} - \frac{1}{2}\, \frac{\dd E}{\dd v}, \\
+%
+&\sum \frac{\dd x}{\dd u}\Add{·} \frac{\dd^{2}x}{\dd u\, \dd v}
+  = \frac{1}{2}\, \frac{\dd E}{\dd v},
+&&\sum \frac{\dd x}{\dd v}\Add{·} \frac{\dd^{2}x}{\dd u\, \dd v}
+  = \frac{1}{2}\, \frac{\dd G}{\dd u}, \\
+%
+&\sum \frac{\dd x}{\dd u}\Add{·} \frac{\dd^{2}x}{\dd v^{2}}
+  = \frac{\dd F}{\dd v} - \frac{1}{2}\, \frac{\dd G}{\dd u};
+&&\DPtypo{}{\sum} \frac{\dd x}{\dd v} · \frac{\dd^{2}x}{\dd v^{2}}
+  = \frac{1}{2}\, \frac{\dd G}{\dd v};
+\end{aligned}
+\right.
+\]
+et l'on voit qu'en associant ces équations aux équations~\Eq{(2)},
+on obtiendra effectivement les expressions de toutes les dérivées
+du second ordre.
+
+Pour faciliter ce calcul, nous introduirons les cosinus
+directeurs de la normale:
+\[
+\Tag{(6)}
+l = \frac{A}{H},\qquad
+m = \frac{B}{H},\qquad
+n = \frac{C}{H};
+\]
+et nous substituerons à la forme $\sum A\, d^{2}x$ la forme
+\[
+\Tag{(7)}
+\sum l · d^{2}x
+  = \frac{1}{H} \sum A · d^{2}x
+  = L · du^{2} + 2 · M · du · dv + N · dv^{2}
+\]
+de sorte qu'on aura
+\[
+\Tag{(8)}
+L = \frac{E'}{H},\qquad
+M = \frac{F'}{H},\qquad
+N = \frac{G'}{H};
+\]
+et que les équations~\Eq{(2)} seront remplacées par
+%% -----File: 072.png---Folio 64-------
+\[
+\Tag{(9)}
+\sum l\, \frac{\dd^{2} x}{\dd u^{2}} = L,\qquad
+\sum l\, \frac{\dd^{2} x}{\dd u\, \dd v} = M,\qquad
+\sum l\, \frac{\dd^{2} x}{\dd v^{2}} = N.
+\]
+Cela fait, si on pose:
+\begin{alignat*}{3}
+\frac{\dd^{2} x}{\dd u^{2}}
+  &= L'\, \frac{\dd x}{\dd u} &&+ L''\, \frac{\dd x}{\dd v} &&+ L''' l, \\
+\frac{\dd^{2} y}{\dd u^{2}}
+  &= L'\, \frac{\dd y}{\dd u} &&+ L''\, \frac{\dd y}{\dd v} &&+ L''' m, \\
+\frac{\dd^{2} z}{\dd u^{2}}
+  &= L'\, \frac{\dd z}{\dd u} &&+ L''\, \frac{\dd z}{\dd v} &&+ L''' n,
+\end{alignat*}
+$L', L'', L'''$ étant des coefficients à déterminer, on aura pour
+les calculer les conditions
+\[
+\sum \frac{\dd x}{\dd u}\, \frac{\dd^{2} x}{\dd u^{2}} = E L' + F L'',\quad
+\sum \frac{\dd x}{\dd v}\, \frac{\dd^{2} x}{\dd u^{2}} = F L' + G \DPtypo{L'}{L''},\quad
+\sum l\, \DPtypo{\frac{dx}{du}}{\frac{\dd^2 u}{\dd x^2}} = L''';
+\]
+d'où on conclut d'abord $L''' = L$; et ensuite, en se servant
+des formules~\Eq{(5)}, des équations qui donneront $L'$~et~$L''$.
+
+En opérant de même pour les autres dérivées, on obtient
+les résultats suivants
+\[
+\Tag{(10)}
+\left\{
+\begin{alignedat}{5}
+\frac{\dd^{2} x}{\dd u^{2}}
+  &= L'\, &&\frac{\dd x}{\dd u} &&+ L''\, &&\frac{\dd x}{\dd v} &&+ L·l, \\
+\frac{\dd^{2} x}{\dd u\, \dd v}
+  &= M'\, &&\frac{\dd x}{\dd u} &&+ M''\, &&\frac{\dd x}{\dd v} &&+ M·l, \\
+\frac{\dd^{2} x}{\dd v^{2}}
+  &= N'\, &&\frac{\dd x}{\dd u} &&+ N''\, &&\frac{\dd x}{\dd v} &&+ N·l,
+\end{alignedat}
+\right.
+\]
+avec les équations auxiliaires:
+\[
+\Tag{(11)}
+\left\{
+\begin{alignedat}{4}
+E L' &+ F L'' &&= \frac{1}{2}\, \frac{\dd E}{\dd u}, &
+F L' &+ G L'' &&= \frac{\dd F}{\dd u} - \frac{1}{2}\, \frac{\dd E}{\dd v}, \\
+E M' &+ F M'' &&= \frac{1}{2}\, \frac{\dd E}{\dd v}, &
+F M' &+ G M'' &&= \frac{\dd G}{\dd u}, \\
+E N' &+ F N'' &&= \frac{\dd F}{\dd v} - \frac{1}{2}\, \frac{\dd G}{\dd u},
+\quad &
+F N' &+ G N'' &&= \frac{1}{2}\, \frac{\dd G}{\dd v},
+\end{alignedat}
+\right.
+\]
+d'où on déduirait les valeurs des coefficients $L', L''$, $M', M''$, $N',
+N''$. On remarquera qu'elles ne dépendent que des coefficients
+$E, F, G$ de l'élément linéaire $ds^{2} = \Phi(du, dv)$, et des dérivées
+premières de ces coefficients.
+%% -----File: 073.png---Folio 65-------
+
+Enfin, les mêmes équations~\Eq{(10)} subsisteront pour les
+autres coordonnées $y, z$; il n'y aura qu'à y laisser les mêmes
+coefficients, et à y remplacer la lettre~$x$ par la lettre~$y$ ou
+la lettre~$z$, en même temps qu'on changera~$l$ en~$m$ ou en~$n$.
+
+Nous concluons de là que si on \DPchg{connait}{connaît}, pour un système
+de valeurs de $u, v$, les valeurs de $x, y, z$ et de leurs dérivées
+premières, on pourra calculer les valeurs de leurs dérivées
+secondes; et, par des différentiations nouvelles, celles
+de toutes leurs dérivées d'ordre supérieur. Et par suite les
+développements en séries de Taylor d'une intégrale quelconque
+ne peuvent contenir d'autres arbitraires que les valeurs initiales
+de
+\[
+x,\ y,\ z,\quad
+\frac{\dd x}{\dd u},\
+\frac{\dd x}{\dd v},\quad
+\frac{\dd y}{\dd u},\
+\frac{\dd y}{\dd v},\quad
+\frac{\dd z}{\dd u},\
+\frac{\dd z}{\dd v};
+\]
+et encore celles-ci doivent être liées par les équations~\Eq{(1)};
+et, lorsque ces valeurs initiales sont données, l'intégrale
+est entièrement déterminée.
+
+Donc, pour prouver que \Eq{(4)}~donne l'intégrale générale, il
+reste seulement à montrer que \Eq{(4)}~peut satisfaire aux conditions
+initiales énoncées. Or, si nous introduisons les cosinus
+directeurs $l', m', n'$; $l'', m'', n''$ des tangentes $MU, MV$ aux deux
+courbes coordonnées qui passent par un point quelconque~$M$ de
+la surface, nous aurons
+\begin{alignat*}{3}
+\frac{\dd x}{\dd u} &= l' \sqrt{E},\qquad &
+\frac{\dd y}{\dd u} &= m' \sqrt{E},\qquad &
+\frac{\dd z}{\dd u} &= n' \sqrt{E}, \\
+\frac{\dd x}{\dd v} &= l'' \sqrt{G},\qquad &
+\frac{\dd y}{\dd v} &= m'' \sqrt{G},\qquad &
+\frac{\dd z}{\dd v} &= n'' \sqrt{G};
+\end{alignat*}
+et les conditions~\Eq{(1)} se réduiront à
+\[
+\sum l'{}^{2} = 1,\qquad
+\sum l''{}^{2} = 1,\qquad
+\sum l'l'' = \frac{F}{\sqrt{EG}} = \cos \omega,
+\]
+%% -----File: 074.png---Folio 66-------
+$\omega$ étant l'angle $\DPchg{\widehat{UMV}}{(MU, MV)}$.
+
+Les conditions initiales signifient donc que l'on se donne
+arbitrairement la position du point~$M$, et la direction des
+tangentes $M U, M V$, sous la réserve que ces directions fassent
+entre elles le même angle qu'elles font au point correspondant
+de~$(S)$. Il y a donc bien une des positions de~$(S)$ qui satisfait
+à ces conditions, et notre résultat se trouve définitivement
+établi.
+
+\Paragraph{Remarque.} Le raisonnement précédent serait en défaut,
+si les courbes coordonnées étaient les lignes minima (à cause
+de $E = G = 0$). Mais il suffit de remarquer que si $\Phi$~et~$\Psi$\DPtypo{,}{}
+sont connues, pour un système de coordonnées $u, v$, on en déduit
+leurs expressions pour un autre système de coordonnées $u, v$, en
+y effectuant directement le changement de variables correspondant.
+Notre théorème est donc vrai pour tout système de coordonnées
+superficielles, dès qu'il est vrai pour un seul.
+
+
+\Section{Les \DPtypo{Conditions}{conditions} d'\DPtypo{Intégrabilité}{intégrabilité}.}
+{2.}{} Les coefficients des formules~\Eq{(10)} satisfont à certaines
+conditions, dites \emph{conditions d'intégrabilité} qu'on
+obtient en écrivant que les dérivées du troisième ordre $\dfrac{\dd^{3} x}{\dd u^{2}\, \dd v}$,
+$\dfrac{\dd^{3} x}{\dd u\, \dd v^{2}}$ ont la même valeur, qu'on les obtienne en différentiant
+l'une ou l'autre des formules~\Eq{(10)}.
+
+Pour pouvoir calculer ces conditions, il est commode d'avoir
+des formules qui donnent les dérivées des cosinus directeurs
+$l, m, n$ de la normale. Ils sont définis par les équations
+\[
+\sum l\, \frac{\dd x}{\dd u} = 0,\qquad
+\sum l\, \frac{\dd x}{\dd v} = 0,\qquad
+\sum l^{2} = 1,
+\]
+qui donnent, par différentiation:
+%% -----File: 075.png---Folio 67-------
+\[
+\Tag{(12)}
+\begin{aligned}
+&\sum \frac{\dd l}{\dd u}\, \frac{\dd x}{\dd u}
+  = - \sum l\, \frac{\dd^{2} x}{\dd u^{2}} = -L, &&
+\sum \frac{\dd l}{\dd v}\, \frac{\dd x}{\dd u}
+  = - \sum l\, \frac{\dd^{2} x}{\dd u\, \dd v} \rlap{${} = -M$,} \\
+&\sum \frac{\dd l}{\dd u}\, \frac{\dd x}{\dd v}
+  = - \sum l\, \frac{\dd^{2} x}{\dd u\, \dd v} = -M, &&
+\sum \frac{\dd l}{\dd v}\, \frac{\dd x}{\dd v}
+  = - \sum l\, \frac{\dd^{2} x}{\dd v^{2}} = -N\Add{,} \\
+&\sum \frac{\dd l}{\dd u}\, l = 0, &&
+\sum \frac{\dd l}{\dd v}\, \DPtypo{v}{l} = 0.
+\end{aligned}
+\]
+Si donc on pose, en suivant la même méthode qu'au paragraphe
+précédent,
+\begin{align*}
+\frac{\dd l}{\dd u} &= P'\, \frac{\dd x}{\dd u} + P''\, \frac{\dd x}{\dd v} + Pl, \\
+\frac{\dd \DPtypo{n}{m}}{\dd u} &= P'\, \frac{\dd y}{\dd u} + P''\, \frac{\dd y}{\dd v} + Pm, \\
+\frac{\dd n}{\dd \DPtypo{v}{u}} &= P'\, \frac{\dd z}{\dd u} + P''\, \frac{\dd \DPtypo{x}{z}}{\dd v} + Pn,
+\end{align*}
+on trouvera:
+\[
+\sum \frac{\dd x}{\dd u}\, \frac{\dd l}{\dd u} = EP' + FP'',\quad
+\sum \frac{\dd x}{\dd v}\, \frac{\dd l}{\dd u} = FP' + GP'',\quad
+\sum l\, \frac{\dd l}{\dd u} = P;
+\]
+c'est-à-dire qu'on peut écrire, en tenant compte des formules~\Eq{(12)},
+\[
+\Tag{(13)}
+\left\{
+\begin{aligned}
+\frac{\dd l}{\dd u} &= P'\, \frac{\dd x}{\dd u} + P''\, \frac{\dd x}{\dd v},
+\quad\text{et de même:} \\
+\frac{\dd l}{\dd v} &= Q'\, \frac{\dd x}{\dd u} + Q''\, \frac{\dd x}{\dd v},
+\end{aligned}
+\right.
+\]
+les coefficients $P', P'', Q', Q''$ étant définis par:
+\[
+\Tag{(14)}
+\left\{
+\begin{alignedat}{2}
+EP' + FP'' &= -L,\qquad & FP' + GP'' &= -M, \\
+EQ' + FQ'' &= -M,\qquad & FQ' + GQ'' &= -N,
+\end{alignedat}
+\right.
+\]
+et qu'on aura les mêmes formules pour $m, n$ en changeant~$x$ en~$y$,
+et en~$z$, respectivement.
+
+Nous achèverons le calcul, en supposant la surface rapportée
+à ses lignes minima. Les calculs précédents se simplifient
+alors beaucoup. Si nous appliquons directement les formules
+trouvées, nous obtenons:
+\begin{gather*}%[** TN: Rearranged]
+E = 0,\qquad G = 0, \\
+L'' = 0,\quad L' = \frac{\dd \log F}{\dd u},\quad
+M'' = 0,\quad M' = 0, \quad
+N'' = \frac{\dd \log F}{\dd v}, \quad N' = 0; \\
+P'' = - \frac{L}{F},\quad P' = - \frac{M}{F},\qquad
+Q'' = - \frac{M}{F},\quad Q' = - \frac{N}{F};
+\end{gather*}
+%% -----File: 076.png---Folio 68-------
+c'est-à-dire
+\begin{align*}
+\Tag{(15)}
+&\left\{
+\begin{aligned}
+\frac{\dd^{2} x}{\dd u^{2}}
+  &= \frac{\dd \log F}{\dd u} · \frac{\dd x}{\dd u} + L · l, \\
+\frac{\dd^{2} x}{\dd u\, \dd v}
+  &= M · l, \\
+\frac{\dd^{2} x}{\dd v\DPtypo{}{^{2}}}
+  &= \frac{\dd \log F}{\dd v} · \frac{\dd x}{\dd v} + N · l,
+\end{aligned}
+\right. \\
+\Tag{(16)}
+&\left\{
+\begin{aligned}
+\frac{\dd l}{\dd u}
+  &= -\frac{1}{F} \left(M\, \frac{\dd x}{\dd u}
+                      + L\, \frac{\dd x}{\dd v}\right), \\
+\frac{\dd l}{\dd v}
+  &= -\frac{1}{F} \left(N\, \frac{\dd x}{\dd u}
+                      + M\, \frac{\dd x}{\dd v}\right).
+\end{aligned}
+\right.
+\end{align*}
+Alors, en différentiant la première équation~\Eq{(15)}, il vient:
+\begin{align*}
+\frac{\dd^{3} x}{\dd u^{2}\, dv}
+  &= \left(\frac{\dd^{2} \log F}{\dd u\, \dd v} - \frac{NL}{F} \right)
+       \frac{\dd x}{\dd u}
+   - \frac{LM}{F}\, \frac{\dd x}{\dd v}
+   + \left(\frac{\dd \log F}{\dd u}\, M + \frac{\dd L}{\dd v} \right)l, \\
+\intertext{en différentiant la deuxième équation~\Eq{(15)}, il vient}
+\frac{\dd^{3} x}{\dd u^{2}\, \dd v}
+  &= \frac{-M^{2}}{F} · \frac{\dd x}{\dd u} - \frac{LM}{F}\, \frac{\dd x}{\dd v} + \frac{\dd M}{\dd u}\, l;
+\end{align*}
+et en égalant, on obtient:
+\[
+\Tag{(17)}
+\left( \frac{\dd^{2} \log F}{\dd u\, \dd v} - \frac{LN - M^{2}}{F} \right) \frac{\dd x}{\dd u}
+  + \left( \frac{\dd \log F}{\dd u}\, M + \frac{\dd L}{\dd v} - \frac{\dd M}{\dd u} \right) l = 0.
+\]
+C'est là une condition de la forme
+\begin{alignat*}{3}
+S'\, \frac{\dd x}{\dd u} &+ S''\, \frac{\dd x}{\dd v} &&+ Sl &&= 0, \\
+\intertext{et en reprenant le même calcul, pour $y$~et~$z$, on obtiendrait
+les conditions analogues}
+S'\, \frac{\dd y}{\dd u} &+ S''\, \frac{\dd y}{\dd v} &&+ Sm &&= 0, \\
+S'\, \frac{\dd z}{\dd u} &+ S''\, \frac{\dd z}{\dd v} &&+ Sn &&= 0.
+\end{alignat*}
+On en conclut qu'on a nécessairement $\DPtypo{S'}{S} = S' = S'' = 0$, c'est-à-dire
+ici
+\[
+\Tag{(18)}
+\frac{\dd^{2} \log F}{\dd u\, \dd v} - \frac{LN - M^{2}}{F} = 0,\qquad
+M\, \frac{\dd \log F}{\dd u} + \frac{\dd L}{\dd v} - \frac{\dd M}{\dd u} = 0;
+\]
+et cela est suffisant pour que~\Eq{(17)} ait lieu.
+
+En égalant de même les deux valeurs de $\dfrac{\dd^{3} x}{\dd u\, \dd v^{2}}$, on obtiendra
+%% -----File: 077.png---Folio 69-------
+les conditions qui se déduisent de~\Eq{(18)} en échangeant les
+rôles des variables $u, v$; cela ne modifie que la seconde de ces
+conditions.
+
+Les conditions d'intégrabilité cherchées sont donc:
+\[
+\Tag{(19)}
+\left\{
+\begin{aligned}
+M\, \frac{\dd \log F}{\dd u} &= \frac{\dd M}{\dd u} - \frac{\dd L}{\dd v}\Add{,} \\
+\frac{\dd^{2} \log F}{\dd u\, \dd v} &= \frac{LN - M^{2}}{F}\Add{,} \\
+M\, \frac{\dd \log F}{\dd v} &= \frac{\dd M}{\dd v} - \frac{\dd N}{\dd u}\Add{,}
+\end{aligned}
+\right.
+\]
+et ce sont là, d'après la théorie des équations différentielles,
+les seules conditions d'intégrabilité du système considéré.
+
+
+\Section{Courbure totale.}
+{3.}{} La \Ord{2}{e} des formules précédentes, due à Gauss
+\[
+\frac{\dd^{2} \log F}{\dd u\, \dd v} = \frac{LN - M^2}{F}
+\]
+conduit à une conséquence importante. Reprenons en effet l'équation
+aux rayons de courbure principaux qui est ici
+\[
+H^{2}(LN - M^{2}) + 2SFHM - S^{2}F^{2} =0,
+\]
+où
+\[
+S = \frac{H}{R}.
+\]
+
+On peut l'écrire
+\[
+LN - M^{2} + 2FM · \frac{1}{R} - \frac{F^{2}}{R^{2}} = 0,
+\]
+d'où
+\[
+\frac{1}{R_{1}R_{2}} = - \frac{LN - M}{F},
+\]
+c'est-à-dire
+\[
+\frac{1}{R_{1}R_{2}} = - \frac{1}{F}\, \frac{\dd^{2} \log F}{\dd u\, \dd v}\Add{;}
+\]
+\emph{le produit des rayons de courbure principaux ne dépend que de
+l'élément linéaire; il se conserve donc dans la déformation
+des surfaces}. On donne à $\dfrac{1}{R_{1}R_{2}}$ le nom de \emph{\DPtypo{Courbure}{courbure} totale}.
+
+\Paragraph{Représentation sphérique.} De même que l'on a fait correspondre
+%% -----File: 078.png---Folio 70-------
+à une courbe son indicatrice sphérique, on peut imaginer
+une correspondance entre une surface quelconque et la
+sphère de rayon~$1$, l'homologue d'un point $(u\Add{,} v)$ de la surface
+étant le point $(l, m, n)$. A une aire de la surface correspond une
+aire sphérique. La considération de la limite du rapport de
+ces aires lorsqu'elles deviennent infiniment petites dans toutes
+leurs dimensions va nous conduire à une définition directe
+de la courbure totale.
+
+L'aire sur la surface a pour expression
+\[
+\Area = \iint \sqrt{A^{2} + B^{2} + C^{2}}\, du\, dv = \iint H\, du\, dv,
+\]
+Pour avoir l'aire homologue sur la sphère, il faut d'abord
+calculer l'élément linéaire $dl^{2} + dm^{2} + dn^{2}$. D'après les formules~\Eq{(16)}
+du \Numero~précédent, nous avons
+\begin{align*}
+dl &= \frac{\dd l}{\dd u}\, du + \frac{\dd l}{\dd v}\, dv
+  = - \frac{du}{F} \left(L\, \frac{\dd x}{\dd v} + M\, \frac{\dd x}{\dd u} \right)
+    - \frac{dv}{F} \left(N\, \frac{\dd x}{\dd u} + M\, \frac{\dd x}{\dd v} \right) \\
+  &= -\frac{1}{F}
+    \left[L\, \frac{\dd x}{\DPtypo{d}{\dd v}}\, du
+        + M\, dx + N\, \frac{\dd x}{\dd u}\, dv \right];
+\end{align*}
+d'où
+\begin{gather*}
+\sum dl^{2} = \frac{1}{F^{2}} \left[M^{2}·2F\, du\, dv + 2LMF·du^{2} + 2MNF·dv^{2} + 2LNF·du\, dv \right], \\
+\sum dl^{2} = \frac{2LM}{F}\, du^{2} + 2\, \frac{LN + M^2}{F}\, du\, dv + \frac{2MN}{F}\, dv^{2}.
+\end{gather*}
+Pour la sphère la fonction~$H$ sera donc
+\[
+\sqrt{4\, \frac{LM^{2}N}{F^{2}} - \frac{(LN + M^{2})^{2}}{F^{2}}}
+  = \frac{LN - M^{2}}{iF} = \frac{LN - M^{2}}{H},
+\]
+et l'aire sphérique a pour expression
+\[
+\Area' = \iint \frac{LN - M^{2}}{H}\, du\, dv;
+\]
+ce qui peut s'écrire, en remarquant que
+\begin{gather*}%[** TN: Set first two equations on a single line]
+d\Area = H·du\, dv,\qquad
+\Area' = \iint \frac{LN - M^{2}}{H^{2}} · d\Area
+  = \iint \frac{1}{R_{1} R_{2}}\, d\Area\Add{,} \\
+\intertext{donc}
+d\Area' = \frac{1}{R_{1} R_{2}}\, d\Area;
+\end{gather*}
+%% -----File: 079.png---Folio 71-------
+\emph{le rapport des aires homologues sur la surface et sur la sphère
+a donc pour limite la courbure totale, lorsque ces aires
+deviennent infiniment petites dans toutes leurs dimensions}.
+
+
+\Section{Coordonnées orthogonales et isothermes.}
+{4.}{} Pour éviter l'emploi des imaginaires dans les considérations
+qui \DPtypo{précèdent}{précédent}, nous introduirons un nouveau système
+de coordonnées curvilignes. La surface étant supposée réelle,
+nous choisirons les coordonnées minima de façon que $u, v$ soient
+imaginaires conjugués. Nous poserons donc
+\[
+u = u' + iv',\qquad
+v = u' - iv',
+\]
+$u'\Add{,} v'$ étant des quantités réelles. Nous en tirons
+\[
+du = du' + i\, dv',\qquad
+dv = du' - i\, dv',
+\]
+d'où
+\[
+du\, dv = du'{}^{2} + dv'{}^{2}.
+\]
+L'élément linéaire prend la forme
+\[
+ds^{2} = 2F·du\, dv = 2F (du'{}^{2} + dv'{}^{2});
+\]
+les coordonnées $u'\Add{,} v'$ sont orthogonales; on leur donne le nom
+de \emph{coordonnées orthogonales et isothermes}. On peut dire que
+\emph{ces coordonnées divisent la surface en un réseau de carrés infiniment
+petits}. Considérons en effet les courbes coordonnées
+$u', u' + h, u' + 2h\Add{,} \dots$ et $v', v' + h, v'+ 2h\Add{,} \dots$; si on prend l'un
+des quadrilatères curvilignes obtenus, ses \Card{4} angles sont droits,
+ses côtés sont $\sqrt{2F}\Add{·}du'$~et~$\sqrt{2F}·dv'$, c'est-à-dire~$\sqrt{2F}·h$, aux
+infiniment petits d'ordre supérieur près; ces arcs sont égaux.
+
+Avec ce système de coordonnées particulières, en désignant
+%[** TN: Reworded to follow the typeset edition]
+par $\overline{E}, \overline{F}, \overline{G}, \overline{H}$ les valeurs des fonctions \DPtypo{$\overline{E}, \overline{F}, \overline{G}, \overline{H}$}{analogues à $E, F, G, H$}, nous avons
+\[%[** TN: Not displayed in manuscript, but displayed in typeset edition]
+\overline{E} = 2F,\quad
+\overline{G} = 2F,\quad
+\overline{F} = 0,\quad
+\overline{H}^{2} = \overline{E}\overline{G} - \overline{F}^{2} = 4 F^{2},\quad
+\overline{H} = 2F,
+\]
+donc
+\[
+ds^{2} = \overline{H} (du'{}^{2} + dv'{}^{2}).
+\]
+%% -----File: 080.png---Folio 72-------
+Mais nous avons
+\[
+\frac{\dd \Phi}{\dd u'} = \frac{\dd \Phi}{\dd u} + \frac{\dd \Phi}{\dd v},\qquad
+\frac{\dd \Phi}{\dd v'} = i \left(\frac{\dd \Phi}{\dd u} - \frac{\dd \Phi}{\dd v} \right);
+\]
+d'où
+\[
+\frac{\dd^{2} \Phi}{\dd u'{}^{2}}
+  = \frac{\dd^{2} \Phi}{\dd u^{2}}
+  + 2\, \frac{\dd^{2} \Phi}{\dd u\, \dd v}
+  + \frac{\dd^{2} \Phi}{\dd v^{2}},\qquad
+\frac{\dd^{2} \Phi}{\dd v'{}^{2}}
+  = - \left[ \frac{\dd^{2} \Phi}{\dd u^{2}}
+    - 2\, \frac{\dd^{2} \Phi}{\dd u\, \dd v}
+    + \frac{\dd^{2} \Phi}{\dd v^{2}} \right],
+\]
+et
+\[
+\frac{\dd^{2} \Phi}{\dd u'{}^{2}} + \frac{\dd^{2} \Phi}{\dd v'{}^{2}}
+  = 4\, \frac{\dd^{2} \Phi}{\dd u\, \dd v}\Add{,}
+\]
+\DPtypo{D'où}{d'où} par conséquent
+\[
+4\, \frac{\dd^{2} \log F}{\dd u\, \dd v}
+  = \DPtypo{}{4\,}\frac{\dd^{2} \log \overline{H}}{\dd u\, \dd v}
+  = \frac{\dd^{2} \log \overline{H}}{\dd u'{}^{2}}
+  + \frac{\dd^{2} \log \overline{H}}{\dd v'{}^{2}}.
+\]
+En supprimant les accents, nous avons donc les formules suivantes,
+en coordonnées orthogonales et isothermes:
+\begin{gather*}
+ds^{2} = H (du^{2} + dv^{2}), \\
+\frac{1}{R_{1} R_{2}}
+  = -\frac{1}{2H} \left( \frac{\dd^{2} \log H}{\dd u^{2}} + \frac{\dd^{2} \log H}{\dd v^{2}} \right).
+\end{gather*}
+Nous poserons encore
+\[
+\sum l\, d^{2} x = L\, du^{2} + 2M\, du\, dv + N\, dv^{2}.
+\]
+L'équation aux rayons de courbure principaux sera
+\[
+(LN -M^{2}) - \frac{H}{R} (L + N) + \frac{H^{2}}{R^{2}} = 0,
+\]
+et on aura
+\[
+\frac{1}{R_{1}R_{2}} = \frac{LN - M^{2}}{H^{2}}.
+\]
+
+Calculons la représentation sphérique. Posons
+\begin{alignat*}{3}
+l' &= \frac{1}{\sqrt{H}}\Add{·} \frac{\dd x}{\dd u},\qquad &
+m' &= \frac{1}{\sqrt{H}} · \frac{\dd y}{\dd u},\qquad &
+n' &= \frac{1}{\sqrt{H}} · \frac{\dd z}{\dd u}, \\
+l'' &= \frac{1}{\sqrt{H}}\Add{·} \frac{\dd x}{\dd v},\qquad &
+m'' &= \frac{1}{\sqrt{H}} · \frac{\dd y}{\dd v},\qquad &
+n'' &= \frac{1}{\sqrt{H}} · \frac{\dd z}{\dd v}.
+\end{alignat*}
+De la relation
+\[
+\sum  l^{2} = 1,
+\]
+nous tirons
+\[
+\sum l · \frac{\dd l}{\dd u} = 0.
+\]
+Maintenant
+\[
+L = \sum l\, \frac{\dd^{2} x}{\dd u^{2}}
+  = - \sum \frac{\dd l}{\dd u} · \frac{\dd x}{\dd u}
+  = - \sqrt{H} · \sum l'\, \frac{\dd l}{\dd u};
+\]
+d'où
+\[
+\sum l'\, \frac{\dd l}{\dd u} = - \frac{L}{\sqrt{H}};
+\]
+de même
+\[%[** TN: Set on two lines in original]
+M = \sum l\, \frac{\dd^{2} x}{\dd u\, \dd v}
+  = - \sum \frac{\dd l}{\dd u} · \frac{\dd x}{\dd v}
+  = - \sqrt{H} \sum l''\, \frac{\dd l}{\dd u}, \qquad
+\sum l''\, \frac{\dd l}{\dd u} = - \frac{M}{\sqrt{H}}.
+\]
+D'où \Card{3} équations en $\dfrac{\dd l}{\dd u}, \dfrac{\dd m}{\dd u}, \dfrac{\dd n}{\dd u}$. Multiplions respectivement par
+%% -----File: 081.png---Folio 73-------
+$l'\Add{,} l''\Add{,} l'''$ et ajoutons, il vient
+\begin{align*}
+\frac{\dd l}{\dd u}
+  &= - \frac{L}{H} · \frac{\dd x}{\dd u}
+     - \frac{M}{H} · \frac{\dd x}{\dd v}; \\
+\intertext{de même:}
+\frac{\dd m}{\dd u}
+  &= - \frac{L}{H} · \frac{\dd y}{\dd u}
+     - \frac{M}{H}\Add{·} \frac{\dd y}{\dd v}, \\
+\frac{\dd n}{\dd u}
+  &= - \frac{L}{H}\Add{·} \frac{\dd z}{\dd u}
+     - \frac{M}{H} · \frac{\dd z}{\dd v}.
+\end{align*}
+On obtiendra de même
+\begin{align*}
+\frac{\dd l}{\dd v}
+  &= - \frac{1}{H} \left(M\, \frac{\dd x}{\dd u}
+                       + N\, \frac{\dd x}{\dd v}\right), \\
+\frac{\dd m}{\dd v}
+  &= - \frac{1}{H} \left(M\, \frac{\dd y}{\dd u}
+                       + N\, \frac{\dd y}{\dd v}\right), \\
+\frac{\dd n}{\dd v}
+  &= - \frac{1}{H} \left(M\, \frac{\dd z}{\dd u}
+                       + N\, \frac{\dd z}{\dd v}\right).
+\end{align*}
+Alors, sur la sphère, les \Card{3} fonctions $E\Add{,} F\Add{,} G$ seront
+\begin{alignat*}{2}
+\scrE &= \sum \left(\frac{\dd l}{\dd u}\right)^{2}\!\!
+  &&= \frac{1}{H^{2}} \sum \left(L\, \frac{\dd x}{\dd u}
+                               + M\, \frac{\dd x}{\dd v}\right)^{2}
+  = \frac{L^{2} + M^{2}}{H}, \\
+%
+\scrF &= \sum \frac{\dd l}{\dd u} · \frac{\dd l}{\dd v}
+  &&= \frac{1}{H^{2}}
+      \sum \left(L\, \frac{\dd x}{\dd u} + M\, \frac{\dd x}{\dd v}\right)
+     \!·\! \left(M\, \frac{\dd x}{\dd u} + N\, \frac{\dd x}{\dd v}\right)
+  = \frac{M (L + N)}{H}, \\
+%
+\scrG &= \sum \left(\frac{\dd l}{\dd v}\right)^{2}\!\!
+  &&= \frac{1}{H^{2}} \sum \left(M\, \frac{\dd x}{\dd u}
+                               + N\, \frac{\dd x}{\dd v}\right)^{2}
+  = \frac{M^{2} + N^{2}}{H};
+\end{alignat*}
+et
+\[
+\scrH^{2} = \scrE\scrG - \scrF^{2}
+  = \frac{(L^{2} + M^{2}) (M^{2} + N^{2})- M^{2} (L + N)^{2}}{H^{2}}
+  = \left( \frac{LN - M^{2}}{H}\right)^{2},
+\]
+et alors l'aire sur la sphère a pour expression
+\[
+\Area' = \iint \frac{LN - M^{2}}{H}\, du\, dv.
+\]
+On retrouve la même expression que précédemment, et on arriverait
+de même à la définition directe de la courbure totale.
+
+\Paragraph{Remarque.} Dans l'expression précédente, $\Area$ a un signe,
+qui est celui de $LN - M^{2}$. Considérons le déterminant des cosinus
+$l, m, n$; $l', m', n'$; $l'', m'', n''$: il est égal, à un facteur positif près
+à
+\[
+\begin{vmatrix}
+l & m & n \\
+\mfrac{\dd l}{\dd u} & \mfrac{\dd m}{\dd u} & \mfrac{\dd n}{\dd u} \\
+\mfrac{\dd l}{\dd v} & \mfrac{\dd m}{\dd v} & \mfrac{\dd n}{\dd v}
+\end{vmatrix}
+= \frac{LN  - M^{2}}{H^{2}}
+\begin{vmatrix}
+l & m & n \\
+\mfrac{\dd x}{\dd u} & \mfrac{\dd y}{\dd u} & \mfrac{\dd z}{\dd u} \\
+\mfrac{\dd x}{\dd v} & \mfrac{\dd y}{\dd v} & \mfrac{\dd z}{\dd v}
+\end{vmatrix}\Add{.}
+\]
+Il résulte de cette formule que, si $\Area\Area' > 0$, le point mobile $x\Add{,}y\Add{,}z$
+%% -----File: 082.png---Folio 74-------
+décrivant le contour qui limite l'aire sur la surface dans le
+sens direct le point $l, m, n$ décrira le contour qui limite l'aire
+homologue sur la sphère aussi dans le sens direct. Si $\Err{\Area'}{\Area\Area'} < 0$,
+les conclusions sont inverses.
+
+
+\Section{Relations entre la courbure totale et la courbure géodésique.}
+{5.}{} La courbure totale est un élément qui reste invariant
+dans la déformation des surfaces. Cherchons s'il y a des relations
+entre elle et les autres éléments invariants dans la
+déformation. Considérons la courbure géodésique. En coordonnées
+orthogonales et isothermes, son expression est
+\[
+\llap{$\dfrac{1}{R_{g}} = $}\dfrac{1}{H\, ds^{3}}
+  \left[ H^{2} (du\, d^{2} v - dv\, d^{2} u)
+  -
+  \begin{vmatrix}
+  \phantom{-}  \mfrac{1}{2}\, \mfrac{\dd H}{\dd u}\, du^{2}
+                 + \mfrac{\dd H}{\dd v}\, du\, dv
+  - \mfrac{1}{2}\, \mfrac{\dd H}{\dd u}\, dv^{2} & H\, du \\
+  - \mfrac{1}{2}\, \mfrac{\dd H}{\dd v}\, du^{2}
+                 + \mfrac{\dd H}{\dd u}\, du\, dv
+  + \mfrac{1}{2}\, \mfrac{\dd H}{\dd v}\, dv^{2} & H\, dv
+\end{vmatrix}
+\right],
+\]
+ou
+\[
+\frac{1}{R_{g}} = \frac{1}{ds^{3}} \left[
+  H (du\, d^{2} v - dv\, d^{2} u)
+  + \frac{1}{2} \left(\frac{\dd H}{\dd u}\, dv - \frac{\dd H}{\dd v}\, du\right) (du^{2} + dv^{2})\right];
+\]
+mais on a
+\[
+ds^{2} = (du^{2} + dv^{2})\, H,
+\]
+et la formule précédente s'écrit
+\begin{align*}
+\frac{ds}{R_{g}}
+  &= \frac{du\, d^{2} v - dv\, d^{2} u}{du^{2} + dv^{2}}
+   + \frac{1}{2}\, \frac{\dd \log H}{\dd u}\, dv - \frac{1}{2}\, \frac{\dd \log H}{\dd v}\, du; \\
+\intertext{ou encore}
+\frac{ds}{R_{g}}
+  &= d \left(\arctg \frac{dv}{du}\right)
+   + \frac{1}{2}\, \frac{\dd \log H}{\dd u}\, dv - \frac{1}{2}\, \frac{\dd \log H}{\dd v}\, du.
+\end{align*}
+Imaginons alors dans le plan tangent les tangentes $MU, MV$ aux
+courbes coordonnées dans le sens des $u, v$ croissants; considérons
+la tangente à une courbe quelconque $MT$ de la surface, et
+soit $(MU, MT) = \phi$. Nous avons
+%[** TN: Set each of next two aligned pairs on a single line]
+\[
+\Cos \phi = \sqrt{H} · \frac{du}{ds}, \qquad
+\sin \phi = \sqrt{H}\Add{·} \frac{dv}{ds};
+\]
+%% -----File: 083.png---Folio 75-------
+d'où
+\[
+\tg \phi = \frac{dv}{du}, \qquad
+\phi = \arctg\DPtypo{.} \frac{dv}{du};
+\]
+et la formule précédente devient
+\[
+\frac{ds}{R_{g}}
+  = d \phi
+  + \frac{1}{2}\, \frac{\dd \log H}{\dd u}\,  dv
+  - \frac{1}{2}\, \frac{\dd \log H}{\dd v}\,  du.
+\]
+Prenons alors sur~$S$ un contour fermé et intégrons le long de
+ce contour dans le sens direct
+\[
+\int \frac{ds}{R_{g}}
+  = \int d \phi
+  + \frac{1}{2} \int \frac{\dd \log H}{\dd u}\,  dv
+  - \frac{1}{2} \int \frac{\dd \log H}{\dd v}\,  du.
+\]
+\Figure[2.75in]{083a}
+Rappelons le \emph{Théorème de Green}. Dans le plan des $u, v$, le
+point $(u\Add{,}v)$ décrit un contour fermé, aussi
+dans le sens direct. Menons les tangentes
+parallèles à l'axe des~$u$; soient $A\Add{,} B$~les
+points de contact. Nous avons ainsi \Card{2} arcs
+$AMB$~et~$ANB$, et si nous désignons par~$\DPtypo{C}{c}$ le
+contour, nous avons
+\[
+\int_c \frac{\dd f}{\dd u}\, dv
+  = \int_{AMB} \frac{\dd f}{\dd u}\, dv
+  + \int_{BNA} \frac{\dd f}{\dd u}\, dv\Add{.}
+\]
+Menons une parallèle à $OU$ qui coupe le contour en deux points
+$M (u_{2})$~et~$N (u_{1})$\Add{.}
+
+Soient $a, b$ les valeurs de~$U$ qui correspondent aux deux points
+$A, B$. Nous avons
+\[
+\int_c \frac{\dd f}{\dd u}\, dv
+  = \int_a^b \left(\frac{\dd f}{\dd u}\right)_{u_{1}u_{2}} \kern -12pt \Err{dv\, dv}{d\gamma}
+  - \int_a^b \left(\frac{\dd f}{\dd u}\right)_{u_{1}u_{2}} \kern -12pt \Err{dv\, du}{d\gamma}
+  = \int_a^b \left[
+    \left(\frac{\dd f}{\dd u}\right)_{2}
+  - \left(\frac{\dd f}{\dd u}\right)_{1}\right] dv.
+\]
+Mais
+\[
+\left(\frac{\dd f}{\dd u}\right)_{2} - \left(\frac{\dd f}{\dd u}\right)_{1}
+  = \int_{u_{1}}^{u_{2}} \frac{\dd^{2} f}{\dd u^{2}} · du,
+\]
+et alors
+\[
+\int_c \frac{\dd f}{\dd u} · dv
+  = \int_a^b dv \int_{u_{1}}^{u_{2}} \frac{\dd^{2} f}{\dd u^{2}} · du
+  = \iint \frac{\dd^{2} f}{\dd u^{2}}\, du\, dv,
+\]
+%% -----File: 084.png---Folio 76-------
+l'intégrale double étant étendue à toute l'aire limitée par le
+contour. Cette formule subsiste pour un contour simple quelconque.
+De même
+\[
+\int_c \frac{\dd f}{\dd v}\, du
+  = - \iint \frac{\dd^{2} f}{\dd v^{2}} · du\, dv\Add{.}
+\]
+Alors nous aurons
+\[
+\int \frac{ds}{R_{g}}
+  = \int d \phi
+  + \frac{1}{2} \iint \left[\frac{\dd^{2} \log H}{\dd u^{2}}
+                           + \frac{\dd^{2} \log H}{\dd v^{2}}\right] du\, dv,
+\]
+ou
+\[
+\int \frac{ds}{R_{g}}
+  = \int d \phi - \iint \frac{H}{R_{1} R_{2}} · du\, dv
+  = \int d \phi - \iint \frac{d\Area}{R_{1} R_{2}},
+\]
+d'où la \emph{formule d'Ossian Bonnet}
+\[
+\Area' = \iint \frac{d\Area}{R_{1} R_{2}}
+  = \int d \phi - \int \frac{ds}{R_{g}}.
+\]
+
+\Paragraph{Remarque.} L'angle~$\phi$ est l'angle de~$MU$ avec la tangente
+à la courbe. Supposons qu'en chaque point de la surface on
+détermine une direction~$MO$, dont les cosinus directeurs sont
+des fonctions bien déterminées de~$u\Add{,}v$. Soit $\psi = (MO, MU)$ et
+$\psi' = (MO, MT)$.
+
+Nous avons
+\[
+\phi' = \psi + \phi,
+\]
+d'où
+\[
+d \phi' = d \psi + d \phi.
+\]
+Intégrons le long d'un contour fermé quelconque
+\[
+\int d \phi' = \int d \psi + \int d \phi;
+\]
+or\Add{,} $\psi$~est une fonction de~$u\Add{,}v$, et le long d'un contour fermé,
+on a
+\[
+\int d \psi (u, v) = 0;
+\]
+donc
+\[
+\int d \phi' = \int d \phi,
+\]
+et l'on peut substituer à l'angle~$\phi$ l'angle~$\phi'$ précédemment
+défini.
+%% -----File: 085.png---Folio 77-------
+
+
+\Section{Triangles géodésiques.}{}{}
+Nous appellerons \emph{triangle géodésique} la figure formée par
+\Card{3} lignes géodésiques. Le long de chacun des côtés on a
+\[
+\int \frac{ds}{R_{g}} = \int \frac{\sin \theta}{R}\, ds = 0,
+\]
+et la formule d'O.~Bonnet nous donne
+\[
+\Area' = \int d \phi,
+\]
+c'est-à-dire
+\[
+\Area' = \int_{AB} d \phi + \int_{BC} d \phi + \int_{CA} d \phi.
+\]
+
+Les coordonnées
+orthogonales et isothermes
+constituent
+une représentation
+conforme de la surface
+\Figure[5in]{085a}
+sur le plan des~$u\Add{,}v$. Considérons donc sur ce plan la représentation~$a\Add{,} b\Add{,} c$ du triangle~$ABC$. Menons aux extrémités $a, b, c$ les
+tangentes aux côtés dans le sens direct: Soient $T_{1}, T_{2}, T_{3}$, $T'_{1},
+T'_{2}, T'_{3}$ ces tangentes. Si par un point du plan nous menons des
+parallèles à ces tangentes, nous aurons
+\[
+\int_{AB} d \phi = (T'_{1}, T_{2}),\qquad
+\int_{BC} d \phi = (T'_{2}, T_{3}),\qquad
+\int_{CA} d \phi = (T'_{3}, T_{1});
+\]
+or\Add{,} si nous appelons $a, b, c$ les \Card{3} angles du triangle géodésique,
+nous avons
+\begin{multline*}%[** TN: Re-breaking]
+(T'_{1}, T_{2}) + (T'_{2}, T_{3}) + (T'_{3}, T_{1}) \\
+\begin{aligned}
+  &= - \bigl[(T_{1}, T'_{1}) + (T_{2}, T'_{2}) + (T_{3}, T'_{3})\bigr]
+    + \bigl[(T_{1}, T_{2}) + (T_{2}, T_{3}) + (T_{3}, T_{1})\bigr] \\
+  &= 2 \pi - \bigl[(\pi - a) + (\pi - b) + (\pi - c)\DPtypo{}{\bigr]}
+  = a + b + c - \pi,
+\end{aligned}
+\end{multline*}
+d'où la \emph{formule de Gauss}
+\[
+a + b + c - \pi = \Area'.
+\]
+Si en particulier la surface est la sphère de rayon~$1$, on a
+%% -----File: 086.png---Folio 78-------
+la formule qui donne l'aire d'un triangle sphérique.
+
+\Section{Nouvelle expression de la courbure géodésique.}{}{}%
+Considérons un arc de courbe~$AB$; menons en $AB$ les géodésiques
+tangentes à cette courbe, qui se
+coupent en~$C$ sous un angle~$\epsilon$ que nous appellerons
+\emph{angle de contingence géodésique}.
+Le long du contour de ce triangle on a
+\[
+\int d \phi = - \epsilon,
+\]
+et la formule d'O.~Bonnet nous donne
+\[
+- \epsilon - \int_{AB} \frac{ds}{R_{g}} = \iint d\Area'.
+\]
+Supposons que $A$~corresponde au paramètre~$t$, $B$~à~$t + \Delta t$, et que
+$\Delta t$ tende vers~$0$; soit $\Delta s$~l'arc~$AB$. Nous avons
+\[
+-\frac{\epsilon}{\Delta s} - \frac{1}{\Delta s} \int_{AB} \frac{ds }{R_{g}}
+  = \frac{1}{\Delta s} \iint d\Area'.
+\]
+Soit $\left(\dfrac{1}{R_{g}}\right)_{m}$ la valeur moyenne de la courbure géodésique sur l'arc~$AB$;
+nous avons,
+\[
+\frac{1}{\Delta s} \int_{AB} \frac{ds}{R_{g}}
+  = \left(\frac{1}{R_{g}}\right)_{m},
+\]
+et par suite
+\[
+-\frac{\epsilon}{\Delta s} - \left(\frac{1}{R_{g}}\right)_{m}
+  = \frac{1}{\Delta s} \iint dA'\Add{.}
+\]
+\begin{figure}[hbt]
+\centering
+\Input[1.75in]{086a}\hfil\hfil
+\Input[2.25in]{086b}
+\end{figure}
+Si $\Delta s$ tend vers~$0$, $\left(\dfrac{1}{R_{g}}\right)_{m}$ a pour limite la courbure géodésique
+au point~$A$. Je dis que le \Ord{2}{e} membre a pour limite~$0$; il suffit
+de montrer que $\ds\iint d\Area'$ est infiniment petit
+du \Ord{2}{e} ordre au moins. Considérons la représentation
+$a, b, c$ du triangle~$ABC$ sur le
+plan~$U\Add{,}V$.
+
+Nous avons
+%% -----File: 087.png---Folio 79-------
+\[
+\iint d\Area' = \iint \psi(u, v)\, du\, dv
+  = \bigl[\psi (u, v)\bigr]_{m} \iint du\, dv
+\]
+et au signe près $\ds\iint du\, dv$ est égale à l'intégrale curviligne $\ds\int v\, du$.
+Soient $v_{2}\Add{,} v_{1}$ les fonctions~$v$ sur les arcs $bc$~et~$bk$. La partie de
+$\smash{\ds\int v\, du}$ donnée par ces arcs est $\ds\int_{u_{0}}^{u'} (v_{2} - v_{1})\, du$. Or\Add{,} les courbes $ab$~et~$bc$
+étant tangentes en~$b$, $v_{2} - v_{1}$ est infiniment petit du \Ord{2}{e} ordre au
+moins par rapport à $u'- u$ et \textit{à~fortiori} par rapport à $(u'- u_{0})$.
+L'intégrale $\ds\int_{u_{0}}^{u'} (v_2 - v_1)\, du$, qui est égale au produit de $(u' - u_{0})$ par
+la valeur moyenne de $(v_{2} - v_{1})$ sera donc du troisième ordre au moins
+par rapport à $(u'- u_{0})$, et, par suite, par rapport à~$\Delta s$. Le même
+raisonnement s'appliquant aux autres arcs $ac$~et~$ak$, on voit que
+$\ds\iint d\Area'$ est du troisième ordre au moins.
+
+
+\ExSection{IV}
+
+\begin{Exercises}
+\item[18.] \DPchg{Etablir}{Établir} les conditions d'intégrabilité qui lient les invariants
+fondamentaux, en supposant la surface rapportée à ses
+lignes de courbure.
+
+\item[19.] Même question, en supposant la surface \DPtypo{rapportee}{rapportée} à une famille
+de géodésiques et à leurs trajectoires orthogonales. Exprimer,
+en fonction de la quantité~$H$, la courbure totale, et la forme
+différentielle~$\dfrac{ds}{R_{g}} - d\phi$ (voir \hyperref[exercice11]{exercice~11}); et retrouver ainsi
+la formule d'Ossian Bonnet.
+
+\item[20.] En supposant les coordonnées quelconques, trouver celle des
+conditions d'intégrabilité qui donne l'expression de la courbure
+totale.
+\end{Exercises}
+%% -----File: 088.png---Folio 80-------
+
+
+\Chapitre{V}{Surfaces Réglées.}
+
+\Section{Surfaces développables.}
+{1.}{} Pour définir la variation de la droite qui engendre
+la surface réglée, nous nous donnerons la trajectoire d'un
+point~$M$ de cette droite, et la direction de cette droite pour
+\Figure{088a}
+chaque position du point~$M$. Les coordonnées
+d'un point de la surface sont ainsi exprimées
+en fonction de deux paramètres, l'un
+définissant la position du point~$M$ sur sa
+trajectoire~$(K)$, l'autre définissant la position
+du point~$P$ considéré sur la droite~$D$. Soit
+\[
+x = f(v),\qquad y = g(v),\qquad z = h(v),
+\]
+la courbe~$K$. Soient $l(v), m(v), n(v)$ les coefficients de direction
+de la génératrice, et $u$~le rapport du segment~$MP$ au
+segment de direction de la génératrice. Les coordonnées de~$P$
+sont
+\[
+\Tag{(1)}
+x = f(v) + u·l(v),\qquad
+y = g(v) + u·m(v),\qquad
+z = h(v) + u·n(v).
+\]
+
+Cherchons la condition pour que la surface définie par
+les équations précédentes soit développable. Si nous exceptons
+les cas du cylindre et du cône, la condition nécessaire et
+suffisante est que les génératrices soient tangentes à une même
+courbe gauche. On peut donc trouver sur la génératrice~$G$ un
+point~$P$ tel que sa trajectoire soit constamment tangente à~$G$;
+on doit donc avoir, en appelant $x, y, z$ les coordonnées de ce
+point
+\[
+\frac{dx}{l} = \frac{dy}{m} = \frac{dz}{n} = d \rho;
+\]
+%% -----File: 089.png---Folio 81-------
+d'où
+\[
+\Tag{(2)}
+dx = l\, d \rho,\qquad
+dy = m\, d \rho,\qquad
+dz = n\, d \rho,
+\]
+Mais les équations~\Eq{(1)} donnent
+\[
+dx = df + u\, dl + l\, du,\quad
+dy = dg + u\, dm + m\, du,\quad
+dz = dh + u\, dn + n\, du
+\]
+et les équations~\Eq{(2)} s'écrivent
+\begin{alignat*}{5}
+&df &&+ u\, dl &&+ l\,&&(du - d \rho) &&= 0, \\
+&dg &&+ u\, dm &&+ m\,&&(du - d \rho) &&= 0, \\
+&dh &&+ u\, dn &&+ n\,&&(du - d \rho) &&= 0;
+\end{alignat*}
+ou, en posant
+\begin{gather*}%[** TN: Set second group on separate lines, added brace]
+\Tag{(3)}
+d \sigma = du - d \rho, \\
+\Tag{(4)}
+\left\{
+\begin{alignedat}{4}
+&df &&+ u\, dl &&+ l\, &&d \sigma = 0, \\
+&dg &&+ u\, dm &&+ m\, &&d \sigma = 0, \\
+&dh &&+ u\, dn &&+ n\, &&d \sigma = 0,
+\end{alignedat}
+\right.
+\end{gather*}
+$d \sigma$~et~$u$ doivent satisfaire à ces \Card{3} équations linéaires; donc
+on doit avoir
+\[
+\Tag{(5)}
+\begin{vmatrix}
+df & dl & l \\
+dg & dm & m \\
+dh & dn & n
+\end{vmatrix}
+= 0\Add{.}
+\]
+
+Si les \Card{3} déterminants déduits du tableau
+\[
+\begin{Vmatrix}
+dl & dm & dn \\
+l  &  m &  n
+\end{Vmatrix}
+\]
+ne sont pas tous nuls, la condition~\Eq{(5)} est suffisante. Si ces
+\Card{3} déterminants sont nuls, on a
+\[
+\frac{dl}{l} = \frac{dm}{m} = \frac{dn}{n}\Add{,}
+\]
+et l'intégration de ces équations nous montre que $l, m, n$ sont
+proportionnels à des quantités fixes; la surface est alors un
+cylindre. En écartant ce cas, la condition~\Eq{(5)} est nécessaire
+et suffisante.
+
+\Paragraph{Remarque \1.} Pour que le point~$P$ décrive effectivement
+une courbe, il faut que $dx, dy, dz$, et par suite~$d\rho$, ne soient
+%% -----File: 090.png---Folio 82-------
+pas identiquement \DPtypo{nu}{nul}. Si on avait $d \rho = 0$, toutes les génératrices
+passeraient par un point fixe, la surface serait un
+cône. La condition~\Eq{(5)} s'applique donc au cas du cône.
+
+\Paragraph{Remarque \2.} On emploie souvent les équations de la génératrice
+sous la forme
+\[
+x = Mz + P,\qquad
+y = Nz + Q,
+\]
+$M, N, P, Q$, étant fonctions d'un paramètre arbitraire. C'est un
+cas particulier de la représentation générale~\Eq{(1)} dans laquelle
+on fait $h(v) = 0$ et $n(v) = 1$; alors $z = u$, et on peut écrire
+\[
+\Tag{(6)}
+x = f(v) + z·l(v)\Add{,}\qquad
+y = g(v) + z·m(v)\Add{,}
+\]
+les coefficients de direction sont $l, m, 1$. La courbe~$(K)$ est
+alors la section par le plan $z = 0$; dans ce cas la condition
+\Eq{(5)}~prend la forme simple
+\[
+\Tag{(7)}
+\begin{vmatrix}
+df & dl \\
+dg & dm
+\end{vmatrix} = 0,\quad\text{c'est-à-dire}\quad
+\begin{vmatrix}
+dM & dP \\
+dN & dQ
+\end{vmatrix} = 0.
+\]
+
+
+\Section{Propriétés des développables\Add{.}}{}{}%
+Revenons au cas général; supposons que $l, m, n$ soient les
+cosinus directeurs de la génératrice; on a
+\[
+l^{2} + m^{2} + n^{2} = 1,
+\]
+d'où
+\[
+l\, dl + m\, dm + n\, dn = 0.
+\]
+Multiplions alors les équations~\Eq{(4)} respectivement par $dl, dm,
+dn$, et ajoutons, il vient
+\[
+u = - \frac{\sum dl\, df}{\sum dl^{2}}.
+\]
+Supposons en outre que $MG$ soit normale à la courbe~$(K)$. Il
+est toujours possible de trouver sur une surface réglée des
+trajectoires orthogonales des génératrices. Il suffit que
+%% -----File: 091.png---Folio 83-------
+l'on ait
+\[
+\sum l\, dx =0,
+\]
+ou
+\[
+\sum l\, df + u \sum l\, dl + \sum l^{2}\, du =0;
+\]
+ou, comme ici $\sum l^{2} = 1$, $\sum l\,dl = 0$\Add{,}
+\[
+\sum l\, df + du = 0;
+\]
+la détermination des trajectoires orthogonales se fait donc
+au moyen d'une quadrature. Si donc nous supposons $(K)$~normale
+à la génératrice, nous avons
+\[
+\sum l\, df = 0.
+\]
+Multiplions alors les équations~\Eq{(4)} respectivement par $l, m, n$
+et ajoutons, il vient $d\sigma = 0$, d'où $d \rho = du$, et les équations~\Eq{(2)}
+deviennent
+\[
+\Tag{(3')}
+dx = l\, du,\qquad dy = m\, du,\qquad dz = n\, du.
+\]
+\DPtypo{mais}{Mais}, $l, m, n$ étant les cosinus directeurs de la tangente à
+l'arête de rebroussement~$(R)$, $u$~représente l'arc de cette
+courbe compté dans le sens positif choisi sur la génératrice
+à partir d'une origine arbitraire~$I$; et comme $u$~représente
+aussi la longueur~$MP$, on a
+\[
+d·MP = d·(\arc IP);
+\]
+d'où
+\[
+MP = \arc IP + \cte[].
+\]
+On peut toujours choisir l'origine des arcs telle que la constante
+soit nulle. Alors $MP = \arc IP$. La courbe~$(K)$ est une
+développante de la courbe~$(R)$. \emph{Sur une surface développable,
+les trajectoires orthogonales des génératrices sont des développantes
+de l'arête de rebroussement.}
+
+Les formules~\Eq{(4)} donnent alors
+%% -----File: 092.png---Folio 84-------
+\[
+\Tag{(4')}
+df + u\, dl = 0,\qquad dg + u\, dm = 0,\qquad dh + u\, dn = 0.
+\]
+
+\Section{Développées des courbes gauches.}
+{2.}{} Supposons qu'on se donne la courbe~$(K)$, et cherchons
+à mener à cette courbe une normale en chacun de ses points
+de façon à obtenir une surface développable. Nous prendrons
+pour variable~$v$ l'arc~$s$ de la courbe~$(K)$. Considérons le trièdre
+de Serret au point~$M$ de la courbe. Soit $MG$ la normale
+%[** Original diagram uses \alpha, \beta, \gamma; changed to match text]
+\Figure{092a}
+cherchée; elle est dans le plan normal à
+la courbe, pour la définir, il suffira
+donc de se donner l'angle $(MN, MG) = \chi$. Le
+point à l'unité de distance sur~$MG$ a pour
+coordonnées par rapport au trièdre de Serret
+$0, \cos \chi, \sin \chi$; si donc $l, m, n$ sont les cosinus directeurs
+de~$MG$, nous avons
+\begin{align*}%[** TN: Set on one line in original]
+l &= a' \cos \chi + a'' \sin \chi, \\
+m &= b' \cos \chi + b'' \sin \chi, \\
+n &= c' \cos \chi + c'' \sin \chi.
+\end{align*}
+Or\Add{,} $v$~étant l'arc de la courbe~$(K)$, on a
+\[
+df = a\, dv,\qquad dg = b\, dv,\qquad dh = c\, dv;
+\]
+les formules~\Eq{(4')} donnent
+\[
+a\, dv + u\left[(-a'\sin \chi + a''\cos \chi)\, d \chi
+  - \cos \chi \left(\frac{a}{R} + \frac{a''}{T}\right) dv
+  + \sin \chi\, \frac{a'}{T}\, dv\right] = 0;
+\]
+ou
+\[
+a \left(1 - \frac{u \cos \chi}{R}\right)
+  + a' u \sin \chi \left(-\frac{\DPtypo{dx}{d\chi}}{dv} + \frac{1}{T}\right)
+  + a''u \cos \chi \left( \frac{\DPtypo{dx}{d\chi}}{dv} - \frac{1}{T}\right)
+  = 0\Add{.}
+\]
+On a \Card{2} équations analogues avec $b, b', b''$ et $c, c', c''$; nous avons
+ainsi \Card{3} équations linéaires et homogènes par rapport aux coefficients
+de $a, a', a''$. Le déterminant est~$1$, donc les inconnues
+sont toutes nulles; et comme $u$~n'est pas constamment nul, on a
+%% -----File: 093.png---Folio 85-------
+\[
+1 - \frac{u \cos \chi}{R} = 0,\qquad
+\sin \chi \left( - \frac{d \chi}{dv} + \frac{1}{T}\right) = 0,\qquad
+\cos \chi \left(\frac{d \chi}{dv} - \frac{1}{T}\right) = 0\Add{.}
+\]
+Les \Card{2} dernières donnent, en remplaçant $v$ par l'arc~$\DPtypo{S}{s}$
+\[
+\Tag{(1)}
+\frac{d \chi}{ds} = \frac{1}{T},
+\]
+et la \Ord[f]{1}{e} donne
+\[
+\Tag{(2)}
+u = \frac{R}{\cos \chi}\Add{.}
+\]
+Il y a donc une infinité de solutions: $\chi$~se détermine par une
+quadrature.
+
+\Illustration[2.25in]{093a}
+La formule~\Eq{(2)} nous montre que
+\[
+R = u \cos \chi;
+\]
+donc la projection du point~$P$, où la normale~$MG$
+rencontre son enveloppe, sur la
+normale principale, est le centre de courbure.
+\emph{Le point de contact de la normale
+avec son enveloppe est sur la droite polaire.
+Les développées d'une courbe sont
+sur la surface polaire.}
+
+Considérons \Card{2} solutions $\chi\Add{,} \chi'$ de l'équation~\Eq{(1)}, la différence
+$\chi - \chi'$ est constante; les deux normales $MG, MG'$ se coupent
+sous un angle constant. Donc, \emph{lorsque une normale à une courbe
+décrit une surface développable, si on la fait tourner dans
+chacune de ses positions d'un angle constant autour de la tangente,
+la droite obtenue décrit encore une développable}.
+
+Le plan osculateur à une développée est le plan tangent
+à la développable correspondante: c'est le plan~$GMT$, ce plan
+est normal au plan~$BMC$, plan tangent à la surface polaire.
+%% -----File: 094.png---Folio 86-------
+\emph{Donc les développées sont des géodésiques de la surface polaire.}
+
+Considérons la normale principale~$P''$ en $P$ à la développée,
+elle est dans le plan osculateur $GMT$, elle est perpendiculaire
+à la tangente $MP$, donc parallèle à~$MT$. \emph{Les normales principales
+aux développées d'une courbe sont parallèles aux tangentes à la
+courbe. Le plan normal à la courbe est le plan rectifiant de
+toutes ses développées.}
+
+En partant d'une courbe~$(G)$, et remarquant que la courbe
+donnée~$(K)$ en est la développante, on pourra énoncer les propriétés
+précédentes de façon à obtenir les propriétés des développantes
+d'une courbe.
+
+
+\Section{Lignes de courbure.}
+{3.}{} Considérons sur une surface~$(S)$ une ligne de courbure~$(K)$,
+et la développable circonscrite à~$(S)$ le long de~$(K)$. La
+direction d'une génératrice $MG$ de cette développable est conjuguée
+de la tangente~$MT$ à la ligne de courbure, et par conséquent
+est perpendiculaire à~$MT$, c'est-à-dire normale à~$(K)$. Cette génératrice~$MG$
+est donc constamment tangente à la développée d'une
+ligne de courbure, et nous voyons que \emph{les normales à une ligne
+de courbure tangentes à la surface engendrent une développable}.
+
+\begin{wrapfigure}[10]{O}{2.25in}
+\Input[2.25in]{094a}
+\end{wrapfigure}
+Faisons tourner $\DPtypo{MG'}{MG}$ d'un angle droit
+autour de la tangente, nous obtenons une
+droite~$MG'$ qui, étant perpendiculaire aux
+\Card{2} tangentes à la surface $MT, MG$, sera la
+normale à la surface. Donc \emph{les normales à
+la surface en tous les points d'une ligne
+%% -----File: 095.png---Folio 87-------
+de courbure engendrent une développable}.
+
+Considérons le point~$P'$ où la droite $MG'$ touche son enveloppe;
+c'est le point où la droite polaire de la ligne de
+courbure rencontre la normale à la surface. Or, d'après le
+Théorème de Meusnier, les droites polaires de toutes les courbes
+de la surface tangentes en~$M$ rencontrent la normale en~$M$
+en un même point, qui est le centre de courbure de la section
+normale correspondante: $P'$~est donc le centre de courbure de
+la section principale $G'MT$, c'est l'un des centres de courbure
+principaux de la surface au point~$M$.
+
+Reprenons alors les formules~\Eq{(4')} du §1, que nous écrirons
+\[
+dx + u\, dl = 0,\quad dy + u\, dm = 0,\quad dz + u\, dn = 0;
+\]
+$l, m, n$ sont ici les cosinus directeurs de la normale, $u$~est le
+rayon de courbure principal~$R$; et pour un déplacement sur une
+ligne de courbure, nous avons les \emph{formules d'Olinde Rodrigues}
+\[
+dx + R\, dl =0,\qquad dy + R\, dm = 0,\qquad dz + R\, dn = 0.
+\]
+
+Les \emph{Théorèmes de Joachimsthal} se déduisent aisément de ce
+qui précède. Supposons que l'intersection~$(K)$ de \Card{2} surfaces $(S)\Add{,}
+(S_{1})$ soit une ligne de courbure pour chacune d'elles. Soient
+$MG, MG_{1}$ les normales aux \Card{2} surfaces en un point~$M$ de~$(K)$. Elles
+engendrent deux développables, donc enveloppent deux développées
+de~$(K)$, et par suite leur angle est constant. \emph{Réciproquement},
+si l'intersection $(K)$ de $(S)\Add{,} (S_{1})$ est ligne de courbure
+de~$(S_{1})$, et si l'angle des \Card{2} surfaces est constant tout
+le long de~$(K)$, la normale $MG_{1}$\DPtypo{,}{} à~$(S_{1})$ engendre une développable,
+et comme $\Err{MG}{MG_{1}}$~fait avec~$MG$ un angle constant, elle engendre
+aussi une développable, donc $(K)$~est une ligne de courbure
+sur~$(S)$.
+%% -----File: 096.png---Folio 88-------
+
+La condition~\Eq{(5)} pour qu'une droite engendre une surface
+développable est ici
+\[
+\begin{vmatrix}
+dx & dl & l \\
+dy & dm & m \\
+dz & dn & n
+\end{vmatrix}
+= 0,
+\]
+ou
+\[%[** TN: Added elided entries]
+\begin{vmatrix}
+\mfrac{\dd x}{\dd u} · du + \mfrac{\dd x}{\dd v} · dv &
+\mfrac{\dd l}{\dd u}\Add{·} du + \mfrac{\dd l}{\dd v}\Add{·} dv & l \\
+\mfrac{\dd y}{\dd u} · du + \mfrac{\dd y}{\dd v} · dv &
+\mfrac{\dd m}{\dd u}\Add{·} du + \mfrac{\dd m}{\dd v}\Add{·} dv & m \\
+\mfrac{\dd z}{\dd u} · du + \mfrac{\dd z}{\dd v} · dv &
+\mfrac{\dd n}{\dd u}\Add{·} du + \mfrac{\dd n}{\dd v}\Add{·} dv & n
+\end{vmatrix}
+= 0.
+\]
+Multiplions par
+\[
+\begin{vmatrix}
+\mfrac{\dd x}{\dd u} & \mfrac{\dd x}{\dd v} & l \\
+\mfrac{\dd y}{\dd u} & \mfrac{\dd y}{\dd v} & m \\
+\mfrac{\dd z}{\dd u} & \mfrac{\dd z}{\dd v} & n
+\end{vmatrix}
+\neq 0;
+\]
+Nous obtenons
+\[
+\begin{vmatrix}
+E\, du + F\, dv & -L\, du - M\, dv & 0 \\
+F\, du + G\, dv & -M\, du - N\, dv & 0 \\
+0 & 0 & 1
+\end{vmatrix}
+= 0;
+\]
+et nous retrouvons ainsi \emph{l'équation différentielle des lignes
+de courbure}
+\[
+\begin{vmatrix}
+E\, du + F\, dv & L\, du + M\, dv \\
+F\, du + G\, dv & M\, du + N\, dv
+\end{vmatrix}
+= 0.
+\]
+
+La même méthode, appliquée à l'équation~\Eq{(6)}
+\[
+\begin{vmatrix}
+dx & dl \\
+dy & dm
+\end{vmatrix}
+= 0
+\]
+donne facilement l'équation différentielle
+\[
+\begin{vmatrix}
+dx + p\, dz & dp \\
+dy + q\, dz & dq
+\end{vmatrix}
+= 0.
+\]
+
+%% -----File: 097.png---Folio 89-------
+
+\Section{Développement d'une surface développable sur un plan.}
+{4.}{Toute surface développable est applicable sur un plan\Add{.}}
+
+Considérons d'abord le cas du cylindre, dont les équations
+sont
+\begin{align*}
+x &= f(v) + u·l, & y &= g(v) + u·m, &  z &= h(v) + u·n; \\
+dx &= f'(v)\, dv + l·du, &
+dy &= g'(v)\, dv + m·du, &
+dz &= h'(v)\, dv + n·du.
+\end{align*}
+Nous avons
+\[
+ds^{2} = \sum f'{}^{2}(v)·dv^{2} + 2 \sum lf'(v)·du\, dv + \sum l^{2}·du^{2}.
+\]
+Nous pouvons supposer que la directrice: $x = f(v)$, $y = g(v)$, $z = h(v)$
+est une section droite, ce qui donne $\sum lf' = 0$; que $l, m, n$
+sont cosinus directeurs: $\sum l^{2} = 1$; enfin que $v$~est l'arc sur
+la section droite: $\sum f'{}^{2} = 1$. Alors on a
+\[
+\Tag{(1)}
+ds^{2} = du^{2} + dv^{2};
+\]
+\DPtypo{On}{on} a l'élément linéaire d'un plan. \emph{Un cylindre est applicable
+sur un plan}, $\Phi$~et~\Eq{(1)} donne la loi du développement.
+
+Voyons maintenant le cas du cône
+\[
+x = u·l(v),\qquad y = u·m(v),\qquad z = u·n(v);
+\]
+$u$~est la longueur prise sur la génératrice à partir du sommet;
+supposons que $l, m, n$ soient cosinus directeurs de la génératrice,
+$v$~étant l'arc de courbe sphérique intersection du cône
+avec la sphère $u = 1$. Alors
+\begin{alignat*}{3}%[** TN: Set on one line in original]
+dx &= ul'(v)\, &&dv + l(v)\, &&du, \\
+dy &= um'(v)\, &&dv + m(v)\, &&du, \\
+dz &= u\DPtypo{n}{n'}(v)\, &&dv + n(v)\, &&du;
+\end{alignat*}
+et
+\[
+\Tag{(2)}
+ds^{2} = u^{2}\, dv^{2} + du^{2}.
+\]
+C'est l'élément linéaire d'un plan en coordonnées polaires. Un
+\emph{cône est applicable sur un plan}, $\Phi$~\Add{et}~\Eq{(2)} donne la loi du développement.
+
+Passons enfin au cas général
+\[
+x = f(v) + u·l(v),\qquad y = g(v) + u·m(v),\qquad z = h(v) + u·n(v)\Add{.}
+\]
+%% -----File: 098.png---Folio 90-------
+Nous supposerons que la courbe $x = f(v)$, $y = g(v)$, $z = h(v)$
+soit l'arête de rebroussement, $v$~l'arc sur cette courbe, $l, m, n$
+les cosinus directeurs de la tangente en un point, et $u$~la
+distance comptée sur cette tangente à partir du point de contact.
+Alors $l = f'= a$; $m = g'= b$; $n = h'= c$;
+et
+\begin{gather*}%[** TN: Aligned last three equations]
+l' = \frac{da}{dv} = \frac{a'}{R},\qquad
+m' = \frac{db}{dv} = \frac{b'}{R},\qquad
+n' = \frac{dc}{dv} = \frac{c'}{R}; \\
+\begin{alignedat}{3}
+dx &= a\, dv &&+ u\, \frac{a'}{R}\, dv &&+ a\, du, \\
+dy &= b\, dv &&+ u\, \frac{b'}{R}\, dv &&+ b\, du, \\
+dz &= c\, dv &&+ u\, \frac{c'}{R}\, dv &&+ c\, du\Add{;}
+\end{alignedat}
+\end{gather*}
+et
+\[
+ds^{2} = \bigl[d(u + v)\bigr]^{2} + \frac{u^{2}}{R^{2}}\, dv^{2}.
+\]
+Cet élément reste le même si $R$~garde la même expression en
+fonction de~$v$. Donc \emph{l'élément linéaire est le même pour toutes
+les surfaces développables dont les arêtes de rebroussement
+sont des courbes dont le rayon de courbure a la même expression
+en fonction de l'arc}:
+\[
+R = \Phi (v).
+\]
+Nous pouvons déterminer une courbe plane dont le rayon de courbure
+s'exprime en fonction de l'arc par l'équation précédente.
+Nous prendrons pour coordonnées dans le plan de cette courbe
+l'arc~$\DPtypo{S}{s}$ de la courbe, et la distance comptée sur la tangente
+à partir du point de contact et on aura pour l'élément linéaire
+du plan la forme précédente. La développable sera donc applicable
+sur ce plan. Quand la développable est donnée, on détermine
+par des opérations algébriques son arête de rebroussement,
+et par une quadrature l'arc de cette arête de rebroussement.
+On a alors
+\[
+R = \Phi(s)\Add{.}
+\]
+%% -----File: 099.png---Folio 91-------
+Il faut construire une courbe plane satisfaisant à cette condition.
+Si $\alpha$~est l'angle de la tangente avec~$Ox$, on a
+\[
+R = \frac{ds}{d \alpha};
+\]
+d'où
+\[%[** TN: Set on two lines in original]
+\frac{ds}{d \alpha} = \Phi (s), \qquad
+\alpha = \int \frac{ds}{\Phi (s)};
+\]
+et alors
+\[
+dx = \cos \alpha · ds,\qquad dy = \sin \alpha · ds;
+\]
+$x, y$ se déterminent au moyen de \Card{3} quadratures. La courbe que
+l'on obtient est le développement de l'arête de rebroussement.
+
+
+\Section{Réciproque.}
+{}{Réciproquement toute surface applicable sur un plan est
+une surface développable.}
+
+Soit la surface
+\[
+x = f(u,v),\qquad y = g(u,v),\qquad z = h(u,v),
+\]
+que nous supposons applicable sur un plan. Nous avons, en
+choisissant convenablement les coordonnées $u, v$:
+\[
+ds^{2} = E\Add{·} du^{2} + 2 F·du\, dv + G\Add{·} dv^{2} = du^{2} + dv^{2};
+\]
+d'où
+\[
+\sum \left(\frac{\dd x}{\dd u}\right)^{2} = 1,\qquad
+\sum \frac{\dd x}{\dd u} · \frac{\dd x}{\dd v} = 0,\qquad
+\sum \left(\frac{\dd x}{\dd v}\right)^{2} = 1.
+\]
+Différentions ces relations successivement par rapport à $u, v$,
+nous avons
+\begin{align*}
+&\sum \frac{\dd x}{\dd u} · \frac{\dd^{2} x}{\dd u^{2}} = 0, &
+&\sum \frac{\dd^{2} x}{\dd u^{2}} · \frac{\dd x}{\dd v}
+  + \sum \frac{\dd x}{\dd u} · \frac{\dd^{2} x}{\dd u\, \dd v} = 0, &
+&\sum \frac{\dd x}{\dd v} · \frac{\dd^{2} x}{\dd u\, \dd v} = 0, \\
+%
+&\sum \frac{\dd x}{\dd u}\Add{·} \frac{\dd^{2} x}{\dd u\, \dd v} = 0, &
+&\sum \frac{\dd^{2} x}{\dd u\, \dd v} · \frac{\dd x}{\dd v}
+  + \DPtypo{}{\sum} \frac{\dd x}{\dd u} · \frac{\dd^{2} x}{\dd v^{2}} = 0, &
+&\sum \frac{\dd x}{\dd v} · \frac{\dd^{2} x}{\dd v^{2}} = 0;
+\end{align*}
+d'où nous tirons:
+\[
+\sum \frac{\dd^{2} x}{\dd \DPtypo{u}{u^{2}}} · \frac{\dd x}{\dd v} = 0,\qquad
+\sum \frac{\dd x}{\dd u} · \frac{\dd^{2} x}{\dd \DPtypo{v}{v^{2}}} = 0.
+\]
+Considérons les \Card{2} fonctions $\dfrac{\dd x}{\dd u}$ et $\dfrac{\dd y}{\dd u}$. Leur déterminant fonctionnel
+est:
+%% -----File: 100.png---Folio 92-------
+\[
+\begin{vmatrix}
+\mfrac{\dd^{2} x}{\dd u^{2}} & \mfrac{\dd^{2} x}{\dd u\, \dd v} \\
+\mfrac{\dd^{2} y}{\dd u^{2}} & \mfrac{\dd^{2} y}{\dd u\, \dd v}
+\end{vmatrix}\Add{.}
+\]
+Or, considérons les équations
+\begin{alignat*}{4}
+&X\, \frac{\dd^{2} x}{\dd u^{2}} &&+ Y\, \frac{\dd^{2} y}{\dd u^{2}} &&+ Z\, \frac{\dd^{2} z}{\dd u^{2}} &&= 0, \\
+&X\, \frac{\dd^{2} x}{\dd u\, \dd v} &&+ Y\, \frac{\dd^{2} y}{\dd u\, \dd v} &&+ Z\, \frac{\dd^{2} z}{\dd u\, \dd v} &&= 0;
+\end{alignat*}
+d'après les relations précédemment écrites, ce système admet
+\Card{2} solutions distinctes
+\begin{alignat*}{3}
+X &= \frac{\dd x}{\dd u},\qquad &
+Y &= \frac{\dd y}{\dd u},\qquad &
+Z &= \frac{\dd z}{\dd u}; \\
+X &= \frac{\dd x}{\dd v}, &
+Y &= \frac{\dd y}{\dd v}, &
+Z &= \frac{\dd z}{\dd v}.
+\end{alignat*}
+Ces solutions ne sont pas proportionnelles, sans quoi les
+courbes $u = \cte$ et $v = \cte$ seraient constamment tangentes.
+Donc les \Card{3} déterminants déduits du tableau
+\[
+\begin{Vmatrix}
+\mfrac{\dd^{2} x}{\dd u^{2}}     & \mfrac{\dd^{2} y}{\dd u^{2}}     & \mfrac{\dd^{2} z}{\dd u^{2}} \\
+\mfrac{\dd^{2} x}{\dd u\, \dd v} & \mfrac{\dd^{2} y}{\dd u\, \dd v} & \mfrac{\dd^{2} z}{\dd u\, \dd v}
+\end{Vmatrix}
+\]
+\DPtypo{Sont}{sont} nuls; or\Add{,} ce sont les déterminants fonctionnels des \Card{3} quantités
+$\dfrac{\dd x}{\dd u}$, $\dfrac{\dd y}{\dd u}$, $\dfrac{\dd z}{\dd u}$ prises \Card{2} à \Card{2}, donc ces \Card{3} quantités sont
+fonctions de l'une d'entre elles, c'est-à-dire d'une seule
+variable~$t$. De même $\dfrac{\dd x}{\dd v}$, $\dfrac{\dd y}{\dd v}$, $\dfrac{\dd z}{\dd v}$ sont fonctions d'une même variable~$\theta$.
+De plus la relation
+\[
+\sum \frac{\dd x}{\dd u} · \frac{\dd x}{\dd v} = 0
+\]
+montre que $\theta$ par exemple s'exprime en fonction de~$t$.
+Les \Card{6} dérivées partielles sont donc fonctions d'une même variable;
+le plan tangent à la surface dépend d'un seul paramètre.
+La surface est développable.
+
+%[** TN: Roman numerals in original]
+\Paragraph{Remarque \1.} Dans le développement les lignes géodésiques
+%% -----File: 101.png---Folio 93-------
+se conservent; or\Add{,} les géodésiques du plan sont des droites.
+\emph{Les lignes géodésiques de la surface développable sont donc
+les lignes qui dans le développement de cette surface sur un
+plan, correspondent aux droites de ce plan.}
+
+En particulier, considérons la surface rectifiante d'une
+courbe, enveloppe du plan rectifiant. Cette courbe est une
+géodésique de sa surface rectifiante, puisque son plan osculateur
+est perpendiculaire au plan tangent; elle se développe
+donc suivant une droite lorsqu'on effectue le développement
+de la surface rectifiante sur un plan. \emph{De là le nom de plan
+rectifiant.}
+
+\Paragraph{Remarque \2.} Il résulte de là que la recherche des géodésiques
+d'une surface développable se ramène à son développement,
+et exige par conséquent \Card{4} quadratures.
+
+\Paragraph{Remarque \3.} La détermination des lignes de courbure,
+développantes de l'arête de rebroussement, revient à une quadrature.
+
+
+\Section{Lignes géodésiques d'une surface développable.}
+{5.}{} Nous avons trouvé les lignes géodésiques d'une surface
+développable en considérant le développement de cette
+surface sur un plan. On peut les chercher directement.
+Soit l'arête de rebroussement
+\[
+\Tag{(1)}
+x = f(s),\qquad y = g(s),\qquad z = h(s),
+\]
+$s$~désignant l'arc. Si $a, b, c$ sont les cosinus directeurs de la
+tangente, et $u$~une longueur comptée sur cette tangente à partir
+du point de contact, la surface est représentée par
+\[
+x = f + u · a,\qquad y = g + u · b,\qquad z = h + u · c;
+\]
+%% -----File: 102.png---Folio 94-------
+%[** TN: Seeming error in erratum, changed in accord with surrounding text]
+en désignant par $\Err{u, v, w}{\DPtypo{u', v', w'}{a', b', c'}}$,
+les dérivées de $\DPtypo{u v w}{a\Add{,} b\Add{,} c}$ par rapport \Err{$a, s$}{à~$s$},
+on a
+\[%[** TN: Added elided equations]
+\frac{dx}{ds} = a + u\, \frac{a'}{R} + au',\qquad
+\frac{dy}{ds} = b + u\, \frac{b'}{R} + bu',\qquad
+\frac{dz}{ds} = c + u\, \frac{c'}{R} + cu';
+\]
+ou
+\begin{gather*}%[** TN: Added elided equations in first set]
+\frac{dx}{ds} = a(1 + u') + a'\, \frac{u}{R},\quad
+\frac{dy}{ds} = b(1 + u') + b'\, \frac{u}{R},\quad
+\frac{dz}{ds} = c(1 + u') + c'\, \frac{u}{R}; \\
+%
+\frac{d^{2}x}{ds^{2}} = a\left(u'' - \frac{u}{R^{2}}\right)
+  + a'\, \frac{1}{R} \left(1 + 2u' - u\, \frac{R'}{R}\right)
+  + a''\, \frac{-u}{RT},
+\end{gather*}
+et les analogues.
+
+L'équation des lignes géodésiques est, en remarquant que la
+normale à la surface n'est autre que la binormale à l'arête de
+rebroussement
+\[
+\begin{vmatrix}
+\mfrac{d^{2}x}{ds^{2}} & \mfrac{d^{2}y}{ds^{2}} & \mfrac{d^{2}z}{ds^{2}} \\
+\mfrac{dx}{ds}         & \mfrac{dy}{ds}         & \mfrac{dz}{ds} \\
+a''                   & b''                   & c''
+\end{vmatrix}
+= 0,
+\]
+ou
+\[
+\begin{vmatrix}
+a \left(u'' - \mfrac{u}{R^{2}}\right) + \mfrac{a'}{R} \left(1 + 2u' - u\, \mfrac{R'}{R}\right) + a''\, \mfrac{- u}{RT} & \dots & \dots \\
+a (1 + u') + a'\, \mfrac{u}{R}  & \dots & \dots \\
+a''                             & \dots & \dots \\
+\end{vmatrix}
+= 0,
+\]
+ou, en décomposant en déterminants simples,
+\[
+\frac{1}{R} \left(1 + 2u' - u\, \frac{R'}{R}\right) (1 + u')
+\begin{vmatrix}
+a'  & b'  & c' \\
+a   & b   & c  \\
+a'' & b'' & c''
+\end{vmatrix}
++ \frac{u}{R} \left(u'' - \frac{u}{R^{2}}\right)
+\begin{vmatrix}
+a   & b   & c   \\
+a'  & b'  & c'  \\
+a'' & b'' & c''
+\end{vmatrix}
+= 0;
+\]
+ou enfin
+\[
+\frac{u}{R} \left(u'' - \frac{u}{R^{2}}\right)
+  - \frac{1}{R} (1 + u') \left(1 + 2u' - u\, \frac{R'}{R}\right) = 0,
+\]
+c'est-à-dire
+\[
+\Tag{(2)}
+u · u'' - 2u'{}^{2}
+  - u'\left(3 - u\, \frac{R'}{R}\right)
+  - \frac{u^{2}}{R^{2}} + u · \frac{R'}{R} - 1 =0.
+\]
+Telle est l'équation différentielle qui détermine~$u$.
+%% -----File: 103.png---Folio 95-------
+
+Cherchons la nature de l'intégrale générale. Si nous développons
+la surface sur un plan, la courbe~\Eq{(1)} sera représentée
+par une courbe
+\[
+x = F(s),\qquad y = G(s),
+\]
+dont le rayon de courbure sera encore~$R$. Le point homologue du
+point $(u,s)$ de la surface sera
+\[
+x = F + uF',\qquad y = G + uG'.
+\]
+Les droites du plan sont
+\[
+A(F + uF') + B(G + uG') + C = 0,
+\]
+d'où
+\[
+u = - \frac{AF + BG + C}{AF' + BG'};
+\]
+en remarquant que le dénominateur est la dérivée du numérateur
+nous sommes donc conduits à poser
+\[
+u = -\frac{w}{w'}\DPtypo{.}{,}
+\]
+et à prévoir que l'équation en~$w$ sera linéaire, homogène du
+\Ord{3}{e} ordre. Effectivement
+\begin{align*}
+u' &= -1 + \frac{ww''}{w'{}^{2}}, \\
+\intertext{et}
+u'' &= \frac{ww'''}{w'{}^{2}} + \frac{w''}{w'} - \frac{2ww''{}^{2}}{w'{}^{3}};
+\end{align*}
+\Eq{(2)}~devient alors
+\begin{multline*}
+\DPtypo{}{-}\frac{w}{w'} \left(\frac{ww'''}{w'{}^{2}} + \frac{w''}{w'} - \frac{2ww''{}^{2}}{w'{}^{3}}\right)
+  - 2\left(-1 + \frac{ww''}{w'{}^{2}}\right)^{2}
+  - 3\left(-1 + \frac{ww''}{w'{}^{2}}\right) \\
+  - \frac{R'}{R}\, \frac{w}{w'} \left(-1 + \frac{ww''}{w'{}^{2}}\right)
+  \DPtypo{- \frac{1}{R} · \frac{w}{w} - - \frac{w}{w} - 1}
+         {- \frac{1}{R^{2}} · \frac{w^{2}}{w'{}^{2}} - \frac{R'}{R} · \frac{w}{w'} - 1}
+  = 0;
+\end{multline*}
+ou, après simplification
+\[
+w''' + \frac{R'}{R}\, w'' + \frac{1}{R^{2}}\, w' = 0.
+\]
+Posons
+\[
+w' = \theta\Add{,}
+\]
+il vient
+\[
+\Tag{(3)}
+\theta'' + \frac{R'}{R}\, \theta' + \frac{1}{R^{2}}\, \theta = 0\Add{,}
+\]
+%% -----File: 104.png---Folio 96-------
+équation linéaire du \Ord{2}{e} ordre en~$\theta$. Faisons \DPchg{disparaitre}{disparaître} le \Ord{2}{e} terme
+par le changement de variable
+\[
+\alpha = \phi (s),
+\]
+d'où
+\begin{align*}
+\theta' &= \frac{d \theta}{d \alpha} · \phi', \\
+\theta''
+  &= \frac{d^{2} \theta}{d \DPtypo{\alpha}{\alpha^{2}}} · \phi'{}^{2}
+   + \frac{d \theta}{d \alpha}\, \phi'';
+\end{align*}
+\Eq{(3)}~devient
+\[
+\frac{d^{2} \theta}{d \alpha^{2}}\, \phi'{}^{2}
+  + \frac{d \theta}{d \alpha} \left( \phi'' + \frac{R'}{R}\, \phi'\right)
+  + \frac{1}{R^{2}}\, \theta = 0.
+\]
+Il faut alors choisir la fonction~$\phi$ de façon que l'on ait
+\[
+\phi'' + \frac{R'}{R}\, \phi' = 0 ,
+\]
+ou
+\[
+\frac{\phi''}{\phi'} = -\frac{R'}{R}.
+\]
+On peut prendre
+\[
+\phi' = \frac{1}{R},
+\]
+et poser
+\[
+ds = R · \alpha.
+\]
+Nous obtenons alors l'équation
+\[
+\frac{d^{2} \theta}{d \alpha^{2}} + \theta = 0,
+\]
+dont l'intégrale générale est
+\[
+\theta = w' = A \cos \alpha + B \sin \alpha = \frac{dw}{ds};
+\]
+d'où
+\[
+w = A \int \cos \alpha · ds + B \int \sin \alpha · ds + c,
+\]
+et enfin
+\[
+u = -\frac{\ds A \int \cos \alpha · ds + B \int \sin \alpha · ds + c}
+          {A \cos \alpha + B \sin \alpha},
+\]
+avec
+\[
+\alpha = \int \frac{ds}{R}.
+\]
+On peut se dispenser d'introduire l'arc~$s$ explicitement. Donc
+les lignes géodésiques d'une surface développable s'obtiennent
+par \Card{3} quadratures au plus. On constate de plus que les deux
+méthodes conduisent aux mêmes calculs.
+%% -----File: 105.png---Folio 97-------
+
+
+\Section{Surfaces réglées gauches. Trajectoires orthogonales des génératrices.}
+{6.}{} Soit la surface
+\[
+x = f(v) + u·l(v),\qquad y = g(v) + u·m(v),\qquad z = h(v) + u·n(v);
+\]
+les génératrices étant des géodésiques, il en résulte que
+\emph{les trajectoires orthogonales des génératrices déterminent
+sur ces génératrices des segments égaux}. Pour obtenir ces
+trajectoires orthogonales, il faut déterminer $u$ en fonction de~$v$
+de façon que l'on ait
+\[
+\sum l\, dx = 0.
+\]
+Pour simplifier nous supposerons que $l, m, n$ soient cosinus directeurs;
+on a alors
+\[
+\sum l^{2} = 1,\qquad \sum l\,dl = 0;
+\]
+et l'équation différentielle devient
+\[
+\sum l · df + du = 0,
+\]
+d'où
+\[
+u = - \int \sum \DPtypo{.l\, df}{l·df}.
+\]
+La détermination des trajectoires orthogonales des génératrices
+d'une surface réglée dépend d'une quadrature.
+
+\Paragraph{Remarque.} On peut rattacher ce fait à la formule qui
+donne la variation d'un segment de droite. Prenons sur la
+droite $MM_{1}$ une direction positive; soit $r$~la distance $MM_{1}$, prise
+\Figure[2.5in]{105a}
+en valeur absolue. Soient $(x, y, z)$ et
+$\DPtypo{}{(}x_{1}, y_{1}, z_{1})$, les coordonnées des deux extrémités,
+qui décrivent deux courbes données. Nous
+avons
+\[
+r^{2} = (x_{1} - x)^{2} + (y_{1} - y)^{2} + (z_{1} - z)^{2}\Add{,}
+\]
+d'où
+\[
+r\, dr = (x_{1} - x)(dx_{1} - dx) + (y_{1} - y)(dy_{1} - dy) % +
++ (z_{1} - z)(dz_{1} - dz)\DPtypo{.}{,}
+\]
+%% -----File: 106.png---Folio 98-------
+d'où
+\begin{multline*}
+dr = \left(\frac{x_{1} - x}{r}\, dx_{1}
+         + \frac{y_{1} - y}{r}\, dy_{1}
+         + \frac{z_{1} - z}{r}\, dz_{1}\right) \\
+   - \left(\frac{x_{1} - x}{r}\, dx
+         + \frac{y_{1} - y}{r}\, dy
+         + \frac{z_{1} - z}{r}\, dz\right).
+\end{multline*}
+Soient $\alpha, \beta, \gamma$, $\alpha_{1}, \beta_{1}, \gamma_{1}$ les tangentes aux courbes en $MM_{1}$ dirigées
+dans le sens des arcs croissants. Soient $\lambda\Add{,} \mu\Add{,} \nu$ les cosinus directeurs
+de la droite~$MM_{1}$. Nous avons
+\[
+dr = ds_{1} (\lambda \alpha_{1} + \mu \beta_{1} + \nu \gamma_{1})
+   - ds (\lambda \alpha + \mu \beta + \nu \gamma);
+\]
+soient $\theta, \theta_{1}$ les angles de $MM_{1}$ avec les \Card{2} tangentes; nous avons
+enfin la \emph{formule importante}
+\[
+dr = ds_{1} · \cos \theta_{1} - ds \Add{·} \cos \theta.
+\]
+
+Supposons la droite $MM_{1}$ tangente à la \Ord{1}{ère} courbe et normale
+à la \Ord{2}{e}, $\theta = 0$, $\theta_{1} = ±\frac{\pi}{2}$\Add{;} nous avons
+\[
+dr = - ds
+\]
+et nous retrouvons ainsi les propriétés des développantes et
+des développées.
+
+Supposons la droite normale aux deux courbes, $\theta = ±\frac{\pi}{2}$,
+$\theta_{1} = ±\frac{\pi}{2}$, alors $dr = 0$, $r = \cte$, et nous retrouvons les \DPtypo{propriétes}{propriétés}
+des trajectoires orthogonales des génératrices.
+
+
+\Section{Cône directeur. Point central. Ligne de striction.}
+{7.}{} On appelle \emph{cône directeur} de la surface le cône
+\[
+x = u · l(v),\qquad y = u · m(v),\qquad z = u · n(v).
+\]
+
+Si ce cône se réduit à un plan, ce plan s'appelle \emph{plan
+directeur}, et les génératrices sont toutes parallèles à ce
+plan.
+
+Le plan tangent en un point quelconque de la surface a
+pour coefficients les déterminants déduits de:
+%% -----File: 107.png---Folio 99-------
+\[
+\Tag{(1)}
+\begin{Vmatrix}
+l           & m           & n \\
+df + u\, dl & dg + u\, dm & dh + u\, dn
+\end{Vmatrix}\Add{.}
+\]
+Le plan tangent au cône directeur le long de la génératrice
+correspondant à celle qui passe par le point considéré a pour
+coefficients les déterminants déduits de
+\[
+\begin{Vmatrix}
+l  & m  & n \\
+dl & dm & dn
+\end{Vmatrix}\Add{.}
+\]
+Ces plans sont parallèles si $u$~est infini. On a alors sur la
+surface le plan tangent au point à l'infini sur la génératrice
+qu'on appelle \emph{plan asymptote}. \emph{Les plans asymptotes sont parallèles
+aux plans tangents au cône directeur le long des génératrices
+correspondantes.}
+
+\emph{Dans une surface à plan directeur, tous les plans asymptotes
+sont parallèles au plan directeur.}
+
+Pour que les deux plans tangents soient rectangulaires,
+il faut que la somme des produits des déterminants précédents
+soit nulle
+\[
+\begin{vmatrix}
+\sum l^{2} & \sum l\, df + u \sum l\, dl \\
+\sum ldl   & \sum dl·df + u \sum dl^{2}
+\end{vmatrix}
+= 0.
+\]
+Nous avons une équation du \Ord{1}{er} degré en~$u$. \emph{Il existe donc en
+général sur toute génératrice un point où le plan tangent est
+perpendiculaire au plan tangent au cône directeur, c'est-à-dire
+au plan asymptote. C'est le point central}, le plan tangent
+en ce point s'appelle \emph{plan central}.
+
+Le lieu des points centraux s'appelle \emph{ligne de striction}.
+
+Nous supposerons pour simplifier $\sum l^{2} = 1$, ce qui écarte
+le cas des surfaces réglées à génératrices isotropes. L'équation
+%% -----File: 108.png---Folio 100-------
+qui donne \DPtypo{le~$u$}{l'$u$} du point central se réduit à
+\[
+u \sum dl^{2} + \sum dl · df = 0;
+\]
+le point central existe donc toujours, sauf si l'on a
+\[
+\sum dl^{2} = 0.
+\]
+Dans ce cas la courbe sphérique base du cône directeur est
+une courbe minima de la sphère, c'est-à-dire, une génératrice
+isotrope. Le cône est alors un plan tangent au cône asymptote
+de la sphère, qui est un cône isotrope, c'est un plan isotrope.
+Les surfaces considérées sont des \emph{surfaces réglées à plan directeur
+isotrope}. Toutes sont imaginaires, sauf le \DPchg{paraboloide}{paraboloïde}
+de révolution.
+
+\Paragraph{Remarque.} Le plan tangent est indéterminé si tous les
+déterminants du tableau~\Eq{(1)} sont nuls. Alors $K$~étant un certain
+facteur, on a
+\[
+df + u\, dl + Kl = 0,\qquad
+dg + u\, dm + Km = 0,\qquad
+dh + u\, dn + Kn = 0\DPtypo{;}{,}
+\]
+ce qui donne
+\[
+\begin{vmatrix}
+df & dl & l \\
+dg & dm & m \\
+dh & dn & n
+\end{vmatrix} = 0\DPtypo{:}{;}
+\]
+la surface est développable. Pour trouver le point où le plan
+tangent est indéterminé multiplions par $dl, dm, dn$ et ajoutons,
+il vient
+\[
+u \sum dl^{2} + \sum dl · df = 0;
+\]
+c'est le point de contact de la génératrice et de l'arête de
+rebroussement. C'est ce qui explique que la formule précédente
+qui donne la ligne de striction pour une surface réglée quelconque,
+donne l'arête de rebroussement pour une surface développable.
+%% -----File: 109.png---Folio 101-------
+
+
+\Section{Variations du plan tangent le long d'une génératrice.}
+{8.}{} Proposons-nous de chercher l'angle des plans tangents
+à une surface réglée en 2 points d'une même génératrice.
+A cet effet, traitons d'abord le problème suivant: on a une
+droite~$\Delta$, de cosinus directeurs $\alpha\Add{,} \beta\Add{,} \gamma$, et \Card{2} droites qui la rencontrent
+$D (p\Add{,} q\Add{,} r)$ et $D' (p'\Add{,} q'\Add{,} r')$. Calculons l'angle~$V$ des \Card{2} plans
+$D \Delta$~et~$D' \Delta$.
+
+\Figure[2.25in]{109a}
+Considérons un trièdre trirectangle
+auxiliaire dont l'un des axes soit~$\Delta$;
+soient $\alpha'\Add{,} \beta'\Add{,} \gamma'$, $\alpha''\Add{,} \beta''\Add{,} \gamma''$ les cosinus directeurs
+des autres axes, et soient dans ce
+système $u\Add{,} v\Add{,} w$ et $u'\Add{,} v'\Add{,} w'$ les coefficients de
+direction de~$\Delta \Delta'$. Nous avons
+\[
+\tg V = \frac{vw' - wv'}{vv' + ww'}\Add{.}
+\]
+Mais on a
+\[
+\begin{alignedat}{3}
+  u  &= \alpha p  &&+ \beta q  &&+ \gamma r,  \\
+  u' &= \alpha p' &&+ \beta q' &&+ \gamma r',
+\end{alignedat}
+\quad
+\begin{alignedat}{3}
+  v  &= \alpha' p  &&+ \beta' q  &&+ \gamma' r, \\
+  v' &= \alpha' p' &&+ \beta' q' &&+ \gamma' r',
+\end{alignedat}
+\quad
+\begin{alignedat}{3}
+  w  &= \alpha'' p  &&+ \beta'' q  &&+ \gamma'' r \\
+  w' &= \alpha'' p' &&+ \beta'' q' &&+ \gamma'' r',
+\end{alignedat}
+\]
+d'où
+\begin{align*}
+vw' - wv' &=
+\begin{vmatrix}
+\begin{alignedat}{3}
+  &\alpha' p  &&+ \beta' q  &&+ \gamma' r \\
+  &\alpha' p' &&+ \beta' q' &&+ \gamma' r'
+\end{alignedat}
+&
+\begin{alignedat}{3}
+  &\alpha'' p  &&+ \beta'' q  &&+ \gamma'' r \\
+  &\alpha'' p' &&+ \beta'' q' &&+ \gamma'' r'
+\end{alignedat}
+\end{vmatrix} \\
+&=
+\begin{Vmatrix}
+\alpha'  & \beta'  & \gamma' \\
+\alpha'' & \beta'' & \gamma''
+\end{Vmatrix}
+\begin{Vmatrix}
+p  & q  & r \\
+p' & q' & r'
+\end{Vmatrix}
+=
+\begin{vmatrix}
+\alpha & \beta & \gamma \\
+p      & q     & r \\
+p'     & q'    & r'
+\end{vmatrix}.
+\end{align*}
+D'ailleurs
+\[
+uu' + vv' + ww' = pp' + qq' + rr',
+\]
+d'où
+\[
+vv' + ww' = pp' + qq' + rr' - \sum \alpha p · \sum \alpha p'.
+\]
+Alors
+\[
+\tg V = \frac{
+\begin{vmatrix}
+\alpha & \beta & \gamma \\
+p      & q     & r \\
+p'     & q'    & r'
+\end{vmatrix}}{\sum pp' - \sum \alpha p · \sum \alpha p'}
+= \frac{D \sum \alpha^{2}}
+       {\sum \alpha^{2} · \sum pp' - \sum \alpha p · \sum \alpha p'}.
+\]
+Sous cette forme, on peut alors introduire les coefficients
+%% -----File: 110.png---Folio 102-------
+directeurs $l, m, n$ de la direction~$\Delta$
+\[
+\Tag{(1)}
+\tg V = \frac{\raisebox{-\baselineskip}{$\sqrt{l^{2} + m^{2} + n^{2}}$}\,
+\begin{vmatrix}
+l  & m  & n \\
+p  & q  & r \\
+p' & q' & r'
+\end{vmatrix}}{\sum l^{2} · \sum pp' - \sum lp · \sum lp'}.
+\]
+
+Appliquons cette formule à l'angle des plans tangents en
+\Card{2} points $M$,~$M'$ d'une même génératrice. On peut prendre pour directions
+$D$,~$D'$ les directions tangentes aux courbes $u = \cte$:
+\begin{align*}
+p  &= df + u\, dl,  & q  &= dg + u\, dm,  & r  &= dh + u\, dn; \\
+p' &= df + u'\, dl, & q' &= dg + u'\, dm, & r' &= dh + u'\, dn;
+\end{align*}
+le déterminant de la formule~\Eq{(1)} devient
+\[
+\begin{vmatrix}
+l & df + u\, dl & df + u'\, dl \\
+m & dg + u\, dm & dg + u'\, dm \\
+n & dh + u\, dn & dh + u'\, dn
+\end{vmatrix}
+=
+\begin{vmatrix}
+l & dl & df \\
+m & dm & dg \\
+n & dn & dh
+\end{vmatrix} (u - u');
+\]
+et
+\[
+\tg V = \frac{\raisebox{-\baselineskip}{$(u' - u) \sqrt{l^{2} + m^{2} + n^{2}}$}\,
+\begin{vmatrix}
+df & dg & dh \\
+dl & dm & dn \\
+l  & m  & n
+\end{vmatrix}}
+{\begin{vmatrix}
+\sum l^{2}           & \sum l (df + u\, dl) \\
+\sum l(df + u'\, dl) & \sum (df + u\, dl) (df + u'\, dl)
+\end{vmatrix}}.
+\]
+Nous poserons
+\[
+D =
+\begin{vmatrix}
+df & dg & dh \\
+dl & dm & dn \\
+l  & m  & n
+\end{vmatrix}.
+\]
+
+Pour simplifier ce résultat, nous prendrons pour $l, m, n$
+les cosinus directeurs de la génératrice ($\sum l^{2} = 1$, $\sum l\, dl = 0$);
+nous supposerons que la courbe $x = f(v)$, $y = g(v)$, $z = h(v)$
+soit trajectoire orthogonale des génératrices, $\sum l\, df = 0$.
+Nous déterminerons~$u$ par la relation
+\[
+u \sum dl^{2} + \sum dl · df = 0
+\]
+ce qui revient à prendre pour l'un des points le point central.
+%% -----File: 111.png---Folio 103-------
+Le dénominateur devient
+\[
+\sum df^{2} + u \sum dl\Add{·} df
+  = \frac{\sum df^{2} · \sum dl^{2} - \left(\sum dl · df\right)^{2}}{\sum dl^{2}};
+\]
+et alors
+\[
+\tg V = \frac{(u' - u) D · \sum dl}{\sum df^{2} · dl^{2} - \left(\sum dl · df\right)^{2}}.
+\]
+Posons
+\[
+K = \frac{\sum dl^{2} · \sum df^{2} - \left(\sum dl · df\right)^{2}}
+         {D · \sum dl^{2}}\DPtypo{:}{;}
+\]
+en remarquant que $u'- u = CM$, on a
+\[
+\Tag{(2)}
+\tg V = \frac{CM}{K},
+\]
+\emph{formule de Chasles}. D'où les conséquences bien connues suivantes,
+et qui ne sont en défaut que pour des génératrices singulières:
+
+\ParItem{\Primo.} \emph{\DPtypo{lorsque}{Lorsque} $M$~décrit la génératrice d'un bout à l'autre,
+le plan tangent~$(P)$ en~$M$ tourne autour de la génératrice toujours
+dans le même sens, et la rotation totale qu'il effectue
+est de~$180°$.} En deux points différents, les plans tangents
+sont différents.
+
+\ParItem{\Secundo.} \emph{La division des points~$M$ et le faisceau des plans~$(P)$
+sont en correspondance homographique.}
+
+\ParItem{\Tertio.} Comme trois couples définissent une homographie,
+\emph{deux surfaces réglées qui ont une génératrice commune, et
+qui sont tangentes en trois points de cette génératrice, sont
+tangentes en tous les autres points de cette génératrice},
+c'est-à-dire se raccordent tout le long de cette génératrice.
+L'expression de~$K$ peut se simplifier; on a:
+%% -----File: 112.png---Folio 104-------
+\[
+D =
+\begin{vmatrix}
+\sum df^{2} & \sum dl·df\DPtypo{.}{} & \sum l·df \\
+\sum dl·df  & \sum dl^{2} & \sum l\Add{·}dl \\
+\sum l·df   & \sum l\Add{·}dl & \sum l^{2}
+\end{vmatrix}
+= \sum dl^{2}·\sum df^{2} - \left(\sum dl\Add{·}df\right)^{2},
+\]
+d'où
+\[
+\Tag{(3)}
+K = \frac{D}{\sum dl^{2}}.
+\]
+Dans le cas général, on trouve de même
+\[
+\Tag{(4)}
+K = \frac{D · \sum l^{2}}{\sum l^{2} · \sum dl^{2} - \left(\sum l\Add{·} dl\right)^{2}}.
+\]
+$K$~est le \emph{paramètre de distribution}; il est rationnel. La formule~\Eq{(2)}
+montre que, si $M$~se déplace dans une direction quelconque
+sur la génératrice, le plan tangent tourne, par rapport
+à cette direction, dans le sens positif de rotation, si
+$K$~est positif; et tourne dans le sens négatif, si $K$~est négatif.
+
+La signe de~$K$ correspond donc à une propriété géométrique
+de la surface. D'après \Eq{(3)}~ou~\Eq{(4)}, \emph{le paramètre de distribution
+est nul pour une surface développable}.
+
+\Paragraph{Remarque.} Soient sur une même génératrice \Card{2} points $M$\Add{,}~$M'$
+où les plans tangents soient rectangulaires. On a
+\[
+\tg V · \tg V' = -1,
+\]
+d'où, en vertu de~\Eq{\DPtypo{(7)}{(2)}}\Add{,}
+\[
+CM · CM' = -K^{2};
+\]
+\emph{les points d'une génératrice où les plans tangents sont rectangulaires
+forment une involution dont $C$ est le point central}.
+
+\Paragraph{Exemple \1.} \emph{Surface engendrée par les binormales d'une
+courbe gauche\Add{.}}
+
+Soit la courbe
+\[
+x = f(s),\qquad
+y = g(s),\qquad
+z = h(s);
+\]
+%% -----File: 113.png---Folio 105-------
+avec les notations habituelles, nous avons
+\begin{alignat*}{3}
+df &= a\, ds,  &dg &= b\, ds, & dh &= c\, ds,\\
+l &= a'',       &m &= b'',    &  n &= c'',\\
+\intertext{et}
+dl &= \frac{a'}{T}\, ds, \qquad &
+dm &= \frac{b'}{T}\, ds, \qquad &
+dn &= \frac{c'}{T}\, ds.
+\end{alignat*}
+Le point central est ici défini par $u = 0$; la courbe est ligne
+de striction. Le paramètre de distribution est
+\[
+K = T^{2}
+\begin{vmatrix}
+a  &   b   &  c \\
+\mfrac{a'}{T}  &   \mfrac{b'}{T} & \mfrac{c'}{T} \\
+a'' &   b''  &  c''
+\end{vmatrix} = T ;
+\]
+le paramètre de distribution est égal au rayon de torsion de
+la courbe au point correspondant. La courbe est ligne de
+striction, trajectoire orthogonale des génératrices et géodésique.
+
+\Paragraph{Exemple \2.} \emph{Surface engendrée par les normales principales
+à une courbe.}
+On a ici
+\begin{alignat*}{3}
+df &= a\, ds,   &  dg &= b\, ds,    & dh &= c\, ds,\\
+ l &= a',   &  m &= b',   &  n &= c',\\
+dl &= - \frac{a}{R} - \frac{a''}{T}\, ds, \qquad &
+dm &= - \frac{b}{R} - \frac{b''}{T}\, ds, \qquad &
+dn &= - \frac{c}{R} - \frac{c''}{T}\, ds;
+\end{alignat*}
+le point central est défini par
+\[
+u = \frac{\displaystyle\sum a\left(\frac{a}{R} + \frac{a''}{T}\right)}
+         {\displaystyle\sum  \left(\frac{a}{R} + \frac{a''}{T}\right)^{2} }
+  = \frac{\dfrac{1}{R}}{\dfrac{1}{R^{2}} + \dfrac{1}{T^{2}}}
+  = \frac{RT^{2}}{R^{2} + T^{2}} = MC,
+\]
+et on a:
+\[
+K = - \frac{R^{2}T^{2}}{R^{2} + T^{2}}
+\begin{vmatrix}
+a   &   b  &    c \\
+\mfrac{a}{R} + \mfrac{a''}{T} &
+\mfrac{b}{R} + \mfrac{b''}{T} &
+\mfrac{c}{R} + \mfrac{c''}{T} \\
+a'  &   b'  &   c'
+\end{vmatrix} = \frac{R^{2}T}{R^{2} + T^{2}}.
+\]
+%% -----File: 114.png---Folio 106-------
+Cherchons le plan tangent au centre de courbure~$O$. Nous avons
+\[
+\tg V = \frac{CO}{K}
+  = \frac{MO - MC}{K}
+  = \frac{1}{K} \left(R - \frac{RT^{2}}{R^{2} + T^{2}}\right)
+  = \frac{1}{K} · \frac{R^{2}}{R^{2} + T^{2}}
+  = \frac{R}{T};
+\]
+pour le point~$M$, qui est sur la courbe on a
+\[
+\tg V = \frac{CM}{K} = - \frac{T}{R},
+\]
+donc
+\[
+\tg V · \tg V'= - 1\Add{.}
+\]
+Les plans tangents en $M$~et~$O$ sont rectangulaires, ce qui est un
+cas particulier d'une proposition que nous verrons plus loin
+(\No12).
+
+\Section{\DPchg{Elément}{Élément} linéaire.}
+{9.}{} Cherchons l'élément linéaire d'une surface réglée:
+\[
+x = f(v) + u·l(v) \qquad
+y = g(v) + u·m(v) \qquad
+z = h(v) + u·n(v)\Add{.}
+\]
+
+En désignant par des accents les dérivées par rapport à~$v$,
+il vient:
+\[
+dx = (f' + ul')\, dv + l\, du, \quad
+dy = (g' + um')\, dv + m\, du, \quad
+dz = (h' + un')\, dv + n\, du
+\]
+et
+\[
+ds^{2} = E\, du^{2} + 2F\, du\, dv + G\, dv^{2},
+\]
+avec:
+\[
+E = \sum l^{2}, \quad
+F = u \sum ll' + \sum lf', \quad
+G = u^{2} \sum l'{}^{2} + 2u \sum l'f' + \sum f'{}^{2}\Add{.}
+\]
+Supposons que $l\Add{,} m\Add{,} n$ soient les cosinus directeurs:
+\begin{gather*}
+\sum l^{2} = 1, \qquad \sum ll' = 0, \\
+E = 1, \qquad F = \sum lf', \qquad
+G = u^{2} \sum l'{}^{2} + 2u \sum l'f' + \sum f'{}^{2}.
+\end{gather*}
+Ces résultats s'obtiennent directement en faisant le changement
+de paramètre
+\[
+\sqrt{E} · u = u';
+\]
+d'où
+\[
+du' = \sqrt{E} · du + u\, \smash[t]{\frac{\dfrac{dE}{dv}}{2 \sqrt{E}}}\,dv.
+\]
+Nous avons alors, en supprimant les accents,
+\[
+ds^{2} = du^{2} + 2F du\,dv + G\,dv^{2}.
+\]
+%% -----File: 115.png---Folio 107-------
+Supposons de plus que la courbe $x = f(v), y = g(v), z = h(v)$
+soit trajectoire orthogonale des génératrices, alors $\sum lf' = 0$,
+$F = 0$, et on a
+\[
+ds^{2} = du^{2} + G\, dv^{2};
+\]
+il est évident que l'élément linéaire doit avoir cette forme,
+car on a un système de coordonnées orthogonales. On arrive
+aussi à cette expression en posant
+\[
+du + F\, dv = du',
+\]
+d'où
+\[
+u' = u + \int F\, dv\Add{,}
+\]
+ce qui exige une quadrature. La variable~$u$ est définie à une
+constante près, c'est une longueur portée sur chaque génératrice
+à partir de la même trajectoire orthogonale. Pour définir
+la variable~$v$, considérons la direction de la génératrice
+$x = l$, $y = m$, $z = n$. Ces équations sont celles de la trace du
+cône directeur sur la sphère de rayon~$1$; nous prendrons pour~$v$
+l'arc de cette courbe; alors $\sum l'{}^{2} = 1$, et
+\[
+G = u^{2} + 2u \sum l'f' + \sum f'{}^{2}.
+\]
+Posons
+\[
+\sum l'f' = G_{0}, \qquad \sum f'{}^{2} = G_{1},
+\]
+nous avons
+\[
+G = u + 2u G_{0} + G_{1};
+\]
+les quantités $G_{0}\Add{,} G_{1}$, ainsi introduites sont liées d'une façon
+simple au point central et au paramètre de distribution. Considérons
+l'involution des points~$M\Add{,} M'$ où les plans tangents sont
+rectangulaires; son point central est le point central de la
+génératrice, et on a, en désignant par~$K$ le paramètre de distribution,
+\[
+CM \cdot CM'= -K^{2}.
+\]
+Le plan tangent en un point~$u$ de la génératrice a pour coefficients
+%% -----File: 116.png---Folio 108-------
+les déterminants déduits du tableau
+\[
+\begin{Vmatrix}
+a         & b        &  c     \\
+f' + ul'  & g' + um' &  h' + un'
+\end{Vmatrix};
+\]
+de même le plan tangent au point~$u'$ aura pour \DPtypo{cofficients}{coefficients} les
+déterminants déduits du tableau
+\[
+\begin{Vmatrix}
+a         & b         & c       \\
+f' + u'l' & g' + u'm' & h' + u'n'
+\end{Vmatrix}.
+\]
+Exprimons que ces plans tangents sont rectangulaires. La somme
+des produits des déterminants précédents, et par suite le
+produit des tableaux, doit être nul, ce qui donne
+\[
+\begin{vmatrix}
+1 & 0                           \\
+0 & G_{1} + (u + u')G_{0} + uu'
+\end{vmatrix} = 0 ;
+\]
+la relation d'involution est donc
+\[
+uu' + (u + u')G_{0} + G_{1} = 0,
+\]
+ou
+\[
+(u + G_{0})(u' + G_{0}) = G_{0}^{2} - G_{1}.
+\]
+$u + G_{0}$ doit représenter~$CM$; si donc $I$~est l'intersection de la
+génératrice avec la trajectoire orthogonale $u = 0$, on a
+\[
+u + G_{0} = CM = IM - IC;
+\]
+mais $IM = u$, donc $G_{0} = -IC$, $-G_{0}$~est l'$u$ du point central; posons
+\[
+P = -G_{0} = -\sum l'f'.
+\]
+De plus
+\[
+G_{0}^{2} - G_{1} = -K^{2},
+\]
+d'où
+\[
+G_{1} = G_{0}^{2} + K^{2} = P^{2} + K^{2} = \sum f'{}^{2};
+\]
+alors
+\[
+G = u^{2} - 2uP + P^{2} + K^{2} = (u - P)^{2} + K^{2}.
+\]
+Finalement, \emph{si $l, m, n$ sont les cosinus directeurs de la génératrice,
+$v$~l'arc de la trace du cône directeur sur la sphère
+de rayon~$1$, $u$~la longueur portée sur la génératrice à partir
+d'une trajectoire orthogonale, on a}
+%% -----File: 117.png---Folio 109-------
+\[
+\Tag{(1)}
+ds^{2} = du^{2} + \bigl[(u - P)^{2} + K^{2}\bigr]\, dv^{2}\Add{,}
+\]
+\emph{$P$ étant l'$u$ du point central et $K$~le paramètre de distribution\Add{.}}
+
+\Paragraph{Remarque.} Ceci peut servir à calculer le paramètre de
+distribution. On~a
+\[
+\begin{vmatrix}
+f' & g' & h' \\
+l' & m' & n' \\
+l  & m  & n
+\end{vmatrix}^{2} =
+\begin{vmatrix}
+G_{1} & G_{0} & 0 \\
+G_{0} & 1     & 0 \\
+0     & 0     & 1
+\end{vmatrix} = G_{1} - G_{0}^{2} = K^{2},
+\]
+et on peut écrire
+\[
+\Tag{(2)}
+K =
+\begin{vmatrix}
+f' & g' & h' \\
+l' & m' & n' \\
+l  & m  & n
+\end{vmatrix}, \qquad P = - \sum l'f'.
+\]
+
+\emph{Réciproquement}, soit une surface dont l'élément linéaire
+soit de la forme
+\[
+ds^{2} = du^{2} + \bigl[(u - P)^{2} + K^{2}\bigr] dv^{2};
+\]
+cherchons s'il y a des surfaces réglées applicables sur cette
+surface; les éléments d'une telle surface réglée seront déterminés
+par les relations
+\[
+\sum l^{2} = 1, \quad
+\sum lf' = 0, \quad
+\sum l'{}^{2}= 1, \quad
+\sum l'f'= -P, \quad
+\sum f'{}^{2} = K^{2} + P^{2};
+\]
+la dernière de ces relations s'écrit, d'après l'expression de~$K$,
+\[
+\sum f' (mn' - nm') = -K.
+\]
+Nous pouvons d'abord nous donner arbitrairement le cône directeur
+de façon à satisfaire $\sum l^{2} = 1$, $\sum l'{}^{2} = 1$. Il reste alors à
+satisfaire à \Card{3} équations linéaires en $f'\Add{,} g'\Add{,} h'$ dont le déterminant
+n'est pas nul; $f', g', h'$ seront alors parfaitement déterminés,
+$f, g, h$ le seront à une constante additive près, ce qui revient
+à ajouter à $x, y, z$ des quantités constantes, c'est-à-dire à
+faire subir à la surface une translation. \emph{Lorsqu'on a une surface
+réglée, il y a donc une infinité de surfaces réglées applicables
+%% -----File: 118.png---Folio 110-------
+sur elle, les génératrices correspondant aux génératrices}
+puisqu'on peut prendre arbitrairement le cône directeur.
+Remarquons que dans l'élément linéaire figure, non pas~$K$,
+mais~$K^{2}$, de sorte qu'en particulier \emph{il existe \Card{2} surfaces réglées
+ayant même cône directeur, des paramètres de distribution
+égaux et de signes contraires et applicables l'une sur
+l'autre}.
+
+Pour avoir explicitement $f, g, h$, résolvons le système des
+équations linéaires
+\[
+\sum lf' = 0, \qquad \sum l'f' = -P, \qquad \sum (mn' - nm')f' = -K;
+\]
+$l\Add{,} m\Add{,} n$, $l'\Add{,} m'\Add{,} n'$ sont ici cosinus directeurs de \Card{2} directions rectangulaires.
+Introduisons une nouvelle direction de cosinus
+$l_{2}\Add{,} m_{2}\Add{,} n_{2}$ formant avec les \Card{2} précédentes un trièdre trirectangle
+\[
+l_{2}= mn' - nm', \qquad m_{2} = nl' - ln', \qquad n_{2} = lm' - ml'.
+\]
+Le système devient
+\[
+\sum lf' = 0, \qquad \sum l'f' = -P, \qquad \sum l_{2} f'= -K;
+\]
+d'où
+\[
+\Tag{(3)}%[** TN: Added brace]
+\left\{
+\begin{alignedat}{2}
+f' &= -Pl' &&- K (mn' - nm'), \\
+g' &= -Pm' &&- K (nl' - ln'), \\
+h' &= -Pn' &&- K (lm' - ml').
+\end{alignedat}
+\right.
+\]
+On a $f, g, h$ par des quadratures.
+
+\Section{La forme \texorpdfstring{$\Psi$}{Psi} et les lignes asymptotiques.}
+{10.}{} Nous avons
+\[
+\Psi(du, dv) = \sum Ad^{2}x =
+\begin{vmatrix}
+d^{2}x  &  d^{2}y  & d^{2}z \\
+\mfrac{\dd x}{\dd u} & \mfrac{\dd y}{\dd u} & \mfrac{\dd z}{\dd u} \\
+\mfrac{\dd x}{\dd v} & \mfrac{\dd y}{\dd v} & \mfrac{\dd z}{\dd v}
+\end{vmatrix} = \begin{vmatrix}
+(f'' + ul'')\,dv^{2} + 2l'\,du\,dv & \dots \\
+1            & \dots \\
+f' + ul'     & \dots
+\end{vmatrix}
+\]
+on a pour~$\Psi$ une expression de la forme
+\[
+\Psi(du, dv) = 2F'\, du\, dv + G'\, dv^{2},
+\]
+%% -----File: 119.png---Folio 111-------
+$F'$~étant fonction de $v$~et~$G'$ un trinôme du \Ord{2}{e} degré en~$u$. Nous
+trouvons naturellement pour lignes asymptotiques les courbes
+$dv = 0$, $v = \cte$ qui sont les génératrices. Les autres lignes
+asymptotiques sont déterminées par l'équation différentielle
+\[
+\frac{du}{dv} = - \frac{G'}{2F'},
+\]
+ou
+\[
+\Tag{(1)}
+\frac{du}{dv} = Ru^{2} + 2Su + T,
+\]
+$R\Add{,} S\Add{,} T$ étant fonctions de~$v$. C'est une \emph{équation de \DPtypo{Ricatti}{Riccati}}. Rappelons
+ses propriétés.
+
+\ParItem[\DPchg{Equation}{Équation} de \DPtypo{Ricatti}{Riccati}.]{\Primo.} \emph{Supposons qu'on ait une solution}~$u_{1}$, de cette équation.
+Posons
+\[
+\Tag{(2)}
+u = u_{1} + \frac{1}{w},
+\]
+d'où
+\[
+du = du_{1} - \frac{dw}{w^{2}}.
+\]
+L'équation~\Eq{(1)} devient
+\[
+\frac{du_{1}}{dv} - \frac{1}{w^{2}} · \frac{dw}{dv}
+  = Ru_{1}^{2} + 2R \frac{u_{1}}{w} + R \frac{1}{w^{2}}
+  + 2Su_{1} + 2S \frac{1}{w} + T;
+\]
+mais $u_{1}$ étant intégrale de~\Eq{(1)}, on a
+\[
+\frac{du_{1}}{dv} = Ru_{1}^{2} + 2Su_{1} + T,
+\]
+de sorte que l'équation devient
+\[
+-\frac{dw}{dv} = 2(Ru_{1} + S) w + R,
+\]
+ou
+\[
+\Tag{(3)}
+\frac{dw}{dv} = Qw - R.
+\]
+C'est une équation linéaire dont \emph{l'intégration s'effectue par
+\Card{2} quadratures.}
+
+\ParItem{\Secundo.} \emph{Supposons qu'on ait \Card{2} intégrales} $u_{2}\Add{,} u_{1}$ de l'équation.
+Posons
+\[
+u_{2} = u_{1} + \frac{1}{w_{0}},
+\]
+d'où
+\[
+w_{0} = \frac{1}{u_{2} - u_{1}};
+\]
+%% -----File: 120.png---Folio 112-------
+$w_{0}$~sera une intégrale de l'équation~\Eq{(3)}. Posons alors
+\[
+\Tag{(4)}
+w = w_{0} + \theta,
+\]
+d'où
+\[
+dw = dw_{0} + d\theta;
+\]
+\Eq{(3)}~devient
+\[
+\frac{dw_{0}}{dv} + \frac{d\theta}{dv} = Qw_{0} + Q\theta - R;
+\]
+ou, comme $w$~est intégrale de~\Eq{(3)}
+\[
+\frac{d\theta}{dv} = Q\theta,
+\]
+équation linéaire sans \Ord{2}{e} membre qui s'intègre immédiatement
+\emph{par une seule quadrature}:
+\begin{gather*}
+\Tag{(5)}
+\frac{d\theta}{\theta} = Q\, dv, \\
+\log \theta = \int Q\, dv, \\
+\theta = e^{Q\, dv}.
+\end{gather*}
+
+\ParItem{\Tertio.} \emph{Supposons qu'on ait \Card{3} intégrales} $u_{1}\Add{,} u_{2}\Add{,} u_{3}$ de l'équation~\Eq{(1)}.
+On a alors \Card{2} intégrales de l'équation~\Eq{(3)}. Soit
+\[
+w_{1} = \frac{1}{u_{0} - u_{1}};
+\]
+$w_{1}$~est intégrale de~\Eq{(3)}, et par suite on a une intégrale~$\theta_{0}$ de~\Eq{(5)}
+\[
+\theta_{0} = w_{1} - w_{0}
+  = \frac{1}{u_{3} - u_{1}} - \frac{1}{u_{2} - u_{1}}
+  = \frac{u_{2} - u_{3}}{(u_{3} - u_{1}) (u_{2} - u_{1})}.
+\]
+Posons
+\begin{gather*}
+\theta = \theta_{0} \psi, \\
+d\theta = \theta_{0} · d\psi + \psi · d\theta_{0};
+\end{gather*}
+\Eq{(5)}~devient
+\[
+\theta_{0}\, \frac{d\psi}{dv} + \psi · \frac{d\theta_{0}}{dv}
+  = Q \psi\theta_{0},
+\]
+ou, comme $\theta_{0}$~est intégrale de~\Eq{(5)}\Add{,}
+\[%[** TN: Set on two lines in original]
+\theta_{0} \frac{d\psi}{dv} = 0, \qquad\qquad
+\frac{d\psi}{dv} = 0.
+\]
+$\psi$~est une constante~$C$, et l'intégrale générale de~\Eq{(5)} est
+%% -----File: 121.png---Folio 113-------
+\[
+\Tag{(6)}
+\theta = C\, \theta_{0}.
+\]
+\emph{L'équation s'intègre complètement par des opérations algébriques.}
+Si nous cherchons l'expression de l'intégrale générale~$u$
+en fonction des intégrales particulières $u_{1}\Add{,} u_{2}\Add{,} u_{3}$, nous avons,
+en vertu de \Eq{(2)}\Add{,}~\Eq{(4)}\Add{,}~\Eq{(6)},
+\[
+u = u_{1} + \frac{1}{w}
+  = u_{1} + \frac{1}{\dfrac{1}{u_{2} - u_{1}} + \theta}
+  = u_{1} + \frac{1}{\dfrac{1}{u_{2} - u_{1}}
+            + C \dfrac{u_{2} - u_{3}}{(u_{3} - u_{1}) (u_{2} - u_{1})}}\Add{,}
+\]
+d'où
+\[
+\frac{1}{u - u_{1}}
+  = \frac{1}{u_{2} - u_{1}}
+  + \frac{C(u_{2} - u_{3})}{(u_{3} - u_{1})(u_{2} - u_{1})}
+  = \frac{u_{3} - u_{1} + C (u_{2} - u_{3})}{(u_{3} - u_{1}) (u_{2} - u_{1})},
+\]
+d'où
+\begin{gather*}
+C(u_{2} - u_{3})
+  = \frac{(u_{3} - u_{1}) (u_{2} - u_{1})}{u - u_{1}} - (u_{3} - u_{1})
+  = \frac{(u_{3} - u_{1}) (u_{2} - u)}{(u - u_{1})}, \\
+\Tag{(7)}
+C = \frac{u - u_{2}}{u - u_{1}} : \frac{u_{3} - u_{2}}{u_{3} - u_{1}},
+\end{gather*}
+ou
+\[
+\Ratio{u}{u_{1}}{u_{2}}{u_{3}} = C.
+\]
+\emph{Ainsi le rapport anharmonique de \Card{4} intégrales quelconques d'une
+équation de Riccati est constant.} En remarquant que, dans
+le cas présent, ces intégrales sont précisément les~$u$ des
+points d'intersection d'une génératrice quelconque avec les
+asymptotiques, on voit que \emph{\Card{4} lignes asymptotiques coupent les
+génératrices suivant un rapport anharmonique constant.}
+
+\Paragraph{Remarque.} L'équation~\Eq{(7)} résolue par rapport à~$u$ donne
+\[
+\Tag{(8)}
+u = \frac{VC + V_{0}}{V_{1}C + V_{2}}\Add{,}
+\]
+$V, V_{0}, V_{1}\Add{,} V_{2}$ étant fonctions de~$v$. La constante arbitraire figure
+donc dans l'intégrale générale par une fraction du \Ord{1}{e} degré.
+Inversement toute fonction de la forme~\Eq{(8)} satisfait à une
+équation de Riccati, car si on élimine la constante~$C$ au
+moyen d'une différentiation, on retrouve une équation
+%% -----File: 122.png---Folio 114-------
+différentielle de la forme~\Eq{(1)}.
+
+\MarginNote{Cas
+particuliers.}
+Si la surface réglée a une directrice rectiligne, cette
+directrice est une asymptotique, on a une solution particulière
+de l'équation de \DPtypo{Ricatti}{Riccati}. La détermination des lignes asymptotiques
+se fait au moyen de \Card{2} quadratures. C'est le cas
+des \emph{surfaces réglées à plan directeur}. Si la surface admet
+\Card{2} directrices rectilignes, ces \Card{2} droites sont des asymptotiques,
+et on a \Card{2} solutions particulières de l'équation de Riccati.
+C'est le cas des \emph{surfaces conoïdes à plan directeur}. Il ne
+faut plus alors qu'une quadrature pour déterminer les lignes
+asymptotiques. En réalité, on peut les obtenir sans quadrature\Add{.}
+Considérons en effet une surface réglée admettant \Card{2} directrices
+rectilignes. On peut effectuer une transformation homographique
+de façon que l'une des directrices s'en aille à l'infini,
+la surface se transforme en un conoïde à plan directeur.
+Soit
+\[
+z = \phi \left(\frac{y}{x}\right)
+\]
+l'équation d'un tel conoïde. Posons
+\[
+x = u, \qquad
+y = uv, \qquad
+z = \Err{(v) \phi}{\phi(v)};
+\]
+les lignes asymptotiques sont telles que le plan osculateur
+\DPtypo{coincide}{coïncide} avec le plan tangent; ses coefficients doivent donc
+satisfaire aux relations
+\[
+A \frac{\dd x}{\dd u} + B \frac{\dd y}{\dd u} + C \frac{\dd z}{\dd u} = 0,\qquad
+A \frac{\dd x}{\dd v} + B \frac{\dd y}{\dd v} + C \frac{\dd z}{\dd v} = 0,
+\]
+ou
+\[
+A + B v = 0, \qquad
+B u + C \phi'(v) = 0;
+\]
+équations satisfaites si l'on prend $C = - u$, $B = \phi'(v)$,
+$A = -v\phi'(v)$. On a alors
+%% -----File: 123.png---Folio 115-------
+\[
+\Psi(du, dv) = A\,d^{2} x + B\, d^{2} y + C\,d^{2} z = 0;
+\]
+mais $A\Add{,} B\Add{,} C$~étant les coefficients du plan tangent, on a
+\[
+A\, dx + B\, dy + C\, dz = 0;
+\]
+en différentiant cette relation, on voit qu'on peut mettre
+l'équation différentielle des lignes asymptotiques sous la
+forme
+\[
+dA·dx + dB·dy + dC·dz = 0,
+\]
+ou
+\begin{gather*}
+-du\bigl[\phi'(v)\,dv + v\phi''(v)\, dv\bigr]
+  + \phi''(v)\, dv (v\, du + u\, dv) - du · \phi'(v) · dv = 0, \\
+u\phi''(v) · dv^{2} - 2\phi'(v)\, du\, dv = 0;
+\end{gather*}
+nous trouvons la solution $v = \cte$ qui nous donne les génératrices,
+et il reste
+\[
+\frac{\phi''(v)\, dv}{\phi '(v)} = \frac{2\, du}{u},
+\]
+d'où
+\[
+L\phi'(v) = Lu^{2} - LC,
+\]
+d'où
+\[
+u^{2} = \DPtypo{C\phi'(v)}{C + \phi'(v)};
+\]
+\emph{on a ainsi les lignes asymptotiques d'un conoïde sans quadrature}.
+
+\Paragraph{Remarque.} S'il y a trois directrices rectilignes, la
+surface est une surface du second degré, et est doublement réglée.
+
+\MarginNote{Calcul de $\Psi$.}
+Cherchons l'expression générale de la forme~$\Psi$. Introduisons
+pour cela les variables \DPtypo{Canoniques}{canoniques}~$u\Add{,} v$ qui nous ont permis
+d'arriver à la forme de l'élément linéaire. Considérons le
+trièdre de Serret de la courbe~$(\Sigma)$ \DPtypo{trace}{tracé} du cône directeur sur
+la sphère de rayon~$1$. La génératrice~$(l\Add{,} m\Add{,} n)$ est dans le plan
+normal à cette courbe: soit $\theta$~son angle avec la normale principale;
+avec les notations habituelles, nous avons:
+%% -----File: 124.png---Folio 116-------
+\[%[** TN: Set on separate lines in original]
+l = a' \cos\theta + a'' \sin\theta, \qquad
+m = b' \cos\theta + b'' \sin\theta, \qquad
+n = c' \cos\theta + c'' \sin\theta;
+\]
+d'où
+\[
+l' = a
+  = \theta'(-a' \sin\theta + a'' \cos\theta)
+  - \cos\theta \left(\frac{a}{R} + \frac{a''}{T}\right)
+  + \sin\theta · \frac{a'}{T},
+\]
+et les analogues; %[** TN: Omitted newline]
+d'où
+\[
+\frac{\Cos\theta}{R} = - 1, \qquad
+\theta' = \frac{1}{T}.
+\]
+Nous avons alors
+\begin{alignat*}{4}
+mn' - nm' = &mc &&- nb &&= a' \sin \theta &&- a'' \cos \theta, \\
+&nl' &&- ln' &&= b' \sin \theta &&- b'' \cos \theta, \\
+&lm' &&- ml' &&= c' \sin \theta &&- c'' \cos \theta;
+\end{alignat*}
+et nous obtenons, au moyen des formules~\Eq{(3)} du \No9,
+\begin{alignat*}{2}%[** TN: Completed last two equations]
+f' + ul' &= (u - P)l' - K(mn'- nm')
+  &&= (u - P)a - K \sin \theta · a' + K \cos \theta · a'', \\
+g' + um' &= (u - P)m' - K(nl'- ln')
+  &&= (u - P)b - K \sin \theta · b' + K \cos \theta · b'', \\
+h' + un' &= (u - P)n' - K(lm'- ml')
+  &&= (u - P)c - K \sin \theta · c' + K \cos \theta · c'',
+\end{alignat*}
+puis
+\begin{multline*}
+f'' + ul''\DPtypo{)}{}
+  = - P'a + \frac{u - P}{R} a' - K' \sin\theta · a'
+  - \frac{K \cos \theta}{T} a'
+  + K \sin \theta \left(\frac{a}{R} + \frac{a''}{T}\right) \\
+  + K'\cos \theta · a'' - \frac{K \sin \theta}{T} a''
+  + K · \cos \theta · \frac{a'}{T},
+\end{multline*}
+ou
+\begin{alignat*}{4}%[** TN: Completed last two equations, fixed typos in original]
+f'' &+ ul'' &&= a \left(\frac{K \sin \theta}{R} - P'\right)
+  &&+ a' \left(\frac{u - P}{R} - K' \sin \theta\right)
+  &&+ a'' · K'\cos \theta, \\
+g'' &+ um'' &&= b \left(\frac{K \sin \theta}{R} - P'\right)
+  &&+ b' \left(\frac{u - P}{R} - K' \sin \theta\right)
+  &&+ b'' · K'\cos \theta, \\
+h'' &+ un'' &&= c \left(\frac{K \sin \theta}{R} - P'\right)
+  &&+ c' \left(\frac{u - P}{R} - K' \sin \theta\right)
+  &&+ c'' · K'\cos \theta,
+\end{alignat*}
+Alors
+\[
+\Psi = \begin{vmatrix}
+  \left[
+    \begin{aligned}
+      2a · du\, dv
+      &+ dv^{2} \biggl[a \left(\mfrac{K \sin \theta}{R} - P'\right) \\
+      &\quad+ a' \left(\mfrac{u - P}{R} - K' \sin \theta\right)
+       + a'' · K'\cos \theta\biggr]
+    \end{aligned}
+  \right] & \dots & \dots \\
+  \vphantom{\bigg|}
+  (u - P) a - K \sin \theta · a' + K \cos \theta · a'' & \dots & \dots \\
+  a' \cos \theta + a'' \sin \theta & \dots & \dots
+\end{vmatrix},
+\]
+%% -----File: 125.png---Folio 117-------
+ou
+\[
+\Psi = \begin{vmatrix}
+2\, du\, dv + \left(\mfrac{K \sin \theta}{R} - P'\right)\, dv^{2}
+  &  \left(\mfrac{u - P}{R} - K' \sin \theta\right)\, dv^{2} &  K' \cos \theta · dv^{2} \\
+u - P & - K \sin\theta & K \cos\theta \\
+0     &     \Cos\theta &   \sin\theta
+\end{vmatrix}\Add{,}
+\]
+c'est-à-dire
+\[
+\Psi = -\left[2\, du\, dv
+  + \left(\frac{K \sin \theta}{R} - P'\right) dv^{2}\right] K
+  - (u - P) \left[- K' + (u-P) \frac{\sin \theta}{R}\right] dv^{2},
+\]
+ou enfin
+\[
+\Psi = - 2K · du\, dv + \left\{(u - P) K' + KP'
+  - \frac{\sin \theta}{R} \left[(u - P)^{2} + K^{2}\right]\right\} dv^{2}.
+\]
+Le seul élément nouveau qui intervient est la courbure géodésique
+$\dfrac{\sin \theta}{R}$ de la courbe~$(\Sigma)$ sur la sphère. Cet élément suffit
+à déterminer~$(\Sigma)$; supposons en effet
+\[
+\frac{\sin \theta}{R} = \Phi (v);
+\]
+nous avons
+\[
+\frac{\Cos \theta}{R} = - 1, \qquad
+\frac{1}{T} = \theta';
+\]
+nous en déduisons
+\[
+\Tag{(1)}
+\tg \theta = - \Phi (v), \qquad
+R = - \Cos \theta, \qquad
+T = \frac{dv}{d \theta};
+\]
+nous avons ainsi tous les éléments de la courbe~$(\Sigma)$.
+
+\Paragraph{Remarque.} Les formules~(\DPtypo{'}{1}) nous permettent de trouver
+la condition pour qu'une courbe soit tracée sur la sphère de
+rayon~$1$. Nous avons en effet
+\[
+\frac{dR}{dv} = +\sin\theta · \frac{d\theta}{dv} = \frac{\sin\theta}{R},
+\]
+d'où
+\[
+R^{2} + T^{2} \left(\frac{dR}{dv}\right)^{2} = 1;
+\]
+%% -----File: 126.png---Folio 118-------
+Ce qui exprime que le rayon de la sphère osculatrice est égal
+à~$1$.
+
+\Section{Lignes de courbure.}
+{11.}{} L'équation différentielle des lignes de courbure est
+\[
+\begin{vmatrix}
+\mfrac{\dd ds^{2}}{\dd\, du} & \mfrac{\dd ds^{2}}{\dd\, dv} \\
+\mfrac{\dd \Psi}{\dd\, du}   & \mfrac{\dd \Psi}{\dd\, dv}
+\end{vmatrix} = 0,
+\]
+ou
+\[
+\begin{vmatrix}
+du     & \bigl[(u - P)^{2} + K^{2}\bigr] dv \\
+- K\, dv &
+- K\, du + \left[(u - P) K' + KP'
+    - \mfrac{\sin\theta}{R} \bigl[(u - P)^{2} + K^{2}\bigr]\DPtypo{}{\right]}dv
+\end{vmatrix} = 0,
+\]
+c'est-à-dire
+\begin{multline*}
+- K du^{2} + \Bigl[(u - P)K' + KP'
+  - \Phi \bigl[(u - P)^{2} + K^{2}\bigr]\Bigr]du\,dv \\ % +
+  + K \left[(u - P)^{2} + K^{2}\right]dv^{2} = 0.
+\end{multline*}
+Telle est l'équation différentielle des lignes de courbure,
+où $\Phi$~représente la courbure géodésique de la courbe~$(\Sigma)$.
+
+\Section{Centre de courbure géodésique.}
+{12.}{} Considérons une trajectoire orthogonale des génératrices,
+par exemple $u = 0$,
+\[
+x = f(v), \qquad
+y = g(v), \qquad
+z = h(v);
+\]
+cherchons son centre de courbure géodésique. C'est le point où
+la droite polaire rencontre le plan tangent. Or\Add{,} la génératrice
+étant normale à sa trajectoire orthogonale\DPtypo{,}{} est l'intersection
+du plan normal et du plan tangent; \emph{le centre de courbure
+géodésique est donc à l'intersection de la droite polaire avec
+la génératrice}. Le plan normal est
+\[
+\sum (x - f) f' = 0;
+\]
+la caractéristique est définie par l'équation précédente et
+%% -----File: 127.png---Folio 119-------
+par
+\[
+\sum (x - f) f'' - \sum f'{}^{2} = 0.
+\]
+Pour déterminer le centre de courbure géodésique, il suffit de
+déterminer l'$u$~du point d'intersection de la droite précédente
+avec la génératrice
+\[
+x = f(v) + ul(v), \qquad
+y = g(v) + um(v), \qquad
+z = h(v) + un(v).
+\]
+La \Ord{1}{ère} équation se réduit à une identité, la \Ord{2}{e} donne
+\[
+u \sum lf'' - \sum f'{}^{2} = 0;
+\]
+mais on a
+\[
+\sum lf' = 0,
+\]
+d'où
+\[
+\sum l'f' + \sum lf'' = 0;
+\]
+et l'équation qui donne l'$u$~du point cherché devient
+\[
+u \sum l'f' + \sum f'{}^{2} = 0,
+\]
+ou
+\[
+- u P + P^{2} + K^{2} = 0;
+\]
+ou enfin:
+\[
+P (u - P) = K^{2}.
+\]
+\DPtypo{si}{Si} $C$~est le point central, $M$~le point considéré sur la trajectoire
+orthogonale, $M'$~le centre de courbure géodésique, l'équation
+précédente donne
+\[
+CM \cdot CM' = - K^{2}.
+\]
+Donc les plans tangents en $M\Add{,} M'$ sont rectangulaires. Ainsi le
+\emph{centre de courbure géodésique en un point~$M$ d'une trajectoire
+orthogonale des génératrices d'une surface réglée est le point
+de la génératrice où le plan tangent est perpendiculaire au
+plan tangent en~$M$}.
+
+Si nous considérons maintenant une courbe~$(\DPtypo{C}{c})$ tracée sur
+une surface quelconque~$(S)$ les normales à~$(c)$ tangentes à~$(\DPtypo{s}{S})$
+engendrent une surface réglée~$(\Sigma)$; les surfaces $(\DPtypo{s}{S})\Add{,} (\Sigma)$ étant
+tangentes tout le long de~$(c)$, la courbe~$(c)$ a même centre de
+%% -----File: 128.png---Folio 120-------
+courbure géodésique sur~$(\DPtypo{s}{S})$ et sur~$(\Sigma)$; ce qui permet de construire
+le centre de courbure géodésique d'une courbe tracée
+sur une surface quelconque.
+
+
+\ExSection{V}
+
+\begin{Exercises}
+\item[21.] Trouver les points de contact des plans isotropes \DPtypo{menes}{menés} par
+une génératrice quelconque d'une surface réglée. Quelles relations
+ont-ils avec le point central et le paramètre de
+distribution?
+
+\item[22.] Trouver les surfaces réglées dont les lignes asymptotiques
+interceptent sur les génératrices des segments égaux.
+
+\item[23.] Trouver les surfaces réglées dont les lignes de courbure
+interceptent sur les génératrices des segments égaux.
+
+\item[24.] Trouver les surfaces réglées dont les rayons de courbure
+principaux sont fonctions l'un de l'autre.
+
+\item[25.] Trouver les lignes de courbure et les lignes géodésiques de
+l'\DPchg{hélicoide}{hélicoïde} développable.
+
+\item[26.] Montrer que les lignes d'une surface~$(S)$ quelconque, pour
+lesquelles: $ds - R_{g}\, d\phi = 0$, sont caractérisées par cette
+propriété que, si l'on mène par chacun des points de l'une
+% [** TN: Regularized, "constant" in original]
+d'elles une tangente à la courbe $v = \const.$, la surface réglée
+ainsi obtenue a pour ligne de striction la courbe considérée
+(voir \hyperref[exercice11]{exercice~11}).
+
+\item[27.] \DPchg{Etant}{Étant} donnée une surface~$S$ et une courbe~$C$ de cette surface,
+on considère la surface \DPtypo{reglée}{réglée}~$G$ engendrée par les normales~$MN$
+menées à~$S$ aux divers points $M$~de~$C$. Le point central de~$MN$
+s'appelle le \emph{métacentre} de~$S$, correspondant au point~$M$ et à
+la tangente~$MT$ de~$C$.
+
+%[** TN: Regularized formatting of parts]
+\Primo. Déterminer ce métacentre, le plan
+asymptote, le paramètre de distribution. Discuter la variation
+du métacentre quand la courbe~$(C)$ varie, en passant toujours
+en~$M$.
+
+\Secundo. Montrer que le métacentre est le centre de
+courbure de la section droite du cylindre circonscrit à~$S$, et
+dont les génératrices sont perpendiculaires au plan asymptote
+de~$G$.
+
+\Tertio. On suppose qu'on ait plusieurs surfaces~$S$, et que
+l'on affecte chacune d'elles d'un coefficient \DPtypo{numerique}{numérique}~$a$.
+On considère comme homologues sur ces diverses surfaces les
+points~$M$ (pris un sur chaque surface) pour lesquels les plans
+tangents à ces diverses surfaces sont parallèles; soit~$M_{0}$ le
+centre des moyennes distances d'un tel système de points~$M$
+homologues, et relatif au système des coefficients~$a$. Soit~$S_{0}$
+la surface lieu des points~$M_{0}$. Montrer qu'elle correspond
+à chacune des surfaces~$S$ par plans tangents parallèles; et que
+si~$I_{0}$ est le métacentre de~$S_{0}$ correspondant aux divers métacentres~$I$
+des surfaces~$S$ qui se trouvent associés dans la
+correspondance \DPtypo{considerée}{considérée}, on a $(\sum a) · M_{0} I_{0} = \sum (a · MI)$.
+
+\item[28.] On donne une courbe gauche~$R$, arête de rebroussement d'une
+développable~$\Delta$. Déterminer toutes les surfaces réglées satisfaisant
+aux conditions suivantes: chacune des génératrices~$G$
+d'une telle surface est perpendiculaire à un plan tangent~$P$ de~$\Delta$,
+et le point de rencontre de~$G$ et de~$P$ est le point central
+de~$G$. Soit alors~$\Sigma$ l'une de ces surfaces réglées, chacun des
+plans isotropes passant par une de ses génératrices enveloppe
+une développable. Montrer que le lieu des milieux des segments
+dont les extrémités décrivent, indépendamment l'un de
+l'autre, les arêtes de rebroussement de ces deux développables
+est une surface minima inscrite dans~$\Delta$.
+\end{Exercises}
+%% -----File: 129.png---Folio 121-------
+
+
+\Chapitre{VI}{Congruences de Droites.}
+
+\Section{Points et plans focaux.}
+{1.}{} On appelle \emph{congruence} un ensemble de droites dépendant
+de \Card{2} paramètres; toutes les droites rencontrant \Card{2} droites
+fixes constituent une congruence; de même les droites passant
+par un point fixe, les normales à une surface; si sur une
+surface on considère une famille de courbes dépendant d'un paramètre,
+l'ensemble de leurs tangentes constitue une congruence.
+
+Une droite d'une congruence pourra se représenter par les
+équations
+\[%[** TN: Set on one line in original; added brace]
+\Tag{(1)}
+\left\{
+\begin{alignedat}{2}
+x &= f(v, w) &&+ u · a(v, w), \\
+y &= g(v, w) &&+ u · b(v, w), \\
+z &= h(v, w) &&+ u · c(v, w).
+\end{alignedat}
+\right.
+\]
+Les équations
+\[
+\Tag{(2)}
+x = f(v, w), \qquad
+y = g(v, w), \qquad
+z = h(v, w)
+\]
+définissent le \emph{support} de la congruence, $a, b, c$ définissent les
+directions des droites de la congruence passant par chaque
+point du support. Ce support sera en général une surface, et
+la congruence sera constituée par les droites de directions
+données passant par tous les points d'une surface. Il peut
+arriver que $f, g, h$ ne dépendent que d'un seul paramètre, le
+support est alors une courbe, et par tout point de la courbe
+passent une infinité de droites de la congruence, qui constituent
+un cône. Enfin $f, g, h$ peuvent se réduire à des constantes,
+et la congruence est constituée par toutes les droites
+%% -----File: 130.png---Folio 122-------
+passant par le point fixe de coordonnées~$f, g, h$.
+
+Supposons qu'on établisse une relation entre~$v, w$; cela
+revient à choisir $\infty^{1}$~droites de la congruence, qui constituent
+une surface réglée de la congruence. On retrouverait ainsi les
+équations générales d'une surface réglée. Considérons toutes
+les surfaces réglées passant par une droite~$D$ de la congruence\Add{.}
+Deux de ces surfaces se raccordent en \Card{2} points de la droite~$D$.
+Nous allons montrer que ces \Card{2} points sont indépendants des
+surfaces réglées que l'on considère. En d'autres termes \emph{sur
+chaque droite~$D$ de la congruence il existe \Card{2} points $F, F'$ auxquels
+correspondent \Card{2} plans $P, P'$ passant par la droite~$D$ et
+tels que toutes les surfaces réglées de la congruence passant
+par la droite~$D$ ont pour plans tangents en~$F, F'$ respectivement
+les plans~$P, P'$}. Ces points~$F, F'$ s'appellent \emph{foyers} ou \emph{points focaux}
+de la droite~$D$, les plans~$P, P'$ sont les \emph{plans focaux} associés
+à~$F, F'$. Pour démontrer la proposition, cherchons le plan
+tangent en un point quelconque de la génératrice~\Eq{(1)}. Les paramètres
+$A\Add{,} B\Add{,} C$ du plan tangent satisfont aux équations
+\begin{gather*}
+\Tag{(3)}
+Aa + Bb + Cc = 0, \\
+\Tag{(3')}
+A(df + u\,da) + B(dg + u\,db) + C(dh + u\,dc) = 0.
+\end{gather*}
+On peut choisir~$u$ de façon que le plan tangent soit indépendant
+des différentielles $dv, dw$, et par suite indépendant de la relation
+existant entre~$v\Add{,} w$, c'est-à-dire indépendant de la surface
+réglée. Développons la \Ord{2}{e} équation~\Eq{(3)}
+\begin{alignat*}{5}
+0 = &\biggl[
+  &&  A\left(\frac{\dd f}{\dd v} + u \frac{\dd a}{\dd v}\right)
+  &&+ B\left(\frac{\dd g}{\dd v} + u \frac{\dd b}{\dd v}\right)
+  &&+ C\left(\frac{\dd h}{\dd v} + u \frac{\dd c}{\dd v}\right)
+  &&\biggr] dv \\
+  + &\biggl[
+  && A\left(\frac{\dd f}{\dd w} + u \frac{\dd a}{\dd w}\right)
+  &&+ B\left(\frac{\dd g}{\dd w} + u \frac{\dd b}{\dd w}\right)
+  &&+ C\left(\frac{\dd h}{\dd w} + u \frac{\dd c}{\dd w}\right)
+  &&\biggr] dw.
+\end{alignat*}
+%% -----File: 131.png---Folio 123-------
+Pour que le plan tangent soit indépendant de $dv, dw$, il suffit
+que l'on ait
+\[
+\Tag{(4)}
+\left\{
+\begin{alignedat}{4}
+  &   A\left(\frac{\dd f}{\dd v} + u\frac{\dd a}{\dd v}\right)
+  &&+ B\left(\frac{\dd g}{\dd v} + u\frac{\dd b}{\dd v}\right)
+  &&+ C\left(\frac{\dd h}{\dd v} + u\frac{\dd c}{\dd v}\right)
+  &&= 0 \\
+  &   A\left(\frac{\dd f}{\dd w} + u\frac{\dd a}{\dd w}\right)
+  &&+ B\left(\frac{\dd g}{\dd w} + u\frac{\dd b}{\dd w}\right)
+  &&+ C\left(\frac{\dd h}{\dd w} + u\frac{\dd c}{\dd w}\right)
+  &&= 0
+\end{alignedat}
+\right.
+\]
+Les relations~\Eq{(4)} et la \Ord[f]{1}{e} des relations~\Eq{(3)} doivent être
+satisfaites pour des valeurs non toutes nulles de~$A\Add{,} B\Add{,} C$, donc
+on doit avoir
+\[
+\Tag{(5)}
+\begin{vmatrix}
+a    &   b   &    c  \\
+\mfrac{\dd f}{\dd v} + u\mfrac{\dd a}{\dd v} &
+\mfrac{\dd g}{\dd v} + u\mfrac{\dd b}{\dd v} &
+\mfrac{\dd h}{\dd v} + u\mfrac{\dd c}{\dd v} \\
+\mfrac{\dd f}{\dd w} + u\mfrac{\dd a}{\dd w} &
+\mfrac{\dd g}{\dd w} + u\mfrac{\dd b}{\dd w} &
+\mfrac{\dd h}{\dd w} + u\mfrac{\dd c}{\dd w}
+\end{vmatrix} = 0.
+\]
+Telle est l'équation qui donne les~$u$ des points focaux; elle
+est du \Ord{2}{e} degré, donc il y a \Card{2} points focaux; le plan focal
+correspondant à chacun d'eux aura pour coefficients les valeurs
+de~$A, B, C$ satisfaisant aux équations \Eq{(3)}~et~\Eq{(4)}.
+
+\MarginNote{Surfaces focales.
+Courbes focales.}
+Le lieu des foyers s'obtiendra sans difficulté. Il
+suffit de tirer~$u$ de~\Eq{(5)} et de porter sa valeur dans~\Eq{(1)}.
+L'équation~\Eq{(5)} étant du \Ord{2}{e} degré donne pour~$u$ \Card{2} valeurs, de
+sorte que le lieu se compose de \Card{2} parties distinctes dans le
+voisinage de la droite~$D$. Considérons l'une de ces parties;
+elle peut être une surface, que l'on appellera \emph{surface focale},
+ou une courbe, que l'on appellera \emph{courbe focale}, ou bien elle
+peut se réduire à un point, et la congruence comprend alors
+toutes les droites passant par ce point. En écartant ce cas,
+on voit que le lieu des foyers peut se composer de \Card{2} surfaces,
+d'une courbe et d'une surface, ou de deux courbes.
+
+\ParItem{\Primo.} Supposons qu'une portion du lieu des foyers soit une
+%% -----File: 132.png---Folio 124-------
+surface~$(\Phi)$. Prenons cette surface comme support de la congruence;
+l'équation~\Eq{(5)} a pour racine $u = 0$, on a donc
+\[
+\begin{vmatrix}
+a                    & b                    & c  \\
+\mfrac{\dd f}{\dd v} & \mfrac{\dd g}{\dd v} & \mfrac{\dd h}{\dd v}  \\
+\mfrac{\dd f}{\dd w} & \mfrac{\dd g}{\dd w} & \mfrac{\dd h}{\dd w}
+\end{vmatrix} = 0.
+\]
+Ceci exprime que la droite~$D$ est dans le plan tangent à la
+surface $(\Phi)$ au point~$M$ ($u = 0$), qui est l'un des foyers, soit~$F$.
+\emph{Ainsi les droites de la congruence sont tangentes à la surface
+focale au foyer correspondant.} Cherchons le plan focal
+correspondant à~$F$. Ses coefficients $A\Add{,} B\Add{,} C$ sont déterminés par
+les équations
+\[
+\left\{
+\begin{aligned}
+Aa + Bb + Cc = 0\Add{,}& \\
+\begin{alignedat}{4}
+&A \mfrac{\dd f}{\dd v} &&+ B \mfrac{\dd g}{\dd v}
+  &&+ C \mfrac{\dd h}{\dd v} &&= 0\Add{,} \\
+&A \mfrac{\dd f}{\dd w} &&+ B \mfrac{\dd g}{\dd w}
+  &&+ C \mfrac{\dd h}{\dd w} &&= 0\Add{,}
+\end{alignedat}&
+\end{aligned}
+\right.
+\]
+d'après la condition précédemment écrite, ces équations se réduisent
+à~$2$, et expriment que \emph{le plan focal correspondant au
+foyer~$F$ est le plan tangent en~$F$ à la surface~$(\Phi)$. Toutes les
+surfaces réglées de la congruence sont circonscrites à la surface
+focale}.
+
+\emph{Si le lieu des foyers $F, F'$ comprend deux surfaces focales~$(\Phi)\Add{,} (\Phi')$,
+les droites de la congruence sont tangentes aux
+\Card{2} surfaces focales, les foyers~$F, F'$ sont les points de contact,
+les plans focaux sont les plans tangents aux surfaces focales
+aux foyers correspondants. Le lieu des foyers coïncide avec
+l'enveloppe des plans focaux.}
+%% -----File: 133.png---Folio 125-------
+
+\emph{Réciproquement}, étant données \Card{2} surfaces quelconques $(\Phi)\Add{,}
+(\Phi')$, leurs tangentes communes dépendent de \Card{2} paramètres. Soit~$F$
+un point de~$(\Phi)$. Considérons le plan tangent en~$F$ à~$(\Phi)$; il
+coupe~$(\Phi')$ suivant une certaine courbe; si nous menons de~$F$ des
+tangentes à cette courbe, ces droites, qui sont tangentes aux
+\Card{2} surfaces $(\Phi)\Add{,} (\Phi')$ sont déterminées quand le point~$F$ est déterminé;
+elles dépendent d'autant de paramètres que le point~$F$,
+donc de \Card{2} paramètres; elles constituent une congruence,
+dont les surfaces réglées sont circonscrites aux surfaces $(\Phi)\Add{,}
+(\Phi')$ qui sont les surfaces focales.
+
+Si les surfaces $(\Phi)\Add{,} (\Phi')$ constituent \Card{2} nappes d'une même
+surface~$(S)$, comme cela arrive en général, la congruence sera
+constituée par les tangentes doubles de la surface~$(S)$.
+
+\ParItem{\Secundo.} Supposons qu'une portion du lieu des foyers soit une
+courbe~$(\phi)$, que nous prendrons pour support de la congruence.
+Nous pouvons alors supposer que $f\Add{,} g\Add{,} h$ ne dépendent que d'un paramètre,
+$v$~par exemple; alors $\dfrac{\dd f}{\dd w}, \dfrac{\dd g}{\dd w}, \dfrac{\dd h}{\dd w}$ sont nuls, et $u = 0$ est
+racine de l'équation~\Eq{(5)}. \emph{Si les droites d'une congruence rencontrent
+une courbe fixe, les points de cette courbe sont des
+foyers pour les droites de la congruence qui y passent.} Cherchons
+le plan focal correspondant. Nous avons
+\[
+\left\{
+\begin{aligned}
+Aa + Bb + Cc &= 0\Add{,} \\
+A \frac{\dd f}{\dd v} + B \frac{\dd g}{\dd v} + C \frac{\dd h}{\dd v} &= 0\Add{,}
+\end{aligned}
+\right.
+\]
+\emph{le plan focal passe par la droite~$D$ et est tangent à la courbe
+focale. Toutes les surfaces réglées de la congruence passent
+par la courbe focale, et en un point~$M$ de cette courbe sont
+tangentes au plan tangent à cette courbe passant par la droite~$D$.}
+%% -----File: 134.png---Folio 126-------
+
+Supposons qu'il y ait une surface focale~$(\Phi)$ et une
+courbe focale~$(\phi')$; \emph{la congruence est constituée par les droites
+rencontrant~$(\phi')$ et tangentes à~$(\Phi)$}. On a immédiatement les
+foyers et les plans focaux, d'après ce qui précède. \emph{Réciproquement,
+les droites rencontrant une courbe~$(\phi')$ et tangentes
+à une surface~$(\Phi)$ constituent une congruence qui admet $(\phi')$~et~$(\Phi)$
+pour lieu de ses foyers.}
+
+Supposons qu'il y ait \Card{2} courbes focales $(\phi)\Add{,} (\phi')$. \emph{La
+congruence est constituée par les droites rencontrant $(\phi)\Add{,} (\phi')$,
+et ses surfaces réglées contiennent les \Card{2} courbes focales. Réciproquement
+les droites rencontrant \Card{2} courbes données constituent
+une congruence qui admet ces \Card{2} courbes comme courbes focales.}
+Si $(\phi)\Add{,} (\phi')$ constituent \Card{2} parties d'une même courbe~$(c)$,
+la congruence est constituée par les droites rencontrant~$(c)$
+en \Card{2} points, c'est-à-dire les cordes de~$(c)$.
+
+\MarginNote{Cas singuliers.}
+Voyons dans quels cas les \Card{2} foyers sont confondus sur
+toutes les droites de la congruence.
+
+Examinons d'abord le cas de \Card{2} surfaces focales confondues\Add{.}
+Prenons cette surface~$(\Phi)$ comme support; en chaque point~$F$ de
+cette surface est tangente une droite~$D$ de la congruence. Si
+on considère ces points focaux et les droites correspondantes,
+on peut trouver sur la surface une famille de courbes tangentes
+en chacun de leurs points à la droite correspondante de la
+congruence. Soit la droite~$D$, elle est tangente à la surface,
+donc ses coefficients directeurs sont:
+\[
+a = \lambda \frac{\dd f}{\dd v} + \mu \frac{\dd f}{\dd w}, \qquad
+b = \lambda \frac{\dd g}{\dd v} + \mu \frac{\dd g}{\dd w}, \qquad
+c = \lambda \frac{\dd h}{\dd v} + \mu \frac{\dd h}{\dd w}\Add{.}
+\]
+%% -----File: 135.png---Folio 127-------
+Soit une courbe de la surface~$(\Phi)$ définie en exprimant~$v\Add{,} w$ en
+fonction d'un paramètre; les coefficients directeurs de la
+tangente sont
+\[
+dx = \frac{\dd f}{\dd v} · dv + \frac{\dd f}{\dd w} · dw, \quad
+dy = \frac{\dd g}{\dd v} · dv + \frac{\dd g}{\dd w} · dw, \quad
+dz = \frac{\dd h}{\dd v} · dv + \frac{\dd h}{\dd w} · dw;
+\]
+pour que cette tangente soit la droite~$D$, il faut que l'on
+ait
+\[
+\frac{dv}{\lambda} = \frac{dw}{\mu}.
+\]
+Pour déterminer l'un des paramètres~$v\Add{,} w$ en fonction de l'autre,
+on a à intégrer une équation différentielle du \Ord{1}{er} ordre. La
+famille de courbes dépend d'un paramètre, soit $w = \cte[]$. On
+aura alors
+\[
+a = \frac{\dd f}{\dd v}, \qquad
+b = \frac{\dd g}{\dd v}, \qquad
+c = \frac{\dd h}{\dd v};
+\]
+et \Eq{(5)}~devient
+\[
+\begin{vmatrix}
+\mfrac{\dd f}{\dd v} &
+\mfrac{\dd g}{\dd v} &
+\mfrac{\dd h}{\dd v} \\
+\mfrac{\dd f}{\dd v} + u \mfrac{\dd^{2} f}{\dd v^{2}} &
+\mfrac{\dd g}{\dd v} + u \mfrac{\dd^{2} g}{\dd v^{2}} &
+\mfrac{\dd h}{\dd v} + u \mfrac{\dd^{2} h}{\dd v^{2}} \\
+\mfrac{\dd f}{\dd w} + u \mfrac{\dd^{2} f}{\dd v \dd w} &
+\mfrac{\dd g}{\dd w} + u \mfrac{\dd^{2} g}{\dd v \dd w} &
+\mfrac{\dd h}{\dd w} + u \mfrac{\dd^{2} h}{\dd v \dd w}
+\end{vmatrix} = 0.
+\]
+\DPtypo{en}{En} retranchant la \Ord{1}{ère} ligne de la \Ord{2}{e}, $u$~vient en facteur:
+pour que les points focaux soient confondus, il faut que le
+déterminant s'annule encore pour $u = 0$, ce qui donne
+\[
+\begin{vmatrix}
+\mfrac{\dd f}{\dd v} &
+\mfrac{\dd g}{\dd v} &
+\mfrac{\dd h}{\dd v} \\
+\mfrac{\dd^{2} f}{\dd v^{2}} &
+\mfrac{\dd^{2} g}{\dd v^{2}} &
+\mfrac{\dd^{2} h}{\dd v^{2}} \\
+\mfrac{\dd f}{\dd w} &
+\mfrac{\dd g}{\dd w} &
+\mfrac{\dd h}{\dd w}
+\end{vmatrix} = 0\Add{,}
+\]
+ou $E' = 0$. Alors l'équation des lignes asymptotiques de la surface~$(\Phi)$,
+qui est
+\[
+E'\,dv^{2} + 2F'\,dv · dw + G'\,dw^{2} = 0\Add{,}
+\]
+%% -----File: 136.png---Folio 128-------
+est satisfaite pour $dw = 0$; les courbes $w = \cte$ sont des asymptotiques
+de la surface~$(\Phi)$. Ainsi \emph{les congruences à surface
+focale double peuvent s'obtenir en prenant les tangentes
+aux lignes asymptotiques d'une surface quelconque}.
+
+Considérons maintenant le cas de \Card{2} courbes focales confondues.
+Prenons cette courbe pour support. Nous pouvons supposer
+que $f\Add{,} g\Add{,} h$ ne dépendent plus de~$w$. Exprimons alors que l'équation~\Eq{(5)}
+admet pour racine double $u = 0$, nous avons
+\[
+\begin{vmatrix}
+a                    & b                    & c \\
+\mfrac{\dd f}{\dd v} & \mfrac{\dd g}{\dd v} & \mfrac{\dd h}{\dd v} \\
+\mfrac{\dd a}{\dd w} & \mfrac{\dd b}{\dd w} & \mfrac{\dd c}{\dd w}
+\end{vmatrix} = 0\Add{.}
+\]
+Les droites~$D$ de la congruence passant par un point~$F$ de la
+courbe~$(\phi)$ engendrent un cône. Le plan tangent à ce cône a
+pour coefficients les déterminants déduits du tableau
+\[
+\begin{Vmatrix}
+a                    & b                    & c \\
+\mfrac{\dd a}{\dd w} & \mfrac{\dd b}{\dd w} & \mfrac{\dd c}{\dd w}
+\end{Vmatrix}\Add{,}
+\]
+et la condition précédente exprime que la tangente~$FT$ à la
+courbe focale est dans le plan tangent au cône; ceci devant
+avoir lieu quelle que soit la génératrice du cône que l'on
+considère, tous les plans tangents au cône passent par~$FT$,
+et le cône se réduit à un plan. \emph{Une congruence à courbe focale
+double est engendrée par les droites qui en chaque point~$F$
+d'une courbe sont situées dans un plan passant par la tangente.}
+Ici l'enveloppe des plans focaux ne coïncide plus avec le
+lieu des points focaux.
+%% -----File: 137.png---Folio 129-------
+
+\Section{Développables de la congruence.}
+{2.}{} Cherchons si l'on peut associer les droites de la
+congruence de façon à obtenir des surfaces développables.
+Reprenons les équations de la droite
+\begin{alignat*}{2}%[** TN: Set on one line in original]
+x &= f(v,w) &&+ u · a(v,w), \\
+y &= g(v,w) &&+ u · b(v,w), \\
+z &= h(v,w) &&+ u · c(v,w);
+\end{alignat*}
+la condition pour que cette droite engendre une surface développable
+est
+\[
+\begin{vmatrix}
+a   &  b   &  c   \\
+da  &  db  &  dc  \\
+df  &  dg  &  dh
+\end{vmatrix} = 0,
+\]
+ou
+\[
+\Tag{(1)}
+\begin{vmatrix}
+a    &    b     &       c  \\
+\mfrac{\dd a}{\dd v} dv + \mfrac{\dd a}{\dd w} dw &
+\mfrac{\dd b}{\dd v} dv + \mfrac{\dd b}{\dd w} dw &
+\mfrac{\dd c}{\dd v} dv + \mfrac{\dd c}{\dd w} dw  \\
+%
+\mfrac{\dd f}{\dd v} dv + \mfrac{\dd f}{\dd w} dw &
+\mfrac{\dd g}{\dd v} dv + \mfrac{\dd g}{\dd w} dw &
+\mfrac{\dd h}{\dd v} dv + \mfrac{\dd h}{\dd w} dw
+\end{vmatrix} = 0.
+\]
+Telle est l'équation différentielle qui exprime que la droite
+de la congruence engendre une surface développable. Elle est
+de la forme
+\[
+A\, dv^{2} + 2 B\, dv · dw + C\, dw^{2} = 0;
+\]
+elle donne \Card{2} valeurs de~$\dfrac{dv}{dw}$, il y a donc \Card{2} familles de développables,
+qu'on appelle \emph{développables de la congruence}. \emph{Par chaque
+droite de la congruence passent \Card{2} développables de la congruence.}
+Cherchons les points de contact de cette droite avec
+les arêtes de rebroussement. Les coordonnées de l'un de ces
+points vérifient les équations
+\[
+\left\{
+\begin{alignedat}{3}
+df &+ u\, da &&+ a\, d\sigma &&= 0, \\
+dg &+ u\, db &&+ b\, d\sigma &&= 0, \\
+dh &+ u\, dc &&+ c\, d\sigma &&= 0;
+\end{alignedat}
+\right.
+\]
+%% -----File: 138.png---Folio 130-------
+ou
+\[
+\left\{
+\begin{aligned}
+  \left(\frac{\dd f}{\dd v} + u \frac{\dd a}{\dd v}\right) dv
++ \left(\frac{\dd f}{\dd w} + u \frac{\dd a}{\dd w}\right) dw
++ a\, d\sigma &= 0, \\
+  \left(\frac{\dd g}{\dd v} + u \frac{\dd b}{\dd v}\right) dv
++ \left(\frac{\dd g}{\dd w} + u \frac{\dd b}{\dd w}\right) dw
++ b\, d\sigma &= 0, \\
+  \left(\frac{\dd h}{\dd v} + u \frac{\dd c}{\dd v}\right) dv
++ \left(\frac{\dd h}{\dd w} + u \frac{\dd c}{\dd w}\right) dw
++ c\, d \sigma &= 0.
+\end{aligned}
+\right.
+\]
+\DPchg{Eliminons}{Éliminons} entre ces équations $dv, dw, d\sigma$, nous avons pour déterminer
+l'$u$~du point de contact de la droite avec l'arête de rebroussement,
+l'équation qui donne les points focaux. Donc \emph{les
+points où une droite~$D$ de la congruence touche les arêtes de
+rebroussement des deux développables de la congruence qui passent
+par cette droite sont les foyers de la droite~$D$}.
+
+\MarginNote{Développables
+et surface
+focale.}
+Supposons que le lieu des points focaux comprenne une surface~$(\Phi)$
+que nous prendrons pour support
+\[
+x = f(v,w), \qquad
+y = g(v,w), \qquad
+z = h(v,w).
+\]
+En chaque point~$F$ de la surface~$(\Phi)$ passe une droite~$D$ de la
+congruence tangente en~$F$ à~$(\Phi)$ et admettant~$F$ pour foyer. Nous
+avons trouvé sur la surface~$(\Phi)$ une famille de courbes tangentes
+aux droites~$D$. La développable qui a pour arête de rebroussement
+une de ces courbes~$(A)$ est une développable de la congruence.
+
+\Illustration[2.25in]{138a}
+Nous avons ainsi une des familles de développables.
+Considérons alors les courbes~$(c)$
+formant avec~$(A)$ un réseau conjugué.
+Considérons la développable enveloppe des
+plans tangents à~$(\Phi)$ tout le long d'une
+courbe~$(c)$; la génératrice de cette développable
+en un point~$F$ de~$(c)$ est la caractéristique
+du plan tangent, c'est la tangente
+%% -----File: 139.png---Folio 131-------
+conjuguée de la tangente à~$(c)$, c'est la droite~$D$. Nous
+avons la \Ord{2}{e} famille de développables en prenant l'enveloppe
+des plans tangents à~$(\Phi)$ en tous les points des courbes~$(c)$
+conjuguées des courbes~$(A)$.
+
+Supposons que les courbes $w = \cte$ soient précisément les
+courbes~$(A)$. On a
+\[
+a = \frac{\dd f}{\dd v}, \qquad
+b = \frac{\dd g}{\dd v}, \qquad
+c = \frac{\dd h}{\dd v};
+\]
+l'équation~\Eq{(1)} devient
+\[
+\begin{vmatrix}%[** TN: Added elided columns]
+\mfrac{\dd f}{\dd v} &
+\mfrac{\dd g}{\dd v} &
+\mfrac{\dd h}{\dd v} \\
+%
+\mfrac{\dd^{2} f}{\dd v^{2}}·dv + \mfrac{\dd^{2} f}{\dd v · \dd w}\, dw &
+\mfrac{\dd^{2} g}{\dd v^{2}}·dv + \mfrac{\dd^{2} g}{\dd v · \dd w}\, dw &
+\mfrac{\dd^{2} h}{\dd v^{2}}·dv + \mfrac{\dd^{2} h}{\dd v · \dd w}\, dw \\
+%
+\mfrac{\dd f}{\dd v} · dv       + \mfrac{\dd f}{\dd w}\, dw &
+\mfrac{\dd g}{\dd v} · dv       + \mfrac{\dd g}{\dd w}\, dw &
+\mfrac{\dd h}{\dd v} · dv       + \mfrac{\dd h}{\dd w}\, dw
+\end{vmatrix} = 0.
+\]
+Retranchons la \Ord{1}{ère} ligne de la \Ord{3}{e}: $dw$~vient en facteur, et
+l'équation prend la forme
+\[
+dw (E'\, dv + F'\, dw) = 0;
+\]
+nous trouvons d'abord $dw = 0$, (courbes~$A$); la relation
+\[
+E'\, dv + F'\, dw = 0
+\]
+définit précisément les courbes~$(c)$ conjuguées des courbes
+$w = \cte[]$. Nous retrouvons les résultats précédents.
+
+\MarginNote{Développables
+et courbe
+focale.}
+Examinons maintenant le cas d'une courbe focale~$(\phi)$ que
+nous prendrons pour support:
+\[
+x = f(v), \qquad
+y = g(v), \qquad
+z = h(v),
+\]
+alors $\dfrac{\dd f}{\dd w}, \dfrac{\dd g}{\dd w}, \dfrac{\dd h}{\dd w}$ sont nuls, et l'équation~\Eq{(1)} devient
+\[%[** TN: Filled in last two columns]
+\begin{vmatrix}
+a & b & c \\
+\mfrac{\dd a}{\dd v}\, dv + \mfrac{\dd a}{\dd w}\, dw &
+\mfrac{\dd b}{\dd v}\, dv + \mfrac{\dd b}{\dd w}\, dw &
+\mfrac{\dd c}{\dd v}\, dv + \mfrac{\dd c}{\dd w}\, dw \\
+\mfrac{\dd f}{\dd v}\, dv &
+\mfrac{\dd g}{\dd v}\, dv &
+\mfrac{\dd h}{\dd v}\, dv
+\end{vmatrix} = 0;
+\]
+%% -----File: 140.png---Folio 132-------
+$dv$~est en facteur. L'une des familles de développables est
+formée par les droites $v = \cte$, c'est-à-dire par toutes les
+droites de la congruence passant par un même point~$F$ de~$(\phi)$.
+Ce sont des cônes.
+
+\MarginNote{Examen des diverse
+cas possibles.}
+Examinons alors tous les cas possibles relativement à la
+nature du lieu des foyers.
+
+\ParItem{\Primo.} Supposons qu'il y ait \Card{2} surfaces focales $(\Phi), (\Phi')$.
+Toute droite~$D$ de la congruence est tangente à $(\Phi), (\Phi')$ aux
+\Card{2} points $F\Add{,} F'$ foyers de~$D$. Considérons une des développables
+ayant pour arête de rebroussement l'une des
+\Figure[3.25in]{140a}
+courbes~$(A)$. Toutes ses génératrices sont
+tangentes à $(\Phi')$, cette développable est
+circonscrite à~$(\Phi')$ le long d'une courbe~$(c')$
+que nous appellerons \emph{courbe de contact}.
+Le plan focal correspondant à~$F$ est
+le plan tangent en~$F$ à la surface~$(\Phi)$. Le
+\Ord{2}{e} plan focal est le plan tangent en~$F'$ à~$(\Phi')$,
+et comme la développable est circonscrite à~$(\Phi')$, ce plan
+tangent est le plan tangent à la développable au point~$F'$,
+c'est-à-dire le long de la génératrice~$D$; c'est le plan osculateur
+à l'arête de rebroussement~$(A)$ au point~$F$. Il y a
+évidemment réciprocité entre $(\Phi), (\Phi')$. L'autre série de développables
+aura pour arêtes de rebroussement les enveloppes
+des droites~$D$ sur la surface~$(\Phi')$. Soient $(A')$ ces arêtes de
+rebroussement, et ces développables seront circonscrites à~$(\Phi)$
+le long des courbes de contact~$(C)$. Nous avons ainsi déterminé
+sur~$(\Phi)\Add{,} (\Phi')$ \Card{2} réseaux conjugués qui se correspondent
+de manière qu'aux courbes~$(A)$ correspondent les courbes~$(c')$
+%% -----File: 141.png---Folio 133-------
+et aux courbes~$(c)$ les courbes~$(A')$, l'une des familles de courbes
+correspondantes étant constituée par des arêtes de rebroussement,
+et l'autre par des courbes de contact. Le \Ord{2}{e} foyer~$F'$
+est le point de contact de la droite~$D$ avec son enveloppe
+quand on se \DPtypo{deplace}{déplace} sur la courbe~$(c)$.
+
+\ParItem{\Secundo.} Supposons une surface focale~$(\Phi)$ et une courbe focale~$(\phi')$.
+Une des séries de développables est constituée par
+\Figure[3.25in]{141a}
+des cônes ayant leurs sommets
+sur~$(\phi')$. Les courbes~$(c)$ sur~$(\Phi)$
+sont les courbes de contact des
+cônes circonscrits à~$(\Phi)$ par les
+divers points de~$(\phi')$. Les plans
+focaux sont le plan osculateur à~$(A)$
+au point~$F$ et le plan tangent
+à~$(\Phi)$ au point~$F$, c'est-à-dire
+le plan tangent à~$(\phi')$ passant
+par~$(D)$, et le plan tangent au cône de la congruence de
+sommet~$F'$, le long de~$(D)$. Les courbes~$(c), (A)$ forment un réseau
+conjugué sur~$(\Phi)$.
+
+\ParItem{\Tertio.} Supposons enfin \Card{2} courbes focales $(\phi)\Add{,} (\phi')$; les deux
+familles de développables sont des cônes passant par l'une des
+courbes et ayant leurs sommets sur l'autre.
+
+\MarginNote{Cas singuliers.}
+Voyons maintenant le cas des foyers confondus.
+
+\ParItem{\Primo.} Il y a une \emph{surface focale double}. Dans ce cas la congruence
+est constituée par les tangentes à une famille d'asymptotiques
+de cette surface. Il n'y a plus qu'une famille de
+développables ayant pour arêtes de rebroussement ces asymptotiques.
+%% -----File: 142.png---Folio 134-------
+Prenons cette surface pour support, et pour courbes $w = \cte$
+ces asymptotiques. L'équation différentielle qui détermine
+les développables est
+\[
+dw (E'\, dv + F'\, dw) = 0.
+\]
+L'équation des lignes asymptotiques est
+\[
+E'\, dv^{2} + 2F'\, dv · dw + G'\, dw^{2} = 0;
+\]
+elle doit être vérifiée pour $dw = 0$; donc $E' = 0$, et l'équation
+qui détermine les développables devient $dw^{2} = 0$, ce qui démontre
+le \DPtypo{resultat}{résultat} précédemment énoncé.
+
+\ParItem{\Secundo.} Il y a une \emph{courbe focale double}~$(\phi)$. Les droites de
+la congruence sont dans des plans tangents aux divers points
+de~$(\phi)$. Une famille de ces développables est donc constituée
+par ces plans. On aperçoit immédiatement deux autres développables,
+l'enveloppe des plans tangents précédents, et la développable
+qui a pour arête de rebroussement la courbe~$(\phi)$. Il
+est facile de voir qu'il n'y en a pas d'autre. Soit la courbe~$(\phi)$:
+\[
+x = f(v), \qquad
+y = g(v), \qquad
+z = h(v);
+\]
+la tangente a pour coefficients directeurs $\dfrac{\dd f}{\dd v}, \dfrac{\dd g}{\dd v}, \dfrac{\dd h}{\dd v}$; donnons-nous
+en chaque point les coefficients directeurs d'une droite
+de la congruence $\alpha(v), \beta(v), \gamma(v)$. Une droite quelconque de la
+congruence aura pour coefficients directeurs
+\[
+a = \frac{\dd f}{\dd v} + w \alpha (v), \qquad
+b = \frac{\dd g}{\dd v} + w \beta (v), \qquad
+c = \frac{\dd h}{\dd v} + w \gamma (v).
+\]
+L'équation des développables est
+\[%[** TN: Filled in last two columns]
+\begin{vmatrix}
+f' + w · \alpha & g' + w · \beta & h' + w · \gamma \\
+(f'' + w \alpha')\, dv + \alpha · dw &
+(g'' + w \beta')\, dv + \beta · dw &
+(h'' + w \gamma')\, dv + \gamma · dw \\
+f'\, dv & g'\, dv & h'\, dv
+\end{vmatrix} = 0;
+\]
+%% -----File: 143.png---Folio 135-------
+$dv$~est en facteur; en retranchant la \Ord{3}{e} ligne de la \Ord{1}{ère}, $w$~est
+en facteur, et l'équation se réduit à
+\[%[** TN: Filled in last two columns]
+w · dv^{2}
+\begin{vmatrix}
+\alpha          & \beta & \gamma \\
+f'' + w \alpha' & g'' + w \beta' & h'' + w \gamma' \\
+f'              & g' & h'
+\end{vmatrix} = 0.
+\]
+Nous trouvons $dv = 0$ correspondant aux plans tangents; $w = 0$
+correspondant à la développable d'arête de rebroussement~$\phi$,
+et enfin
+\[
+\begin{vmatrix}
+\alpha & \beta & \gamma \\
+f''    & g''   & h''    \\
+f'     & g'    & h'
+\end{vmatrix} + w \begin{vmatrix}
+\alpha  & \beta  & \gamma  \\
+\alpha' & \beta' & \gamma' \\
+f'      & g'     & h'
+\end{vmatrix} = 0\Add{.}
+\]
+Le plan tangent considéré en un point de la courbe~$(\phi)$ a pour
+équation
+\[
+\begin{vmatrix}
+x - f  & y - g & z - h \\
+f'     & g'    & h'    \\
+\alpha & \beta & \gamma
+\end{vmatrix} = 0;
+\]
+\DPtypo{Cherchons}{cherchons} son enveloppe. La caractéristique est dans le plan
+\[
+\begin{vmatrix}
+x - f  & y - g & z - h \\
+f''    & g''   & h''   \\
+\alpha & \beta & \gamma
+\end{vmatrix} + \begin{vmatrix}
+x - f   & y - g  & z - h \\
+f'      & g'     & h'    \\
+\alpha' & \beta' & \gamma'
+\end{vmatrix} = 0.
+\]
+La droite~$D$ est
+\[%[** TN: Filled in last two equations]
+x = f + u \left[\frac{\dd f}{\dd v} + w \alpha (v)\right]\!\!, \quad
+y = g + u \left[\frac{\dd g}{\dd v} + w \beta (v)\right]\!\!, \quad
+z = h + u \left[\frac{\dd h}{\dd v} + w \gamma (v)\right]\!\!.
+\]
+Exprimons que cette droite est dans le \Ord{2}{e} plan qui contient la
+caractéristique, nous avons
+\[%[** TN: Filled in last two columns]
+\begin{vmatrix}
+f' + w \alpha & g' + w \beta & h' + w \gamma \\
+f''           & g'' & h'' \\
+\alpha        & \beta & \gamma
+\end{vmatrix} + \begin{vmatrix}
+f' + w \alpha & g' + w \beta & h' + w \gamma \\
+f'            & g' & h' \\
+\alpha'       & \beta' & \gamma'
+\end{vmatrix} = 0,
+\]
+%% -----File: 144.png---Folio 136-------
+condition qui se réduit à
+\[
+\begin{vmatrix}
+f'      & g'     & h'     \\
+f''     & g''    & h''    \\
+\alpha  & \beta  & \gamma
+\end{vmatrix} + w \begin{vmatrix}
+\alpha  & \beta  & \gamma \\
+f'      & g'     & h'     \\
+\alpha' & \beta' & \gamma'
+\end{vmatrix} = 0.
+\]
+C'est précisément l'équation qui définit la \Ord{3}{e} développable,
+qui est donc l'enveloppe des plans qui contiennent les droites
+de la congruence.
+
+\Section{Sur le point de vue corrélatif.}
+{3.}{} Nous avons trouvé comme cas particulier du lieu des
+foyers une courbe. En examinant la question au point de vue
+corrélatif, nous sommes conduits à examiner \emph{le cas où l'enveloppe
+des plans focaux est une surface développable}, soit~$\Phi$.
+Soit~$\Phi'$ l'autre nappe le la surface focale. Les droites de la
+congruence sont tangentes à $\Phi, \Phi'$; or\Add{,} une tangente à la développable~$\Phi$
+doit être dans l'un des plans tangents qui enveloppent
+cette développable; les droites de la congruence sont donc les
+tangentes à~$\Phi'$ qui sont dans les plans tangents à~$\Phi$, ce sont
+les tangentes aux sections de~$\Phi'$ par les plans qui enveloppent~$\Phi$.
+Dans ce cas les arêtes de rebroussement~$(A')$ sur la surface~$\Phi'$
+sont des courbes planes, les développables correspondantes
+étant les plans de ces courbes. Les foyers d'une droite~$D$ sont
+le point de contact avec~$\Phi'$, et le point d'intersection avec
+la caractéristique du plan tangent à la développable~$\Phi$. L'autre
+famille de développables aura ses arêtes de rebroussement
+sur la surface~$\Phi$, et correspondant aux courbes~$(c')$ \DPtypo{conjuguees}{conjuguées}
+des courbes~$(A')$.
+
+\emph{Réciproquement, si les arêtes de rebroussement des développables
+%% -----File: 145.png---Folio 137-------
+situées sur une des nappes de la surface focale
+sont des courbes planes, les développables correspondantes
+seront des plans, et leur enveloppe sera la \Ord{2}{e} nappe de la
+surface focale.}
+
+Pour avoir une congruence de cette espèce on peut prendre
+arbitrairement la développable~$\Phi$, et sur cette développable,
+une famille de courbes quelconque. Les tangentes à ces courbes
+engendrent une congruence de l'espèce considérée, car l'une
+des familles de développables est évidemment constituée par
+les plans tangents à la développable~$\Phi$; les courbes de contact
+sur la développable sont les génératrices, qui peuvent
+être considérées comme conjuguées à toute famille de courbes.
+
+\emph{Supposons les \Card{2} nappes de la surface focale développables\Add{.}}
+Il suffit de partir d'une développable~$\Phi$, de la couper par
+une famille de plans quelconques. Les sections seront les
+courbes~$A$, et les plans de ces sections envelopperont l'autre
+développable focale. On peut dire dans ce cas que l'on a \Card{2} familles
+de plans à un paramètre, les droites de la congruence
+étant les intersections de chaque plan d'une famille avec chaque
+plan de l'autre.
+
+Les \emph{\Card{2} cas singuliers} se correspondent à eux-mêmes au
+point de vue corrélatif. Les asymptotiques d'une surface se
+correspondent à elles-mêmes; car une asymptotique est telle
+que le plan osculateur en l'un de ses points est tangent à la
+surface, et au point de vue corrélatif, un point d'une courbe
+se transforme en plan osculateur et inversement.
+%% -----File: 146.png---Folio 138-------
+
+\MarginNote{Congruences de
+Koenigs.}
+Il y a un \emph{autre cas particulier corrélatif de lui-même},
+c'est le \emph{cas de Koenigs}. On appelle \emph{élément de contact} le système
+constitué par un point~$M$ et un plan passant par ce point.
+Les surfaces et les courbes sont alors engendrées de la même
+façon au moyen des \DPtypo{élément}{éléments} de contact: en chaque point d'une
+surface, il y a un plan tangent et un seul, ce qui donne
+$\infty^{2}$~éléments de contact; sur une courbe, il y a $\infty^{1}$~points, et
+en chaque point $\infty^{1}$~plans tangents, ce qui donne encore $\infty^{2}$~éléments
+de contact; pour les développables, nous avons $\infty^{1}$~plans
+et $\infty^{2}$~points, donnant $\infty^{2}$~éléments de contact. Une droite est
+de même constituée par $\infty^{2}$~éléments de contact, $\infty^{1}$~points sur
+la droite et $\infty^{1}$~plans passant par la droite. Dans la \DPtypo{Théorie}{théorie}
+des congruences, \emph{un foyer et le plan focal correspondant constituent
+un élément de contact}, et les surfaces focales, courbes
+focales, développables focales, ou comme l'on dit plus généralement,
+les \emph{multiplicités focales, sont engendrées par les
+éléments de contact focaux}. Nous voyons alors que nous avons
+considéré tous les cas possibles, sauf celui où l'une des multiplicités
+focales est une droite.
+
+\Illustration{146a}
+La droite peut être considérée comme le lieu de $\infty^{1}$~points
+ou comme l'enveloppe de $\infty^{1}$~plans; c'est donc à la fois une
+courbe et une développable; il en résulte qu'une des familles
+de développables de la congruence est constituée par des cônes
+ayant leurs sommets sur la droite, et l'autre par des plans
+passant par la droite. Si en particulier la congruence a pour
+multiplicités focales une droite~$D$ et une
+surface~$\Phi$, les séries de développables
+seront d'une part les cônes circonscrits à~$\Phi$
+%% -----File: 147.png---Folio 139-------
+par les différents points de~$D$, ce qui donne les courbes de
+contact~$(c)$; et les plans passant par~$D$, qui coupent suivant
+les arêtes de rebroussement~$(A)$, et $(A)\Add{,} (c)$ forment un système
+de courbes conjuguées. On obtient ainsi le \emph{Théorème de Koenigs:
+Les courbes de contact des cônes circonscrits à une surface
+par les divers points d'une droite~$D$, et les sections de
+cette surface par les plans passant par~$D$ constituent un réseau
+conjugué}.
+
+\MarginNote{Congruences
+linéaires.}
+Si les multiplicités focales sont \Card{2} droites, la congruence
+est constituée par les droites rencontrant ces \Card{2} droites.
+C'est une \emph{congruence linéaire}.
+
+Il peut encore arriver qu'il y ait une droite focale double;
+il suffira alors d'associer à chaque point~$A$ de la droite
+un plan~$P$ passant par cette droite, et la congruence sera
+constituée par les droites~$D$ situées dans les plans~$P$ et passant
+par les points~$A$.
+
+\Section{Application. Surfaces de Joachimsthal.}
+{}{Rechercher les surfaces dont les lignes de courbure d'un
+système sont dans des plans passant par une droite fixe~$\Delta$.}
+
+Soit $S$ une surface répondant à la question; considérons
+les tangentes aux lignes de courbure; ces tangentes~$D$ constituent
+une congruence, et comme les lignes de courbure sont
+dans des plans passant par~$\Delta$, ces droites~$D$ rencontrent la
+droite~$\Delta$; $S$~est une des nappes de la surface focale; les développables
+comprennent, d'une part les plans des lignes de
+courbure, et d'autre part les cônes circonscrits à~$S$ par les
+différents points de $\Delta$, dont les courbes de contact constituent
+%% -----File: 148.png---Folio 140-------
+un système conjugué du \Ord{1}{er} système de lignes de courbure,
+et par suite forment le \Ord{2}{e} système de lignes de courbure.
+Si nous considérons ce \Ord{2}{e} système de lignes de courbure, le
+cône circonscrit coupe la surface~$S$ suivant un angle constamment
+nul; la courbe de contact, qui est une ligne de courbure
+de~$S$, est donc aussi une ligne de courbure du cône circonscrit,
+d'après le Théorème de Joachimsthal; c'est donc une trajectoire
+orthogonale des génératrices, donc l'intersection du
+cône avec une sphère ayant son centre au sommet; le \Ord{2}{e} système
+de lignes de courbure est donc constitué par des courbes
+sphériques, et les sphères correspondantes coupent orthogonalement
+la surface~$S$ le long des lignes de courbure. \emph{La surface~$S$
+est donc trajectoire orthogonale d'une famille de sphères
+ayant leurs centres sur $\Delta$.} Cette propriété est caractéristique
+de la surface~$S$; supposons en effet une famille de sphères
+ayant leurs centres sur~$\Delta$, et une surface~$S$ orthogonale
+à chacune de ces sphères tout le long de la courbe d'intersection;
+l'intersection est une ligne de courbure de la sphère,
+et comme l'angle d'intersection de~$S$ et de la sphère est constamment
+droit, c'est une ligne de courbure de~$S$. Si on joint
+le centre~$A$ de la sphère à un point~$M$ de la ligne de courbure,
+cette droite est normale à la sphère, donc tangente à la surface~$S$,
+de sorte que la ligne de courbure est la courbe de
+contact du cône circonscrit à~$S$ par le point~$A$.
+
+Nous sommes ainsi conduits à rechercher les surfaces coupant
+à angle droit une famille de sphères. Considérons les lignes
+de courbure du \Ord{1}{er} système; chacune d'elles est tangente
+à la droite~$D$ correspondante, donc normale à la sphère, et
+%% -----File: 149.png---Folio 141-------
+comme elle est dans un plan passant par~$\Delta$, elle est trajectoire
+orthogonale pour le grand cercle section de la sphère
+par ce plan. Si donc on considère les sections de toutes les
+sphères de la famille par un même plan passant par~$\Delta$, la ligne
+de courbure située dans ce plan sera trajectoire orthogonale
+de la famille de cercles obtenue. Si on considère un autre
+plan, la ligne de courbure dans ce plan sera aussi trajectoire
+orthogonale de la famille de cercles. En rabattant le \Ord{2}{e} plan
+sur le \Ord{1}{er} on aura une autre trajectoire orthogonale de
+la même famille de cercles. \emph{On considère donc une famille de
+cercles ayant leurs centres sur~$\Delta$, on en détermine les trajectoires
+orthogonales, et on fait tourner chacune de ces trajectoires
+orthogonales autour de~$\Delta$ d'un angle qui lui corresponde
+et qui varie d'une manière continue quand on passe d'une
+trajectoire à la trajectoire infiniment voisine.} Le lieu des
+courbes ainsi obtenues est une surface qui sera la surface~$S$
+si la loi de rotation est convenablement choisie. Quelle que
+soit d'ailleurs cette loi on obtient toujours une surface répondant
+à la question; cette surface sera en effet engendrée
+par des courbes qui couperont orthogonalement la famille de
+sphères ayant pour grands cercles les cercles considérés, et
+par conséquent la surface coupera à angle droit toutes ces
+sphères tout le long des courbes d'intersection.
+
+Nous allons donc chercher les trajectoires orthogonales
+d'une famille de cercles ayant leurs centres sur une droite~$\Delta$.
+Cherchons plus généralement les trajectoires orthogonales d'une
+famille de cercles quelconque, que nous définirons en donnant
+%% -----File: 150.png---Folio 142-------
+les coordonnées $a\Add{,}b$ du centre et le rayon~$R$ en fonction
+d'un paramètre~$u$\DPtypo{;}{.} Considérons une trajectoire orthogonale rencontrant
+un des cercles en un point~$M$. Les coordonnées du
+point~$M$ sont, en fonction du paramètre~$u$
+\[
+\Tag{(1)}
+x = a + R \cos\phi, \qquad
+y = b + R \sin \phi,
+\]
+$\phi$~étant une fonction de~$u$ convenablement choisie. Tout revient
+à déterminer cette fonction de~$u$ de façon que la courbe représentée
+par les équations~\Eq{(1)} soit normale à tous les cercles.
+La normale au cercle a pour paramètres directeurs $x - a$, $y - b$;
+elle doit être tangente à la courbe, donc
+\[
+\Tag{(2)}
+\frac{dx}{x-a} = \frac{dy}{y-b}.
+\]
+Or:
+\begin{gather*}
+dx = da + \cos \phi · dR - R \sin \phi · d \phi, \quad
+dy = db + \sin \phi · dR + R \cos \phi · d \phi, \\
+x - a = R \cos \phi, \quad
+y - b = R \sin \phi.
+\end{gather*}
+L'équation~\Eq{(2)} devient
+\[
+\begin{vmatrix}
+da + \cos\phi · dR - R \sin\phi · d\phi &
+db + \sin\phi · dR + R \cos\phi · d\phi \\
+R \cos\phi   &    R \sin\phi
+\end{vmatrix} = 0,
+\]
+ou:
+\[
+\sin\phi · da - \cos\phi · db - R\, d\phi = 0,
+\]
+ou:
+\[
+\Tag{(3)}
+\frac{d\phi}{du} = \frac{a'}{R} \sin\phi - \frac{b'}{R} \cos\phi.
+\]
+Si nous posons
+\[
+\tg \frac{\phi}{2} = w,
+\]
+d'où
+\[
+d\phi = \frac{2\, dw}{1 + w^{2}},
+\]
+l'équation différentielle devient
+\[
+\frac{1}{du}\, \frac{2\, \Err{du}{dw}}{1 + w^{2}}
+  = A \frac{2w}{1 + w^{2}} + B \frac{1 - w^{2}}{1 + w^{2}},
+\]
+$A\Add{,} B$ étant fonctions de~$u$; de sorte que l'équation est de la
+forme
+\[
+\frac{dw}{du} = Aw + \frac{B}{2} (1 - w^{2}).
+\]
+%% -----File: 151.png---Folio 143-------
+C'est une équation de Riccati. Le rapport anharmonique de \Card{4} solutions~$w$
+est constant. Or\Add{,} si $M$~est un point d'une trajectoire
+orthogonale, $\tg \dfrac{\phi}{2}$ est le coefficient angulaire de la
+droite~$AM$. Si on considère \Card{4} trajectoires
+orthogonales $M, M', M'', M'''$, les \Card{4} valeurs de~$u$
+correspondantes sont les coefficients angulaires
+des \Card{4} droites $AM, AM', AM'', AM'''$, et
+le rapport anharmonique des \Card{4} solutions~$u$
+est le rapport anharmonique du faisceau
+$(A, M, M', M'' M''')$. Ce rapport est indépendant de la position du
+point~$A$ sur le cercle. Il en résulte que \emph{\Card{4} trajectoires orthogonales
+d'une famille de cercles coupent tous les cercles de
+la famille suivant le même rapport anharmonique}.
+
+%[** TN: Setting inset illos side-by-side and floating]
+\begin{figure}[hbt]
+\centering
+\Input{151a}\hfil\hfil
+\Input[3in]{151b}
+\end{figure}
+Dans le cas particulier où les cercles ont leurs centres
+sur une droite~$\Delta$, les points $M'\Add{,} M''$ d'intersection du cercle avec~$\Delta$
+correspondent à \Card{2} trajectoires orthogonales; on a donc \Card{2} solutions
+de l'équation de Riccati, et la détermination des trajectoires
+orthogonales se ramène à une quadrature. Pour définir
+la famille, au lieu de se donner $a, b, R$ en fonction d'un paramètre,
+on peut se donner une trajectoire orthogonale~$\Gamma$, on
+aura alors \Card{3} solutions de l'équation de \DPtypo{Ricatti}{Riccati}, et la solution
+la plus générale s'obtiendra en écrivant que son rapport
+anharmonique avec les \Card{3} solutions connues est constant.
+
+Supposons que $(\Delta)$ soit
+l'axe~$\DPtypo{OX}{Ox}$, et donnons
+nous $(\Gamma)$ par ses tangentes~$(T)$. L'une d'elles
+a pour équations:
+%% -----File: 152.png---Folio 144-------
+\begin{align*}
+x &= a + \rho\cos u, \\
+y &= \rho\sin u,
+\end{align*}
+$a$~étant une fonction de~$u$. Pour déterminer le point de contact
+avec~$(\Gamma)$, on a, en différentiant:
+\[
+da - \rho\sin u\, du + \cos u\, d\rho = 0, \quad
+\rho\cos u\, du + \sin u\, d\rho = 0,
+\]
+d'où, pour la valeur de~$\rho$, c'est-à-dire du rayon~$R$ du cercle,
+\[
+R = \rho = \frac{da}{du} \sin u.
+\]
+Une trajectoire orthogonale quelconque est donc représentée
+par
+\[
+\Tag{(4)}
+x = a + \frac{da}{du} \sin u · \cos\phi, \qquad
+y = \frac{da}{du} \sin u · \sin\phi,
+\]
+$\phi$~étant lié à~$u$ par la constance du rapport anharmonique
+$(M, M', M'', M''')$, ce qui donne simplement
+\[
+\Tag{(5)}
+\tg \frac{\phi}{2} = m · \tg \frac{u}{2}.
+\]
+Si maintenant on fait tourner la courbe~\Eq{(4)} d'un angle~$v$ autour
+de~$\DPtypo{ox}{Ox}$, en supposant~$m$ fonction de~$v$, et posant
+\[
+a = f(u), \qquad
+m = g(v),
+\]
+on obtiendra une trajectoire orthogonale quelconque de la famille
+de sphères ayant pour grands cercles les cercles considérés:
+\[
+\Tag{(6)}
+\left\{
+  \begin{aligned}
+x &= f(u) + f'(u)\sin u \cos\phi, \\
+y &= f'(u) \sin u \sin\phi \cos v,
+     \text{ (avec $\tg \tfrac{\phi}{2} = g(v)\tg \tfrac{u}{2}$)}  \\
+z &= f'(u) \sin u \sin\phi \sin v.
+\end{aligned}
+\right.
+\]
+Et en considérant dans ces équations $u$~et~$v$ comme des paramètres
+arbitraires, elles représentent la surface de Joachimsthal
+la plus générale.
+%% -----File: 153.png---Folio 145-------
+
+\Section{Détermination des développables d'une congruence.}
+{4.}{} Nous avons vu que la détermination des développables
+d'une congruence dépend de l'intégration d'une équation \DPtypo{difféférentielle}{différentielle}
+du \Ord{1}{er} ordre et du \Ord{2}{e} degré. Cette intégration
+peut se simplifier dans certains cas.
+
+On obtient les développables sans quadrature si la congruence
+admet \Card{2} courbes focales, ou corrélativement deux développables
+focales. Dans le \Ord{1}{e} cas, on obtient des cônes, et
+dans le \Ord{2}{e}, des plans tangents, comme on l'a vu précédemment.
+
+Si on a une courbe focale, ou corrélativement une développable
+focale, on a immédiatement une des familles de développables
+de la congruence; pour avoir l'autre, on a à intégrer
+une équation différentielle du \Ord{1}{e} ordre et du \Ord{1}{er}~degré.
+
+\Illustration[2.25in]{153a}
+Cette équation a des propriétés particulières dans un cas
+corrélatif de lui-même, \emph{cas où l'on a une courbe focale et une
+développable focale}. Soit $(\alpha)$ l'arête de rebroussement de la
+développable focale~$(\Phi)$; considérons
+une génératrice quelconque~$C$
+de cette développable; les droites
+de la congruence rencontrent
+la courbe focale~$(\phi')$ et sont dans
+les plans tangents à~$(\Phi)$. Considérons
+un plan tangent à~$(\Phi)$ qui
+rencontre $(\phi')$ en~$F'$; toutes les droites du plan tangent qui
+passent par~$F'$ sont des droites de la congruence. Considérons
+les développables de la congruence passant par une de ces
+droites~$D$; il y a d'abord le plan qui enveloppe la développable,
+et qui admet pour courbe de contact la génératrice~$C$. Les
+foyers de la droite~$D$ sont $F'$~sur~$(\phi')$ et $F$~sur~$C$. La \Ord{2}{e} développable
+%% -----File: 154.png---Folio 146-------
+a pour arête de rebroussement une courbe~$(A)$ de~$(\Phi)$
+dont les tangentes vont rencontrer~$(\phi')$. Le problème revient
+donc à \emph{trouver les courbes d'une développable~$(\Phi)$ dont les
+tangentes vont rencontrer une courbe~$(\phi')$}. Nous allons chercher
+directement les développables de la congruence, que nous définirons
+en partant de la courbe~$(\phi')$ et en associant à chacun de
+ses points un certain plan dans lequel seront toutes les droites
+de la congruence passant par ce point; la développable~$(\Phi)$
+sera l'enveloppe de ce plan. Soit la courbe~$(\phi')$
+\[
+x = f(v), \qquad
+y = g(v), \qquad
+z = h(v);
+\]
+pour définir un plan passant par un de ses points, il suffit
+de se donner \Card{2} directions $\alpha(v), \beta(v), \gamma(v)$ et $\alpha_{1}(v), \beta_{1}(v), \gamma_{1}(v)$.
+
+On a ainsi le plan contenant toutes les droites de la
+congruence; les coefficients directeurs d'une telle droite
+sont alors:
+\[
+\bar{a} = \alpha + w \alpha_{1}, \quad
+\bar{b} = \beta  + w \beta_{1}, \quad
+\bar{c} = \gamma + w \gamma_{1}.
+\]
+L'équation aux développables
+\[
+\begin{vmatrix}
+\bar{a}   & \bar{b}   & \bar{c}  \\
+d\bar{a}  & d\bar{b}  & d\bar{c} \\
+df        & dg        & dh
+\end{vmatrix} = 0
+\]
+devient ici
+\[%[** TN: Filled in last two columns]
+dv \begin{vmatrix}
+\alpha + w \alpha_{1}  & \beta + w \beta_{1}  & \gamma + w \gamma_{1}  \\
+f'(v)                  & g'(v)                  & h'(v)               \\
+(\alpha' + w \alpha_{1}')\, dv + \alpha_{1} dw  &
+(\beta'  + w \beta_{1}')\, dv + \beta_{1} dw  &
+(\gamma' + w \gamma_{1}')\, dv + \gamma_{1} dw
+\end{vmatrix} = 0.
+\]
+Nous trouvons $dv = 0, v = \cte$ ce qui nous donne les plans des
+droites de la congruence. L'autre solution s'obtiendra par
+l'intégration de l'équation:
+%% -----File: 155.png---Folio 147-------
+\[%[** TN: Filled in last two columns]
+dw \begin{vmatrix}
+\alpha + w \alpha_{1} & \beta + w \beta_{1} & \gamma + w \gamma_{1} \\
+f'                    & g'                    & h'                   \\
+\alpha_{1}            & \beta_{1}            & \gamma_{1}
+\end{vmatrix} + dv \begin{vmatrix}
+\alpha + w \alpha_{1}   & \beta + w \beta_{1}   & \gamma + w \gamma_{1} \\
+f'                      & g'                      & h'                   \\
+\alpha' + w \alpha_{1}' & \beta' + w \beta_{1}' & \gamma' + w \gamma_{1}'
+\end{vmatrix} = 0,
+\]
+équation de la forme
+\[
+\frac{dw}{dv} = Pw^{2} + Qw + R,
+\]
+$P\Add{,} Q\Add{,} R$ étant fonctions de $v$~seulement. C'est une équation de Riccati.
+
+Cherchons dans quels cas on peut avoir des solutions particulières
+de cette équation. Si la courbe~$(\phi')$ est plane, si
+on coupe~$(\Phi)$ par son plan, la section est une courbe dont les
+tangentes rencontrent~$(\phi')$, c'est une courbe~$(A)$; on a une solution
+particulière, le problème s'achève au moyen de \Card{2} quadratures.
+En particulier si $(\phi')$ est le cercle imaginaire à
+l'infini, on a à déterminer sur~$(\Phi)$ des courbes dont les tangentes
+rencontrent le cercle imaginaire à l'infini, ce sont
+les courbes minima. \emph{La détermination des courbes minima d'une
+développable se ramène à \Card{2} quadratures.}
+
+Corrélativement, si $(\Phi)$ est un cône, considérons le cône
+de même sommet et qui a pour base~$(\phi')$; c'est une développable
+de le \Ord{2}{e} famille; on a une solution particulière, et le problème
+s'achève par \Card{2} quadratures.
+
+Si $(\Phi)$ est un cône et $(\phi')$~une courbe plane, on a \Card{2} solutions
+particulières, donc une seule quadrature.
+
+Supposons encore que les plans~$P$ précédemment définis
+soient normaux à la courbe~$(\phi')$. Nous avons la \emph{congruence des
+normales} à la courbe~$(\phi')$, et la recherche des développables
+conduira à celle des \emph{développées} de~$(\phi')$. Le plan normal à~$(\phi')$
+%% -----File: 156.png---Folio 148-------
+en l'un de ses points~$F'$ est perpendiculaire à la tangente~$F'T$.
+Si on \DPtypo{onsidère}{considère} le cône isotrope~$J$ de sommet~$F'$, le plan normal
+est le plan polaire de la tangente par rapport à ce cône isotrope;
+parmi les normales il y a donc les \Card{2} génératrices de
+contact des plans tangents menés par la tangente au cône isotrope.
+Soit $G$ l'une d'elles, on l'obtient algébriquement; considérons
+la surface réglée qu'elle engendre lorsque $F'$~décrit
+la courbe~$(\phi')$. Le plan asymptote, plan tangent à l'infini sur~$G$,
+est le plan tangent au cône isotrope~$J$ le long de~$G$; la
+surface réglée contient la courbe~$(\phi')$, et le plan tangent au
+point~$F'$ est le plan~$G·F'·T$, qui est encore le plan tangent au
+cône isotrope le long de~$G$. Ce plan tangent est donc le même
+tout le long de la génératrice~$G$, et cette droite engendre une
+surface développable. Ainsi \emph{les droites isotropes des plans
+normaux à une courbe gauche décrivent \Card{2} développables et enveloppent
+\Card{2} développées de la courbe gauche}. Nous avons ainsi
+\Card{2} solutions particulières, et la détermination des développées
+doit s'achever par une seule quadrature.
+
+Effectivement, en supposant que $\DPtypo{v}{w}$~est l'arc~$s$ de~$(\phi')$, que
+$\alpha, \beta, \gamma$; $\alpha_{1}, \beta_{1}, \gamma_{1}$, sont les cosinus directeurs $a'\Add{,} b'\Add{,} c'$ de la normale
+principale et $a''\Add{,} b''\Add{,} c''$ de la binormale, l'équation générale
+se réduit, en désignant par $a, b, c$ les cosinus directeurs de la
+tangente,
+\[%[** TN: Filled in last two columns]
+dw \begin{vmatrix}
+a'  & b'  & c' \\
+a   & b   & c  \\
+a'' & b'' & c''
+\end{vmatrix} + ds \begin{vmatrix}
+a' + wa'' & b' + wb'' & c' + wc'' \\
+a & b & c \\
+-\mfrac{a}{R} - \mfrac{a''}{T} + w\mfrac{a'}{T} &
+-\mfrac{b}{R} - \mfrac{b''}{T} + w\mfrac{b'}{T} &
+-\mfrac{c}{R} - \mfrac{c''}{T} + w\mfrac{c'}{T}
+\end{vmatrix} = 0,
+\]
+%% -----File: 157.png---Folio 149-------
+c'est-à-dire
+\[
+- dw + \frac{ds}{T} (1 + w^{2}) = 0,
+\]
+ou enfin
+\[
+%[** TN: "tang." in original]
+w = \tg \int \frac{ds}{T}.
+\]
+On vérifie bien que l'équation différentielle en~$w$ admet les
+deux solutions: $w = ±i$, qui correspondent aux développables
+isotropes.
+
+Si on remarque de plus que la surface focale de la congruence
+des normales est la surface polaire de~$(\phi')$, c'est-à-dire
+que les points de contact des normales avec les développées
+sont sur la droite polaire, on retrouve tous les résultats
+essentiels sur la détermination des développées.
+
+
+\ExSection{VI}
+
+\begin{Exercises}
+\item[29.] On considère la congruence des tangentes communes aux deux
+surfaces $x^{2} + y^{2} = 2az$, $x^{2} + y^{2} = -2az$. Déterminer les développables
+de cette congruence: étudier leurs arêtes de rebroussement,
+leurs courbes de contact, leurs traces sur le plan
+$z = 0$.
+
+\item[30.] Si les deux multiplicités focales d'une congruence sont des
+développables isotropes (congruence isotrope), toutes les
+surfaces réglées qui passent par une même droite de la congruence
+ont même point central et même paramètre de distribution.
+Le plan perpendiculaire à chaque droite de la congruence
+\DPtypo{mene}{mène} à \DPtypo{egale}{égale} distance des deux points focaux enveloppe
+une surface minima. On peut obtenir ainsi la surface minima la
+plus générale.
+
+\item[31.] On suppose que les droites $D$~et~$D'$ de deux congruences se correspondent
+de manière que deux droites correspondantes soient
+parallèles. Si alors les développables des deux congruences se
+correspondent, les plans focaux de~$D$ sont parallèles à ceux
+de~$D'$; les droites $\Delta, \Delta_{1}$, qui joignent les points focaux
+correspondants se coupent en un point~$M$; le lieu de ce point
+admet $\Delta$~et~$\Delta_{1}$, pour tangentes conjuguées, et les courbes conjuguées
+enveloppées par ces droites correspondent aux développables
+des deux congruences.
+\end{Exercises}
+%% -----File: 158.png---Folio 150-------
+
+
+\Chapitre{VII}{Congruences de Normales.}
+
+\Section{Propriété caractéristique des congruences de normales.}
+{1.}{} Considérons une surface, les coordonnées d'un de ses
+points dépendent de deux paramètres; l'ensemble des normales
+à cette surface dépend de deux paramètres, et constitue
+une congruence. Pour obtenir les développables, il suffit
+de considérer sur la surface les \Card{2} \DPtypo{series}{séries} de lignes de
+courbure, puisque les normales à une surface en tous les
+points d'une ligne de courbure engendrent une surface développable.
+Le plan tangent à une développable passe par la normale~$D$
+et par la tangente à la ligne de courbure correspondante.
+C'est l'un des plans focaux de la droite~$D$. Ainsi \emph{les plans
+focaux sont les plans des sections principales de la surface.
+Les plans focaux d'une congruence de normales sont rectangulaires}.
+Il en résulte qu'une congruence quelconque ne peut pas
+en général être considérée comme formée des normales à une
+%[** TN: Added parentheses between \gamma, \gamma']
+surface. Considérons les \Card{2} lignes de courbure $(\gamma)\Add{,}(\gamma')$ passant
+par un point~$M$ de la surface; à la développable
+de~$(\gamma)$ correspond une arête de
+rebroussement~$(A)$ dont le plan osculateur
+est le plan focal, le point de contact~$F$
+de~$A$ et de la droite~$D$ est un des points
+focaux. On peut considérer l'arête de rebroussement~$(A)$
+comme étant l'enveloppe de
+la droite~$D$ quand le point~$M$ se déplace
+%% -----File: 159.png---Folio 151-------
+sur la courbe~$(\gamma)$; le point~$F$ est alors l'un des centres de
+courbure principaux de la surface au point~$M$. Le plan focal
+associé est le \Ord{2}{e} plan de section principale~$FMT'$. On aura de
+même une \Ord{2}{e} arête de rebroussement~$(A')$ en \DPtypo{considerant}{considérant} la courbe~$(\gamma')$.
+
+%[** TN: Exchanged diagram labels for (T') and (\gamma')]
+\Illustration[1.5in]{158a}
+On verra facilement que ces propriétés des centres de
+courbure principaux et des plans de sections principales subsistent,
+quelle que soit la nature des multiplicités focales
+de la congruence considérée.
+
+\emph{Réciproque}. Prenons une congruence constituée par les
+droites~$D$
+\begin{alignat*}{2}%[** TN: Stacked to accommodate figure]
+x &= f(v,w) &&+ u · a(v,w), \\
+y &= g(v,w) &&+ u · b(v,w), \\
+z &= h(v,w) &&+ u · c(v,w).
+\end{alignat*}
+Cherchons à quelles conditions on peut déterminer sur la droite~$D$
+un point~$M$ dont le lieu soit une surface constamment normale
+à~$D$. Il suffit que l'on puisse déterminer~$u$ en fonction
+de~$v\Add{,}w$ de façon que l'on ait
+\[
+\sum a\,dx = 0,
+\]
+ou
+\[
+\sum a(df + u\, da + a\, du) = 0.
+\]
+On peut supposer que $a\Add{,} b\Add{,}c$ soient les cosinus directeurs; alors
+$\sum a^{2} = 1$, et $u$~représentera la distance du point~$P$ où la droite
+rencontre le support, au point~$M$. On a en même temps $\sum a\, da = 0$
+et la condition précédente devient
+\[
+du + \sum a\, df = 0;
+\]
+\DPtypo{Cette}{cette} équation peut encore s'écrire
+\[
+\Tag{(1)}
+-du = \sum a\, df.
+\]
+Elle exprime que $\sum a\, df$ est une différentielle totale exacte;
+or\Add{,} on a
+\[
+\sum a\, df
+  = \sum a \frac{\dd f}{\dd v}\, dv
+  + \sum a \frac{\dd f}{\dd w}\, dw;
+\]
+%% -----File: 160.png---Folio 152-------
+la condition est donc:
+\[
+\frac{\dd}{\dd w} \sum a\Add{·} \frac{\dd f}{\dd v}
+  = \frac{\dd}{\dd v} \sum a\Add{·} \frac{\dd f}{\dd w},
+\]
+ou:
+\[
+\sum \frac{\dd a}{\dd w}\Add{·} \frac{\dd f}{\dd v}
+  = \sum \frac{\dd a }{\dd v} · \frac{\dd f}{\dd w},
+\]
+ou enfin:
+\[
+\Tag{(2)}
+\sum \left(\frac{\dd a}{\dd w}\Add{·} \frac{\dd f}{\dd v}
+         - \frac{\dd a}{\dd v} · \frac{\dd f}{\dd w}\right) = 0.
+\]
+Nous trouvons une condition unique. Or\Add{,} nous avons trouvé précédemment
+comme condition nécessaire l'orthogonalité des plans
+focaux. Nous sommes donc conduits à comparer les deux conditions\Add{.}
+Les coefficients $A\Add{,} B\Add{,} C$ d'un plan focal vérifient les relations
+\begin{gather*}
+%[** TN: Moved equation number per errata list]
+\Tag{(3)}
+Aa + Bb + Cc = 0, \\
+\left\{
+\begin{alignedat}{4}
+  &   A\left(\frac{\dd f}{\dd v} + u \frac{\dd a}{\dd v}\right)
+  &&+ B\left(\frac{\dd g}{\dd v} + u \frac{\dd b}{\dd v}\right)
+  &&+ C\left(\frac{\dd h}{\dd v} + u \frac{\dd c}{\dd v}\right) &&= 0, \\
+%
+  &   A\left(\frac{\dd f}{\dd w} + u \frac{\dd a}{\dd w}\right)
+  &&+ B\left(\frac{\dd g}{\dd w} + u \frac{\dd b}{\dd w}\right)
+  &&+ C\left(\frac{\dd h}{\dd w} + u \frac{\dd c}{\dd w}\right) &&= 0.
+\end{alignedat}
+\right.
+\end{gather*}
+\DPchg{Eliminant}{Éliminant} $u$ entre les \Card{2} dernières équations, nous avons
+\[
+\Tag{(4)}
+\begin{vmatrix}
+\sum A \mfrac{\dd f}{\dd v} & \sum A \mfrac{\dd a}{\dd v} \\
+\sum A \mfrac{\dd f}{\dd w} & \sum A \mfrac{\dd a}{\dd w}
+\end{vmatrix} = 0.
+\]
+Les coefficients de direction des normales aux plans focaux
+sont définis par \Eq{(3)}\Add{,}~\Eq{(4)}. Si nous considérons $A\Add{,} B\Add{,} C$ comme coordonnées
+courantes, \Eq{(3)}~représente un plan passant par l'origine,
+\Eq{(4)}~un cône ayant pour sommet l'origine; et les génératrices
+d'intersection sont précisément les normales cherchées.
+Exprimons que ces deux droites sont rectangulaires; le plan~\Eq{(3)}
+est perpendiculaire à la droite~$(a\Add{,} b\Add{,} c)$, qui est sur le cône~\Eq{(4)},
+car on a, puisque $\sum a^{2} = 1$ et $\sum a\, da = 0$
+\[
+\sum a \frac{\dd a}{\dd v} = 0, \qquad
+\sum a \frac{\dd a}{\dd w} = 0;
+\]
+donc les \Card{2} normales sont perpendiculaires à la droite~$(a\Add{,} b\Add{,} c)$;
+si elles sont rectangulaires, c'est que le cône~\Eq{(4)} est capable
+d'un trièdre trirectangle inscrit, ce qui donne la condition
+%% -----File: 161.png---Folio 153-------
+\[
+\sum \left(\frac{\dd f}{\dd v} · \frac{\dd a}{\dd w}
+         - \frac{\dd f}{\dd w} · \frac{\dd a}{\dd v}\right) = 0;
+\]
+ce qui est précisément la condition~\Eq{(2)}. De sorte que \emph{la condition
+nécessaire et suffisante pour que la congruence soit une
+congruence de normales, c'est que les plans focaux soient rectangulaires}.
+
+Supposons satisfaite la condition~\Eq{(2)}. Pour obtenir la
+surface normale à toutes les droites de la congruence, il suffit
+de calculer~$u$ en fonction de~$v\Add{,}w$, ce qui se fait par l'équation~\Eq{(1)}
+\[
+du = d\Phi(v, w),
+\]
+d'où
+\[
+u = \Phi(v, w) + \cte[.]
+\]
+Il y a donc une infinité de surfaces répondant à la question;
+si un point~$M$ décrit une surface~$(S)$ et un point~$M'$ une surface~$(S')$
+répondant à la question, on a $u = PM$, $u' = PM'$, la distance~$MM'$
+sera une quantité constante. Les surfaces~$(S)\Add{,} (S')$ sont appelées
+\emph{surfaces parallèles} et \emph{une famille de surfaces parallèles
+admet pour chaque normale mêmes centres de courbure principaux
+et mêmes multiplicités focales}; ces multiplicités focales
+constituent la \emph{développée} de l'une quelconque de ces surfaces.
+
+\Section{Relations entre une surface et sa développée.}
+{2.}{} Considérons une nappe de la développée d'une
+surface~$(S)$. Supposons d'abord que ce soit une surface~$(\Phi)$.
+Considérons une droite~$D$ de la congruence des normales à~$(S)$;
+cette droite est tangente en~$F$ à l'arête de rebroussement~$(A)$
+qui appartient à~$(\Phi)$; les plans focaux associés à~$D$ sont
+le plan osculateur à~$(A)$ et le plan tangent à~$(\Phi)$. Pour que la
+%% -----File: 162.png---Folio 154-------
+congruence soit une congruence
+de normales, il faut et il suffit
+que le plan osculateur à~$(A)$
+soit normal à~$(\Phi)$, donc que $(A)$
+soit une géodésique de~$(\Phi)$. \emph{La
+congruence des normales à la
+surface~$(S)$ est constituée par
+les tangentes à une famille de
+géodésiques de sa développée~$(\Phi)$\Add{.}
+Et réciproquement les tangentes
+à une famille de $\infty^{1}$~géodésiques d'une surface quelconque~$(\Phi)$
+constituent une congruence de normales.} Soit $M$~le point où la
+droite~$D$ coupe la surface~$(S)$; lorsque la droite~$D$ roule sur
+l'arête de rebroussement~$(A)$, le point~$M$ décrit une ligne de
+courbure~$(\gamma)$ de~$(S)$. A chaque point~$M$ de~$(S)$ correspond un
+point~$F$ de~$(\Phi)$; il y a correspondance point par point entre
+les \Card{2} surfaces; à la famille de lignes de courbure~$(\gamma)$ de~$(S)$
+correspond une famille de géodésiques de~$(\Phi)$. Voyons maintenant
+les courbes de contact~$(c)$ de~$(\Phi)$; considérons la tangente~$F\theta$
+à~$(c)$, c'est la caractéristique du plan tangent à~$(\Phi)$
+lorsque le point~$M$ décrit~$(\gamma)$; or\Add{,} ce plan tangent à~$(\Phi)$ est le
+\Ord{2}{e} plan focal, c'est le plan perpendiculaire au plan~$FMT$ passant
+par~$FM$, c'est donc le plan normal à~$(\gamma)$ au point~$M$. Donc~$F\theta$
+est la caractéristique du plan normal à~$(\gamma)$, c'est la droite
+polaire de~$(\gamma)$. \emph{Les courbes de contact de~$(\Phi)$ sont les
+courbes tangentes aux droites polaires des différents points
+des courbes~$(\gamma)$.} $F\theta$~étant dans le plan normal à~$(\gamma)$ rencontre
+la tangente à la \Ord{2}{e} section principale; elle passe au centre
+%% -----File: 163.png---Folio 155-------
+de courbure géodésique de~$(\gamma)$ sur~$(S)$.
+
+\begin{figure}[hbt]
+\centering
+\Input[1.75in]{162a}\hfil\hfil
+\Input[2.5in]{163a}
+\end{figure}
+\MarginNote{Surface Canal\Add{.}}
+Supposons que l'une des nappes de la développée se réduise
+à une courbe~$(\phi)$. La droite~$D$ rencontre~$(\phi)$ en l'un des
+points focaux~$F$. L'une des développables passant par~$D$ est un
+cône de soumet~$F$; l'une des lignes
+de courbure~$(\gamma)$ de~$(S)$ passant
+par~$M$ est située sur un
+cône de sommet~$F$. Or\Add{,} $(\gamma)$~est
+constamment normale à~$D$, c'est
+donc une trajectoire orthogonale
+des génératrices du cône; c'est
+l'intersection de ce cône avec une sphère de centre~$F$. Cette
+sphère en chaque point~$M$ a pour normale la droite~$D$, elle est
+donc tangente à la surface~$(S)$ tout le long de la courbe~$(\gamma)$.
+A chaque point~$F$ de~$(\phi)$ correspond une sphère ayant ce point
+pour centre et tangente à~$(S)$ tout le long de la ligne de
+courbure correspondante. \emph{La surface~$(S)$ est l'enveloppe d'une
+famille de sphères dépendant d'un paramètre.} Nous l'appellerons
+une \emph{surface canal}. La réciproque est vraie, comme on le
+verra plus loin. La courbe~$(\gamma)$ est alors l'intersection d'une
+sphère avec une sphère infiniment voisine; c'est un cercle.
+Le cône~$F$ est de révolution, l'axe de ce cône est la position
+limite de la ligne des centres, c'est la tangente~$Fu$ à~$(\phi)$.
+Considérons la tangente~$MT$ à~$(\gamma)$: $MT$~tangente en un point du
+cercle est orthogonale à~$Fu$, $Fu$ est donc dans le \Ord{2}{e} plan de
+section principale. \emph{Les congruences considérées sont donc formées
+des génératrices de $\infty^{1}$~cônes de révolution, dont les axes
+sont tangents à la courbe lieu des sommets} de ces cônes. Et
+%% -----File: 164.png---Folio 156-------
+\emph{réciproquement toute congruence ainsi constituée est une congruence
+de normales}, car les plans focaux sont les plans tangents
+et les plans méridiens de ces cônes, et sont par conséquent
+rectangulaires.
+
+\MarginNote{Cyclide de Dupin.}
+Voyons si les \Card{2} nappes de la développée peuvent se réduire
+à \Card{2} courbes $(\phi)\Add{,} (\phi')$. Les développables de la congruence sont
+les cônes ayant leur sommet sur l'une des courbes et passant
+par l'autre. Tous les cônes~$F$ de révolution doivent passer par la
+courbe~$(\phi')$. Cette courbe~$(\phi')$ est telle qu'il passe par
+cette courbe une infinité de cônes de révolution. De même~$(\phi)$;
+$(\phi), (\phi')$ ne peuvent donc être que des biquadratiques gauches ou
+leurs éléments de décomposition. Aucune de ces courbes ne peut
+être une biquadratique gauche, sans quoi par chacune d'elles
+il passerait \Card{4} cônes du \Ord{2}{e} degré seulement. Voyons si l'une
+d'elles peut être une cubique gauche; les cônes du \Ord{2}{e} degré
+passant par~$(\phi')$ ont leurs sommets sur~$\Err{(\phi)}{(\phi')}$: les \Card{2} courbes $(\phi)\Add{,}
+(\phi')$ seraient confondues. Voyons donc s'il peut exister des cubiques
+gauches telles que les cônes du \Ord{2}{e} degré qui les contiennent
+soient de révolution. Un tel cône aurait pour axe
+la tangente~$Fu$; or\Add{,} il contient cette
+tangente, donc il se décompose. Donc ni~$(\phi)$
+ni~$(\phi')$ ne peuvent être des cubiques
+gauches. Supposons donc que~$(\phi')$ soit une
+conique; le lieu des sommets des cônes de
+révolution passant par cette conique est
+une autre conique, qui est dite focale de la \Ord{1}{ère}. Il y a réciprocité
+entre ces coniques, et les cônes de révolution ont
+%% -----File: 165.png---Folio 157-------
+pour axes les tangentes aux focales. Donc \emph{les droites rencontrant
+\Card{2} coniques focales l'une de l'autre constituent une congruence
+de normales}. Les surfaces normales à ces droites s'appellent
+\emph{Cyclides de Dupin. Leurs \Card{2} systèmes de lignes de courbure
+sont des cercles}.
+
+\Illustration[1.25in]{164a}
+Supposons en particulier que $(\phi')$ soit un cercle; alors
+le lieu les sommets des cônes de révolution passant par $(\phi')$~est
+l'axe~$(\phi)$ de ce cercle, et nous voyons que toutes les
+droites qui s'appuient sur $(\phi)\Add{,} (\phi')$ sont normales à une famille
+de surfaces. Ces surfaces sont des \emph{tores} de révolution autour
+de l'axe~$(\phi)$, le lieu du centre du cercle méridien étant le
+cercle~$(\phi')$.
+
+Supposons que $(\phi')$ soit une droite, la surface est l'enveloppe
+d'une famille de sphères ayant leurs centres sur cette
+droite. C'est une surface de révolution autour de~$(\phi')$; la \Ord{1}{ère} nappe
+de la développée est la droite~$(\phi')$, la \Ord{2}{e} est engendrée
+par la rotation de la développée de la méridienne principale;
+pour que ce soit une courbe, il faut que la développée soit un
+point, donc que la méridienne soit un cercle, et nous retombons
+sur le cas du tore.
+
+\MarginNote{Cas singulier.}
+Cherchons enfin si les \Card{2} nappes de la développée peuvent
+être confondues. Alors les \Card{2} familles de lignes de courbure de
+la surface~$(S)$ sont confondues. C'est le cas des \emph{surfaces réglées
+à génératrices isotropes}. Les \Card{2} nappes se réduisent à
+une seule courbe, comme on le verra au paragraphe suivant.
+%% -----File: 166.png---Folio 158-------
+
+\Section{\DPchg{Etude}{Étude} des surfaces enveloppes de sphères.}
+{3.}{} Nous avons été amenés dans ce qui précède à considérer
+les surfaces enveloppes de sphères. Nous allons
+maintenant étudier les réciproques des propriétés précédentes.
+
+Considérons une surface~$(S)$ enveloppe de $\infty^{1}$ sphères~$(\Sigma)$.
+Chaque sphère coupe la sphère infiniment voisine suivant un
+cercle, et les normales à~$(S)$ en tous les points de ce cercle
+passent par le centre de la sphère. Le lieu des centres des
+sphères est une courbe rencontrée par les normales à~$(S)$,
+c'est une des nappes de la développée. D'autre part, la sphère~$(\Sigma)$
+étant tangente à la surface~$(S)$ tout le long du cercle caractéristique,
+ce cercle est une ligne de courbure de la
+surface~$(S)$, d'après le Théorème de Joachimsthal. \emph{Les surfaces
+enveloppes de sphères ont une famille de lignes de courbure
+circulaires. Réciproquement toute surface ayant une famille
+de lignes de courbure circulaires est une enveloppe de sphères\Add{.}}
+Considérons une ligne de courbure circulaire~$(K)$; toute sphère
+passant par~$(K)$ coupe la surface~$(S)$ sous un angle constant,
+d'après le Théorème de Joachimsthal. Or\Add{,} il est possible de
+trouver une sphère passant par~$(K)$ et tangente à~$(S)$ en l'un
+des points de ce cercle; cette sphère sera alors tangente à~$(S)$
+en tous les points du cercle~$(K)$, et toute ligne de courbure
+circulaire est courbe de contact d'une sphère avec la
+surface. La surface est l'enveloppe des sphères ainsi déterminées.
+
+Soit une sphère de centre $(a,b,c)$ et de rayon~$r$, $a,b,c,r$
+étant fonctions d'un même paramètre.
+%% -----File: 167.png---Folio 159-------
+La sphère a pour équation
+\[%[** TN: Omitted large brace grouping next two equations]
+(x - a)^{2} + (y - b)^{2} + (z - c)^{2} - r^{2} = 0;
+\]
+la caractéristique est en outre définie par l'équation
+\[
+(x - a)\, da + (y - b)\, db + (z - c)\, dc + r\, dr = 0;
+\]
+On vérifie bien que c'est un cercle dont le plan est perpendiculaire
+à la direction~$da, db, dc$, de la tangente au lieu des
+centres des sphères.
+
+Nous venons de considérer les surfaces dont une famille
+de lignes de courbure est constituée par des cercles. Voyons
+si les \Card{2} familles de lignes de courbure peuvent être circulaires.
+La surface correspondante pourra être considérée de
+\Card{2} façons différentes comme l'enveloppe de $\infty^{1}$~sphères. Les \Card{2} nappes
+de la développée sont des courbes. Nous obtenons la
+Cyclide de Dupin, que nous allons étudier à un point de vue
+nouveau.
+
+\MarginNote{Correspondance
+entre les
+droites et les
+sphères.}
+Les droites et les sphères sont des éléments géométriques
+dépendant de \Card{4} paramètres. Ce fait seul permet de prévoir
+qu'il y aura une espèce de correspondance entre l'étude des
+systèmes de droites et celle des systèmes de sphères. Cette
+correspondance trouve son expression analytique dans une
+transformation, due à Sophus Lie, et que nous exposerons plus
+tard. Mais nous la verrons se manifester auparavant dans diverses
+questions. C'est ainsi que l'on peut considérer dans la
+géométrie des sphères les enveloppes de $\infty^{1}$~sphères comme correspondant
+aux surfaces réglées, lieux de $\infty^{1}$~droites; la cyclide
+de Dupin correspond alors aux surfaces doublement réglées,
+%% -----File: 168.png---Folio 160-------
+donc aux surfaces réglées du \Ord{2}{e} degré. Nous allons voir
+l'analogie se développer dans l'étude qui suit.
+
+%[** TN: Several {1}-like superscripts rendered as prime accents]
+Soit $(\Sigma)$ une sphère de la \Ord{1}{ère} famille, $(\Sigma')$~une sphère
+de la \Ord{2}{e} famille, $(\Sigma)$~touche~$(S)$ suivant un cercle~$(K)$, $\Sigma'$~touche~$(S)$
+suivant un cercle~$(K')$. La surface~$(S)$ étant engendrée
+par le cercle~$(c)$ ou par le cercle~$(c')$, il en résulte que ces
+\Card{2} cercles ont au moins un point commun~$M$\DPtypo{.}{}; soient $O\Add{,} O'$ les centres
+des sphères $(\Sigma)\Add{,} (\Sigma')$, $OM$~et~$O'M$ sont normales aux sphères
+$(\Sigma)\Add{,} (\Sigma')$ et par suite normales en~$M$ à la surface. Donc elles
+coïncident, $O\Add{,} M\Add{,} O'$~sont sur une même droite; les sphères $(\Sigma)\Add{,} (\Sigma')$
+sont tangentes en~$M$. \emph{Une sphère de l'une des familles est
+tangente à une sphère quelconque de l'autre famille.} (Deux génératrices
+de systèmes différents d'une quadrique se rencontrent).
+
+Considérons \Card{3} sphères fixes $(\Sigma), (\Sigma_{1}), (\Sigma_{2})$ d'une des familles.
+Elles sont tangentes à toutes les sphères de l'autre famille,
+et par suite \emph{la surface est l'enveloppe des sphères
+tangentes à \Card{3} sphères fixes}. (Une quadrique est le lieu d'une
+droite rencontrant \Card{3} droites fixes). Les \Card{3} sphères $(\Sigma), (\Sigma_{1}), (\Sigma_{2})$
+se coupent en \Card{2} points qui peuvent être considérés comme des
+sphères de rayon nul tangentes à $(\Sigma), (\Sigma_{1}), (\Sigma_{2})$; donc il y a \Card{2} sphères
+de rayon nul dans chaque famille de sphères enveloppées
+par la cyclide. Les sphères de l'autre famille devant
+être tangentes à ces \Card{2} sphères de rayon nul passent par leurs
+centres. Ces deux points sont sur le lieu des centres des
+sphères, donc sur les coniques focales; \emph{si donc nous considérons
+les \Card{2} coniques focales, les sphères d'une des familles
+ont leurs centres sur l'une des coniques et passent par \Card{2} points
+%% -----File: 169.png---Folio 161-------
+fixes de l'autre, symétriques par rapport au plan de
+la \Ord{1}{ère}.} Il est alors facile, avec cette génération, de trouver
+l'équation de la cyclide.
+
+\Section{\DPchg{Equation}{Équation} de la
+\DPtypo{Cyclide}{cyclide} de
+Dupin.}
+{}{\normalfont\Primo.} Supposons d'abord que l'une des coniques soit une
+ellipse, par exemple: l'autre est une hyperbole. Prenons pour
+axes $\DPtypo{ox, oy}{Ox, Oy}$ les axes de l'ellipse, dont l'équation dans son
+plan est:
+\[
+\Tag{(E)}
+\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - 1 = 0.
+\]
+
+\Illustration{169a}
+\noindent L'hyperbole focale est dans le plan
+$y = 0$. Elle a pour équations
+\[
+\Tag{(H)}
+y = 0, \qquad
+\frac{x^{2}}{a^{2} - b^{2}} - \frac{z^{2}}{b^{2}} - 1 = 0\Add{.}
+\]
+Un point~$\omega$ de l'ellipse~\Eq{(E)} a pour
+coordonnées
+\[
+x = a \cos\phi, \qquad
+y = b \sin\phi, \qquad
+z = 0.
+\]
+Soit sur l'hyperbole~\Eq{(H)} le point fixe~$A$ de coordonnées
+\[
+x_{0}, \qquad
+y_{0} = 0, \qquad
+z_{0}^{2} = b^{2} \left(\frac{x_{0}^{2}}{a^{2} - b^{2}} - 1\right).
+\]
+L'équation d'une sphère~$\Sigma$ ayant pour centre~$\omega$ et passant par
+le point~$A$ sera
+\[
+\DPtypo{(x - a \cos\phi^{2})}{(x - a \cos\phi)^{2}}
+  + (y - b \sin\phi)^{2} + z^{2}
+  = (x_{0} - a \cos\phi)^{2} + b^{2} \sin^{2} \phi
+  + b^{2} \left(\frac{x_{0}^{2}}{a^{2} - b^{2}} - 1\right),
+\]
+ou
+\[
+x^{2} + y^{2} + z^{2} - 2ax \cos\phi - 2by \sin\phi
+  = x_{0}^{2} + b^{2} \frac{x_{0}^{2}}{a^{2} - b^{2}}
+  - b^{2} - 2ax_{0} \cos\phi,
+\]
+ou
+\[
+2a (x - x_{0}) \cos\phi + 2by \sin\phi
+  = x^{2} + y^{2} + z^{2} + b^{2} - \frac{a^{2} x_{0}^{2}}{c^{2}},
+\]
+en posant comme d'habitude
+\[
+c^{2} = a^{2} - b^{2}.
+\]
+L'équation de la sphère est de la forme
+\[
+A \cos\phi + B \sin\phi = C\Add{,}
+\]
+%% -----File: 170.png---Folio 162-------
+la condition pour qu'il y ait une racine double, c'est-à-dire,
+l'équation de l'enveloppe, est
+\[
+A^{2} + B^{2} = C^{2}\Add{.}
+\]
+Donc la cyclide a pour équation:
+\[
+4a^{2} (x - x_{0})^{2} + 4b^{2} y^{2}
+  = \left(x^{2} + y^{2} + z^{2} + b^{2} - \frac{a^{2} x_{0}^{2}}{c^{2}}\right)^{2}.
+\]
+
+\ParItem{\Secundo.} Supposons maintenant qu'une des coniques soit une
+parabole. L'autre est aussi une parabole. Prenons les axes ordinaires,
+nous avons pour équations
+des \Card{2} coniques
+\begin{align*}
+\Tag{(P)}
+z &= 0, \qquad y^{2} = 2px, \\
+\Tag{(P')}
+y &= 0, \qquad x^{2} + z^{2} = (x - p)^{2}.
+\end{align*}
+
+%\Illustration{170a}
+\begin{wrapfigure}[11]{O}{2.125in}
+\smash[t]{\raisebox{-1.75in}{\Input{170a}}}
+\end{wrapfigure}
+\noindent Le centre~$C$ de la sphère sur la parabole~$P$
+a pour coordonnées
+\[
+x = 2p \lambda^{2}, \qquad
+y = 2p \lambda, \qquad
+z = 0.
+\]
+Le point fixe~$A$ sur la parabole~$P'$ a pour coordonnées
+\[
+x_{0}, \qquad
+y_{0} = 0, \qquad
+z_{0}^{2} = (x_{0} - p)^{2} - x_{0}^{2}.
+\]
+L'équation de la sphère est
+\[
+(x - 2p \lambda^{2})^{2} + (y - 2p \lambda)^{2} + z^{2}
+  = (x_{0} - 2p \lambda^{2})^{2} + 4p^{2} \lambda^{2}
+  + (x_{0} - p)^{2} - x_{0}^{2},
+\]
+ou
+\[
+x^{2} + y^{2} + z^{2}
+  - (x_{0} - p)^{2} - 4p \lambda y - 4p (x - x_{0}) \lambda^{2} = 0,
+\]
+et l'équation de l'enveloppe, c'est-à-dire de la cyclide, est
+\[
+\bigl[x^{2} + y^{2} + z^{2} - (x_{0} - p)^{2}\bigr] (x - x_{0}) + py^{2} = 0.
+\]
+L'ordre de la surface, qui est on général~$4$, s'abaisse ici
+à~$3$.
+
+\MarginNote{Surface canal
+isotrope.}
+Parmi les surfaces réglées, nous avons considéré les surfaces
+développables, où chaque génératrice rencontre la génératrice
+infiniment voisine. Le cas correspondant pour les enveloppes
+de sphères sera celui où chaque sphère est tangente
+à la sphère infiniment voisine. Pour qu'il en soit ainsi, il
+faut que le plan radical des \Card{2} sphères leur soit tangent.
+%% -----File: 171.png---Folio 163-------
+Prenons la sphère
+\[
+\Tag{(1)}
+(x - a)^{2} + (y - b)^{2} + (z - c)^{2} - r^{2} = 0.
+\]
+Le plan radical de cette sphère et de la sphère infiniment
+voisine est
+\[
+\Tag{(2)}
+(x - a)\, da + (y - b)\, db + (z - c)\, dc + r\, dr = 0;
+\]
+pour qu'il soit tangent à la sphère~\Eq{(1)} il faut et il suffit
+que sa distance au centre~$(a\Add{,} b\Add{,} c)$ soit égale à~$±r$, donc que
+l'on ait
+\[
+\frac{r\, dr}{\sqrt{da^{2} + db^{2} + dc^{2}}} = ±r,
+\]
+ou
+\[
+\Tag{(3)}
+da^{2} + db^{2} + dc^{2} = dr^{2}.
+\]
+
+\Illustration[1.75in]{171a}
+\noindent Donc $r$~n'est autre que l'arc~$S$ de la
+courbe~$(c)$ lieu des centres des sphères,
+cet arc étant compté à partir
+d'une origine arbitraire. Cherchons
+le point de contact de la sphère
+avec la sphère infiniment voisine.
+Les coordonnées satisfont aux équations
+\[
+\frac{x - a}{da} = \frac{y - b}{db} = \frac{z - c}{dc}
+  = \frac{-r\, dr}{\Err{dr}{dr^2}} = -\frac{r}{dr} = - \frac{s}{ds};
+\]
+d'où
+\[
+x = a - s \frac{da}{ds} = a - s \alpha, \quad
+y = b - s \beta, \quad
+z = c - s \gamma,
+\]
+$\alpha\Add{,} \beta\Add{,} \gamma$ étant les cosinus directeurs de la tangente. On obtient
+ainsi le point~$I$, qui décrit une développante~$(r)$ de la courbe~$(c)$.
+L'intersection d'une sphère avec la sphère infiniment
+voisine n'est autre que l'intersection de cette sphère avec
+un de ses plans tangents: c'est un couple de droites isotropes
+se coupant au point~$I$. \emph{L'enveloppe se compose de deux surfaces
+réglées à génératrices isotropes. Nous l'appellerons une
+surface canal isotrope. Réciproquement une surface réglée à
+%% -----File: 172.png---Folio 164-------
+génératrices isotropes est une nappe d'une enveloppe de sphères.}
+Considérons en effet une génératrice~$D$ et le cercle de
+l'infini. Par la génératrice isotrope~$D$ passent une infinité
+de sphères; ces sphères contiennent la droite~$D$ et le cercle
+imaginaire à l'infini, ce qui donné \Card{7} conditions; elles dépendent
+de \Card{2} paramètres arbitraires. Nous pouvons faire en sorte
+que la sphère soit tangente à la surface considérée $(S)$ en \Card{2} points
+à distance finie de la droite~$D$; la sphère est alors
+déterminée; mais de plus elle est tangente à la surface~$(S)$ au
+point à l'infini sur~$D$, donc elle se raccorde avec~$(S)$ tout le
+long de la génératrice~$D$. La surface~$(S)$ fera partie de l'enveloppe
+de ces sphères.
+
+Sur une telle surface, les \Card{2} systèmes de lignes de courbure
+sont confondus avec les génératrices isotropes, les \Card{2} nappes
+de la développée sont confondues avec la courbe~$(C)$. La
+courbe~$(r)$ joue ici un rôle analogue à l'arête de rebroussement
+des surfaces développables. En effet, pour une développable,
+il y a un élément de contact (point de l'arête de rebroussement
+et plan osculateur en ce point) commun à une génératrice
+et à la génératrice infiniment voisine. Ici, c'est
+l'élément de contact constitué par le point~$I$ et le plan tangent
+à la sphère en ce point, plan normal à~$I\omega$, qui est commun
+à la sphère~$(\Sigma)$ et à la sphère infiniment voisine. Le point~$I$
+est un ombilic de la surface~$(S)$. La ligne~$(r)$ en est une ligne
+double, c'est un lieu d'ombilics. Nous l'appellerons la
+\emph{ligne ombilicale} de la surface canal isotrope.
+
+%[** TN: In the manuscript, S_{0} is a tiny 0 set directly below the S]
+\Section{Lignes de courbure et lignes asymptotiques.}
+{4.}{} Considérons une surface~$(S_{0})$ et une ligne asymptotique.
+Les tangentes en chacun des points de cette ligne
+%% -----File: 173.png---Folio 165-------
+engendrent une développable, et l'élément de contact commun à
+une génératrice et à la génératrice infiniment voisine, comprenant
+un point de la ligne et le plan osculateur, qui est
+tangent à~$(S_{0})$, est un élément de contact de~$(S_{0})$. Considérons
+maintenant une ligne de courbure~$(\DPtypo{r}{\Gamma})$: la normale en chaque
+point engendre une développable. Soit $(c)$
+l'arête de rebroussement; $OI$~est égal à
+l'arc de~$(c)$; si donc nous considérons les
+sphères de centres~$\DPtypo{o}{O}$ et de rayons~$OI$, chacune
+de ces sphères touche la sphère infiniment
+voisine, et l'élément de contact
+$(I,P)$ commun à ces \Card{2} sphères est un élément de contact de la
+surface~$(S_{0})$.
+
+%[** TN: Fixed diagram label (T) -> (\Gamma) as per errata list]
+\Illustration[1.5in]{173a}
+Appelons \emph{sphère de courbure} de~$(S_{0})$ toute sphère ayant
+pour centre un centre de courbure principale et pour rayon le
+rayon de courbure principal correspondant. Et nous pourrons
+dire:
+
+%[** TN: Not marked for italicization in the original]
+\emph{Les sphères de courbure de~$(S_{0})$ qui correspondent à une
+même ligne de courbure~$(\Gamma)$, enveloppent une surface canal isotrope,
+ayant~$(\Gamma)$ pour ligne ombilicale.}
+
+\emph{Réciproquement}, si une surface canal isotrope~$(S)$ est
+circonscrite à la surface~$(S_{0})$ le long de sa ligne ombilicale,
+celle-ci étant ligne de courbure pour~$(S)$ sera ligne de courbure
+pour~$(S_{0})$, d'après le théorème de Joachimsthal.
+
+Les choses s'énoncent d'une manière plus nette en substituant
+à la notion de courbe la notion de \emph{bande}. Une bande
+sera, par définition, formée de $\infty^{1}$~éléments de contact appartenant
+à une même multiplicité: le lieu des points (de ces
+%% -----File: 174.png---Folio 166-------
+éléments de contact) sera une courbe, et les plans (de ces
+éléments de contact) seront tangents à la courbe aux points
+correspondants. Une bande appartenant à une surface sera formée
+des points d'une courbe tracée sur la surface, associés
+aux plans tangents à la surface en ces points. On appellera
+\emph{bande de rebroussement} d'une surface développable le lieu des
+éléments de contact communs à chaque génératrice et à la génératrice
+infiniment voisine. Et on appellera \emph{bande ombilicale}
+d'une surface canal isotrope le lieu des éléments de contact
+communs à chacune des sphères inscrites à la surface et à la
+sphère infiniment voisine.
+
+Appelant de même \emph{bandes asymptotiques}, \emph{bandes de courbure}
+les lieux des éléments de contact d'une surface appartenant
+aux lignes asymptotiques ou aux lignes de courbure de cette
+surface, on concluera:
+
+\emph{Une bande asymptotique d'une surface est la bande de rebroussement
+d'une développable; une bande de courbure d'une
+surface est la bande ombilicale d'une surface canal isotrope.
+Et réciproquement}: toute bande de rebroussement (d'une développable),
+qui appartient à une surface~$(S_{0})$, est bande asymptotique
+de~$(S_{0})$; toute bande ombilicale (d'une surface canal
+isotrope), qui appartient à une surface~$(S_{0})$, est bande de
+courbure pour~$(S_{0})$.
+
+On voit ainsi, qu'au point de vue de la correspondance
+entre droites et sphères, les lignes asymptotiques correspondent
+aux lignes de courbure.
+%% -----File: 175.png---Folio 167-------
+
+\Section{Bandes asymptotiques et \DPtypo{Bandes}{bandes} de \DPtypo{Courbure}{courbure}.}
+{}{Remarque \1.} Sur chaque élément de contact~$(M, P)$
+d'une bande, il y a \emph{deux éléments
+linéaires} à considérer. (Un élément
+linéaire étant formé d'un point et
+d'une droite passant par ce point).
+\Figure{175a}
+C'est \emph{l'élément linéaire tangent} formé
+du point~$M$ de l'élément et de la
+tangente~$(T)$ à la courbe qui sert de \emph{support} à la bande,
+qu'on peut appeler simplement la \emph{courbe de la bande}; et \emph{l'élément
+linéaire caractéristique} formé du point~$M$ et de la caractéristique~$(K)$
+du plan~$(P)$, c'est-à-dire de la génératrice
+rectiligne de la développable enveloppée par les plans~$(P)$,
+ou \emph{développable de la bande}. Ces deux éléments linéaires sont
+corrélatifs, au point de vue de la dualité; une bande est corrélative
+d'une bande.
+
+Dans une \emph{bande asymptotique}, $(T)$~et~$(K)$ sont confondues
+dans une \emph{bande de courbure}, $(T)$~et~$(K)$ sont rectangulaires;
+ces termes ont donc un sens par eux-mêmes, sans supposer une
+surface~$(S_{0})$ à laquelle appartienne la bande considérée. Si la
+bande de rebroussement est donnée, la développable correspondante
+est la développable de la bande. Si la bande de courbure
+est donnée, sa courbe~$(\gamma)$ est ligne de courbure de la développable
+de la bande; et la surface canal isotrope dont la
+bande ombilicale se confond avec cette bande de courbure est
+l'enveloppe des sphères de courbure de la développable, construites
+aux divers points~$M$. Les mots bande ombilicale, bande
+de courbure sont donc équivalents; de même que ceux de bande
+%% -----File: 176.png---Folio 168-------
+asymptotique et bande de rebroussement.
+
+Remarquons encore que, si l'on se donne une bande de
+courbure, la sphère de courbure qui correspond à un élément
+de contact~$(M,P)$ de la bande est définie par la condition
+d'admettre~$(M,P)$ pour un de ses éléments de contact et d'avoir
+son centre sur la droite polaire de la courbe~$(\gamma)$ lieu
+des points~$M$ (Voir \No2~et~\No3). Cette seconde condition exprime
+que la sphère a avec~$(\gamma)$ un contact du second ordre; de
+même que dans une bande asymptotique chaque plan~$(P)$ est osculateur
+à~$(\gamma)$. C'est donc une nouvelle analogie entre les bandes
+de courbure et les bandes asymptotiques.
+
+\Section{Lignes de courbure des enveloppes de sphères.}
+{5.}{} Nous connaissons déjà une des familles de lignes
+de courbure, celle constituée par les caractéristiques
+des sphères. Déterminons la \Ord{2}{e} famille. Soit $(c)$ le lieu
+des centres des sphères. Exprimons
+ses coordonnées en fonction
+de l'arc~$(S)$; l'une des sphères
+de centre~$\omega$ rencontre la sphère
+infiniment voisine suivant un
+cercle~$(K)$ dont le plan est normal
+à la tangente~$\omega T$. Introduisons
+le trièdre de Serret au
+point~$\omega$ de la courbe~$(c)$, et
+définissons par rapport à ce trièdre les coordonnées d'un
+point~$M$ de la surface, c'est-à-dire du cercle~$(K)$. Appelons $\theta$
+l'angle de~$\omega M$ avec~$\omega T$, cet angle est le même pour tous les
+points du cercle~$(K)$. Projetons $M$~en~$P$ sur le plan normal, et
+soit $\phi$~l'angle de~$\omega P$ avec~$\omega N$. Les coordonnées de~$M$ par
+%% -----File: 177.png---Folio 169-------
+rapport au trièdre de Serret sont, en appelant~$r$ le rayon de
+la sphère,
+\[
+\Tag{(1)}
+\xi   = r \cos\theta, \qquad
+\eta  = r \sin\theta \cos\phi, \qquad
+\zeta = r \sin\theta \sin\phi.
+\]
+Par rapport à un système d'axes quelconques, ces coordonnées
+sont, en appelant $x\Add{,}y\Add{,}z$ les coordonnées de~$\omega$
+\[
+\Tag{(2)}
+X = x + a \xi + a' \eta + a'' \zeta, \quad
+Y = y + b \xi + b' \eta + b'' \zeta, \quad
+Z = z + c \xi + c' \eta + c'' \zeta;
+\]
+
+\Illustration[2.5in]{176a}
+\noindent $r, \theta$ sont fonctions de~$S$; les paramètres variables sont $s$~et~$\phi$.
+\DPchg{Ecrivons}{Écrivons} que $(K)$~est le cercle caractéristique, nous avons
+\[
+\left\{
+\begin{gathered}
+\sum (X - x)^{2}- r^{2}= 0, \\
+\sum a(X - x) + r \frac{dr}{ds} = 0.
+\end{gathered}\right.
+\]
+En supposant que le trièdre de coordonnées coïncide avec le
+trièdre de Serret, cette équation devient:
+\[
+\xi + r \frac{dr}{ds} = 0,
+\]
+c'est-à-dire:
+\[
+r \cos\theta + r \frac{dr}{ds} = 0,
+\]
+ou
+\[
+\Tag{(3)}
+\Cos\theta = -\frac{dr}{ds}.
+\]
+$\theta$~est ainsi défini en fonction de~$S$.
+
+Une surface enveloppe de sphères est engendrée par des
+cercles; c'est une surface cerclée. Inversement, on peut chercher
+si une surface cerclée est une enveloppe de sphères. Le
+calcul précédent montre que, pour qu'il en soit ainsi, il
+faut que les axes des cercles engendrent une surface développable,
+et on outre, que l'on ait la condition~\Eq{(3)}.
+
+Cherchons les lignes de courbure. Ce sont les trajectoires
+orthogonales des cercles~$(K)$ définis par $S = \cte[]$. La tangente
+à une ligne quelconque passant par $M$ a pour coefficients
+directeurs
+\begin{alignat*}{7}%[** TN: Filled in last two equations]
+dX &= a\, ds &&+ \xi \frac{a'}{R}\, ds
+  &&+ \Err{\zeta}{\eta}
+          \left(-\frac{a}{R} - \frac{a''}{T} - \dots\right)ds
+  &&+ \zeta \frac{a'}{T}\, ds &&+ a\, d\xi &&+ a'\, d\eta  &&+ a''\, d\zeta, \\
+dY &= b\, ds &&+ \xi \frac{b'}{R}\, ds
+  &&+ \eta  \left(-\frac{b}{R} - \frac{b''}{T} - \dots\right)ds
+  &&+ \zeta \frac{b'}{T}\, ds &&+ b\, d\xi &&+ b'\, d\eta  &&+ b''\, d\zeta, \\
+dZ &= c\, ds &&+ \xi \frac{c'}{R}\, ds
+  &&+ \eta  \left(-\frac{c}{R} - \frac{c''}{T} - \dots\right)ds
+  &&+ \zeta \frac{c'}{T}\, ds &&+ c\, d\xi &&+ c'\, d\eta  &&+ c''\, d\zeta.
+\end{alignat*}
+%% -----File: 178.png---Folio 170-------
+En prenant de nouveau le trièdre de Serret pour trièdre de
+coordonnées, ces coefficients directeurs deviennent:
+\[
+\left(1 - \frac{\eta}{R}\right) ds + d\xi, \qquad
+\left(\frac{\xi}{R} + \frac{\zeta}{T}\right) ds + d\eta, \qquad
+- \frac{\eta}{T} ds + d\zeta.
+\]
+Pour la tangente au cercle~$(K)$, on a $ds = 0$, et les coefficients
+directeurs sont:
+\[
+\dd \xi= 0, \qquad
+\dd \eta = -r \sin\theta \sin\phi\, d\phi, \qquad
+\dd \zeta = r \sin\theta \cos\phi\, d\phi.
+\]
+La condition qui définit les trajectoires orthogonales des
+cercles~$(K)$ est donc
+\[
+-\left[\left(\frac{\xi}{R} + \frac{\zeta}{T}\right) ds + d \eta \right] \sin\phi
+  + \left[-\frac{\eta}{T}\, ds + d\zeta\right] \cos\phi = 0.
+\]
+Telle est l'équation différentielle des lignes de courbure.
+On peut l'écrire
+\begin{align*}%[** TN: Re-breaking]
+-\frac{r \cos\theta}{T}\, ds\Add{·} \sin\phi
+  &- \frac{ds}{T} \left(\zeta \sin\phi + \eta \cos\phi\right)
+     - d\eta \sin\phi + d\zeta \cos\phi = 0\Add{,}
+\\
+- \frac{r \cos\theta}{R}\, ds\Add{·} \sin\phi
+  &\begin{aligned}[t]- \frac{r \sin\theta}{T}\, ds
+    &- \sin\phi \bigl[d(r\sin\theta) · \cos\phi - r\sin\theta\Add{·} \sin\phi · d\phi\bigr] \\
+    &+ \cos\phi \bigl[d(r\sin\theta) \Add{·}\sin\phi + r\sin\theta · \cos\phi · d\phi\bigr] = 0,
+\end{aligned} \\
+-\frac{r \cos\theta}{R}\, ds · \sin\phi
+&-\frac{r \sin\theta}{T}\, ds + r \sin\theta · d\phi = 0,
+\end{align*}
+ou
+\[
+\frac{d\phi}{ds} = \frac{1}{T} + \frac{\cotg\theta · \sin\phi}{R};
+\]
+équation de la forme
+\[
+\frac{d\phi}{ds} = A \sin\phi + \beta.
+\]
+\DPtypo{si}{Si} on prend comme fonction inconnue $\tg \frac{\phi}{2}$, on est ramené à
+une équation de Riccati. Mais l'angle~$\phi$ est l'angle du rayon~$IM$
+avec un rayon fixe. Donc \emph{\Card{4} lignes de courbure du \Ord{2}{e} système
+coupent les cercles caractéristiques en \Card{4} points dont le rapport
+anharmonique est constant}. Nouvelle analogie avec les lignes
+asymptotiques d'une surface réglée.
+
+On a les simplifications connues si on a \textit{à~priori} une
+solution de l'équation. Ainsi si on considère l'enveloppe de
+%% -----File: 179.png---Folio 171-------
+sphères~$(\Sigma)$ ayant leurs centres dans un plan, tous les cercles
+caractéristiques sont orthogonaux à la section de la surface
+par ce plan, qui est alors une ligne de courbure. La détermination
+des lignes de courbure se ramène dans ce cas à \Card{2} quadratures.
+
+\Paragraph{Remarque.} Plus généralement, la détermination des trajectoires
+orthogonales d'une famille de $\infty^{1}$~cercles dépend de
+l'intégration d'une équation de Riccati. D'où des conclusions
+analogues aux précédentes.
+
+\Section{Cas où une des nappes de la développée est une développable.}
+{6.}{} Nous venons de considérer le cas où une des nappes
+de la développée d'une surface est une courbe. Corrélativement,
+considérons maintenant le cas où une des
+nappes de la développée est une surface développable.
+Alors les plans tangents à cette développable constituent une
+des familles de développables de la congruence; un tel plan~$P$
+coupe la surface suivant une courbe normale à toutes les droites
+de la congruence situées dans ce plan et qui sera une
+ligne de courbure. En tout point de cette ligne, la normale à
+la surface est dans le plan~$P$. Donc le plan~$P$ coupe orthogonalement
+la surface~$(S)$ tout le long de la ligne de courbure.
+
+\Illustration[1.875in]{180a}
+Réciproquement, si une surface coupe orthogonalement une
+famille de plans, ses sections par ces plans sont des lignes
+de courbure, d'après le Théorème de Joachimsthal, et ces
+plans, constituant une des familles de développables de la
+congruence des normales, enveloppent une développable, qui
+est une des nappes de la développée de la surface.
+
+Considérons la \Ord{2}{e} ligne de courbure passant par un point~$M$;
+%% -----File: 180.png---Folio 172-------
+sa tangente~$MU$ est perpendiculaire à
+la tangente~$MT$ à la \Ord{1}{ère}~ligne de courbure
+et à la normale~$MN$ à la surface; ces \Card{2} droites
+étant dans le plan~$P$, $MU$~est perpendiculaire
+au plan~$P$. \emph{Les lignes de
+courbure de la \Ord{2}{e} famille sont trajectoires
+orthogonales des plans~$P$.}
+
+Considérons une de ces trajectoires orthogonales~$(K)$; les
+plans~$P$ sont normaux à la courbe~$(K)$: l'une des nappes de la
+développée, celle qui est une développable, est ainsi l'enveloppe
+des plans normaux, ou la surface polaire de la courbe~$(K)$.
+\emph{Toutes les lignes de courbure~$(K)$ non planes ont donc
+même surface polaire, qui est l'enveloppe des plans des lignes
+de courbure planes. L'arête de rebroussement de cette surface
+est le lieu des centres des sphères osculatrices à la courbe~$(K)$.}
+La ligne~$(K)$ étant une ligne de courbure, les normales à
+la surface en tous les points de~$(K)$ forment une développable,
+et par suite enveloppent une développée de la courbe~$(K)$, qui
+est une géodésique de sa surface polaire. Si donc on part des
+plans~$P$, pour avoir les courbes~$(K)$ on est ramené à la recherche
+des géodésiques d'une surface développable, ce qui se réduit
+à des quadratures; et comme la surface cherchée peut
+être considérée comme engendrée par les courbes~$(K)$ on voit
+qu'on obtiendra cette surface par des quadratures.
+
+Partons des plans~$P$, et cherchons leurs trajectoires orthogonales.
+Considérons l'arête de rebroussement~$(A)$ de l'enveloppe
+des plans~$P$, et introduisons le trièdre de Serret en
+%% -----File: 181.png---Folio 173-------
+chaque point~$\omega$ de cette courbe, soit~$\DPchg{(\omega\xi\eta\zeta)}{\Tri{\omega}{\xi}{\eta}{\zeta}}$. Le plan~$P$ est
+le plan osculateur~$\xi\omega\eta$, et nous voulons chercher dans ce
+plan un point~$M(\xi, \eta)$ dont le lieu soit normal à~$P$. Les coordonnées
+de~$M$ sont
+\[
+X = x + a \xi + a' \eta, \qquad
+Y = y + b \xi + b' \eta, \qquad
+Z = z + c \xi + c' \eta,
+\]
+la direction de la tangente au lieu du point~$M$ est définie par
+\begin{alignat*}{5}%[** TN: Filled in last two equations]
+dX &= a · ds &&+ \xi \frac{\DPtypo{\alpha'}{a'}}{R}\, ds
+  &&- \eta \left(\frac{a}{R} + \frac{a''}{T}\right) ds
+  &&+ a\, d\xi &&+ a'\, d\eta, \\
+dY &= b · ds &&+ \xi \frac{b'}{R}\, ds
+  &&- \eta \left(\frac{b}{R} + \frac{b''}{T}\right) ds
+  &&+ b\, d\xi &&+ b'\, d\eta, \\
+dZ &= c · ds &&+ \xi \frac{c'}{R}\, ds
+  &&- \eta \left(\frac{c}{R} + \frac{c''}{T}\right) ds
+  &&+ c\, d\xi &&+ c'\, d\eta,
+\end{alignat*}
+expressions de la forme
+\[
+dX = Aa + Ba' + Ca'', \quad
+dY = Ab + Bb' + Cb'', \quad
+dZ = Ac + Bc' + Cc''.
+\]
+
+\DPchg{Ecrivons}{Écrivons} que cette direction est normale au plan~$\xi\omega\eta$,
+c'est-à-dire parallèle à la binormale $(a'', b'', c'')$. Nous avons
+$A = B = 0$, ou
+\[
+ds - \frac{\eta}{R} · ds + d\xi = 0, \qquad
+\frac{\xi}{R}\, ds + d\eta = 0;
+\]
+ou
+\[
+\frac{d\xi}{ds} = \frac{\eta}{R} - 1, \qquad
+\frac{d\eta}{ds} = - \frac{\xi}{R};
+\]
+$\xi, \eta$ sont donnés par \Card{2} équations différentielles du \Ord{1}{er} ordre.
+Il en résulte que par chaque point du plan~$P$ passe une trajectoire
+orthogonale et une seule. Il existe ainsi une correspondance
+point par point entre les divers plans~$P$, les points
+correspondants étant sur une même trajectoire orthogonale.
+
+\Illustration{181a}
+Considérons dans un plan~$P$ \Card{2} points
+$M, N$; et soit $D$ la droite~$MN$; lorsque
+le plan~$P$ varie, la droite~$D$ engendre
+une surface réglée sur laquelle les
+lieux des points $M$~et~$N$ sont trajectoires
+orthogonales des génératrices;
+or\Add{,} les trajectoires orthogonales interceptent
+sur les génératrices des
+%% -----File: 182.png---Folio 174-------
+segments égaux; il en résulte que si l'on considère \Card{2} positions
+$P, P'$, et les positions $MN, M'N'$ correspondantes, on a $MN =
+M'N'$. La correspondance entre les plans~$P$ transforme une courbe
+du plan~$P$ en une courbe égale. En particulier, les plans~$P$
+contenant les lignes de courbure planes, \emph{toutes ces lignes de
+courbure planes sont égales. La surface~$(S)$ est donc engendrée
+par le mouvement d'une courbe plane de forme invariable}. Pour
+la définir, il suffit de \DPchg{connaitre}{connaître} le mouvement de son plan~$P$.
+
+Pour cela, reprenons les équations
+\[
+\Tag{(1)}
+\frac{d\xi}{ds} - \frac{\eta}{R} + 1 = 0, \qquad
+\frac{d\eta}{ds} + \frac{\xi}{R} = 0,
+\]
+et intégrons-les. Considérons d'abord les équations sans \Ord{2}{e} membre
+\[
+R \frac{d\xi}{ds} - \eta = 0, \qquad
+R \frac{d\eta}{ds} + \xi = 0.
+\]
+Posons
+\[
+\frac{R}{ds} = \frac{1}{d\phi},
+\]
+d'où
+\[
+\Tag{(2)}
+d\phi = \frac{ds}{R};
+\]
+les équations deviennent
+\[
+\frac{d\xi}{d\phi} - \eta = 0, \qquad
+\frac{d\eta}{d\phi} + \xi = 0,
+\]
+équations linéaires sans \Ord{2}{e} membre à coefficients constants,
+dont la solution générale est
+\[
+\Tag{(3)}
+\xi = A \cos\phi + B \sin\phi, \qquad
+\eta = -A \sin\phi + B \cos\phi.
+\]
+Passons alors au système avec \Ord{2}{e} membre
+\[
+\Tag{(4)}
+\frac{d\xi}{d\phi} = \eta - R, \qquad
+\frac{d\eta}{d\phi} = - \xi.
+\]
+Considérons dans~\Eq{(3)} $AB$~comme des fonctions de~$\phi$, et cherchons
+à satisfaire au système~\Eq{(4)}. Nous avons
+\begin{alignat*}{4}
+\frac{d\xi}{d\phi}
+  &= \phantom{-}\eta &&+ \frac{dA}{d\phi} \cos\phi &&+ \frac{dB}{d\phi} \sin\phi
+  &&= \phantom{-}\eta - R, \\
+\frac{d\eta}{d\phi}
+  &= -\xi &&- \frac{dA}{d\phi} \sin\phi &&+ \frac{dB}{d\phi} \cos\phi
+  &&= -\xi;
+\end{alignat*}
+%% -----File: 183.png---Folio 175-------
+d'où
+\[
+ \frac{dA}{d\phi} \cos\phi + \frac{dB}{d\phi} \sin\phi = -R, \quad
+-\frac{dA}{d\phi} \sin\phi + \frac{dB}{d\phi} \cos\phi = 0;
+\]
+d'où
+\[
+\frac{dA}{d\phi} = - R \cos\phi, \qquad
+\frac{dB}{d\phi} = - R \sin\phi;
+\]
+ou, en réintroduisant $s$ d'après la formule~\Eq{(2)},
+\[
+\frac{dA}{ds} = -\cos\phi, \qquad
+\frac{dB}{ds} = -\sin\phi;
+\]
+et
+\[
+A = -\int \cos\phi · ds, \qquad
+B = -\int \sin\phi · ds.
+\]
+Posons
+\[
+x_{0} = \int \cos\phi · ds, \qquad
+y_{0} = \int \sin\phi · ds;
+\]
+alors
+\[
+A = -x_{0}, \qquad
+B = -y_{0}.
+\]
+Nous avons donc une solution particulière
+\[
+\xi = -x_{0} \cos\phi - y_{0} \sin\phi, \qquad
+\eta = x_{0} \sin\phi - y_{0} \cos\phi;
+\]
+et la solution générale est, $x_{1}\Add{,} y_{1}$ désignant \Card{2} constantes arbitraires,
+\[
+\Tag{(5)}
+\left\{
+\begin{alignedat}{2}%[** TN: Set on one line in original; added brace]
+\xi  &= \phantom{-} (x_{1} - x_{0}) \cos\phi &&+ (y_{1} - y_{0}) \sin\phi, \\
+\eta &= -(x_{1} - x_{0}) \sin\phi &&+ (y_{1} - y_{0}) \cos\phi.
+\end{alignedat}
+\right.
+\]
+Nous avons \Card{3} quadratures à effectuer. Interprétons géométriquement
+ces résultats:
+
+Les formules précédentes, résolues en $\Err{x, y}{x_{1}, y_{1}}$, donnent
+\[
+\Tag{(6)}
+x_{1} = x_{0} + \xi \cos\phi - \eta \sin\phi, \quad
+y_{1} = y_{0} + \xi \sin\phi + \eta \cos\phi.
+\]
+
+\Illustration[2.25in]{183a}
+Prenons dans le plan~$P$ deux axes
+fixes $0_{1}x_{1}\Add{,} \DPtypo{}{0_{1}}y_{1}$, et construisons par
+rapport à ces axes la courbe~$(R)$ lieu
+du point $(x_{0}\Add{,} y_{0})$. La courbe~$(R)$ est
+la courbe du plan~$P$ qui a même rayon
+de courbure que l'arête de rebroussement~$(A)$.
+Pour chaque valeur de~$s$,
+le point $(x_{0}\Add{,} y_{0})$ occupa une position~$\omega$
+%% -----File: 184.png---Folio 176-------
+sur la courbe~$(R)$, et $\phi$~est l'angle de la tangente à~$R$ en~$\omega$
+avec~$\Err{0,x}{0x_{1}}$. Considérons un système d'axes $\omega \xi \eta$, où l'axe $\omega \xi$
+est la tangente à~$(R)$ correspondant au sens dans lequel se
+déplace~$\omega$; $\phi$~est l'angle de~$\omega \xi$ avec~$\DPtypo{o}{O}_{1}x_{1}$; $\xi, \eta$ fonctions de~$s$,
+sont les coordonnées d'un point~$M$ fixe par rapport au système~$x_{1} \DPtypo{o}{O}_{1} y_{1}$,
+prises par rapport aux axes~$\xi \omega \eta$, et~$x_{1}\Add{,} y_{1}$ sont
+les coordonnées de ce même point par rapport aux axes~$x_{1} \DPtypo{o}{O}_{1} y_{1}$.
+Pour avoir la trajectoire orthogonale, il suffit de porter le
+plan~$P$ dans l'espace, sur le plan osculateur à la courbe~$(A)$,
+$\omega \xi$~et~$\omega \eta$ coïncidant respectivement avec $\omega \xi$~et~$\omega \eta$; dans ce
+mouvement, les courbes $(R)$~et~$(A)$ coïncideront successivement
+en tous leurs points; les rayons de courbure étant les mêmes
+en grandeur et en signe, les centres de courbure seront confondus.
+Si $S$~varie, la courbe~$(R)$ va rouler sur la courbe~$(A)$,
+et un point quelconque~$M$ invariablement lié à la courbe~$(R)$
+décrira la trajectoire orthogonale. \emph{Le mouvement du plan~$P$
+s'obtiendra donc en faisant rouler la courbe plane $(R)$ sur la
+courbe~$(A)$ de façon que le plan~$P$ \DPtypo{coincide}{coïncide} à chaque instant
+avec le plan osculateur à la courbe~$(A)$.} On peut dire que \emph{le
+plan~$P$ roule sur la développable qu'il enveloppe}, comme nous
+allons l'expliquer.
+
+Considérons l'arête de rebroussement~$(A)$ et une tangente~$\omega \xi$;
+pour développer cette courbe
+sur un plan, il faut construire la
+courbe plane dont le rayon de courbure
+en chaque point ait même expression
+en fonction de l'arc que celui
+de la courbe~$(A)$, c'est précisément
+\Figure[1.5in]{184a}%****
+%% -----File: 185.png---Folio 177-------
+la courbe~$(R)$. La position d'un point~$P$ sur la développable
+est définie par l'arc~$S$, qui fixe le point~$\omega$ sur~$(A)$ et par
+le segment $\omega P = u$. Le point~$P'$ qui correspond à~$P$ dans le développement
+est déterminé par les mêmes valeurs de~$s, u$. Les
+génératrices de la développable viennent se développer suivant
+les tangentes à la courbe~$(R)$. Considérons une courbe~$(\Gamma)$ sur
+la développable, et la courbe correspondante~$(\Gamma_{1})$ dans le plan:
+les arcs homologues sur ces \Card{2} courbes sont égaux, de sorte
+que toute courbe tracée sur le plan roule sur la courbe correspondante
+de la développable. \emph{On peut imaginer que l'on ait
+enroulé sur la développable une feuille plane déformable; le
+mouvement du plan~$P$ consistera alors à dérouler cette feuille
+de façon qu'elle reste constamment tendue.} Un point quelconque
+de la feuille décrira une trajectoire orthogonale des plans
+tangents à la développable. Nous obtenons ainsi en quelque
+sorte la \emph{surface développante d'une développable} par la généralisation
+du procédé qui donne les développantes d'une courbe
+plane.
+
+Nous pouvons enfin examiner le mouvement du plan~$P$ au
+point de vue cinématique. Nous avons
+\[
+\frac{dX}{ds} = -\frac{a''}{T} \eta, \qquad
+\frac{dY}{ds} = -\frac{b''}{T} \eta, \qquad
+\frac{dZ}{ds} = -\frac{c''}{T} \zeta;
+\]
+et par suite les projections de la vitesse sur les axes~$\xi \eta \zeta$
+invariablement liés au plan~$P$ sont
+\[
+V_{\xi} = \sum a · \frac{dX}{ds} = 0, \quad
+V_{\eta} = \sum a' \frac{dY}{ds} = 0, \quad
+V_{\zeta} = \sum a'' \frac{dZ}{ds} = - \frac{1}{T} \eta.
+\]
+\DPtypo{le}{Le} mouvement instantané du plan~$P$ est une rotation autour de~$\omega \xi$
+tangente à~$(A)$, la rotation instantanée étant~$-\dfrac{1}{T}$. \emph{Le
+plan \DPtypo{osoulateur}{osculateur}~$P$ roule sur la courbe~$(A)$ en tournant autour
+de la tangente avec une vitesse de rotation égale à~$-\dfrac{1}{T}$.}
+%% -----File: 186.png---Folio 178-------
+
+La surface~$(S)$ engendrée par le mouvement précédent est
+une \emph{surface moulure}, ou \emph{surface de Monge}. Considérons dans le
+plan~$P$ une courbe~$(c)$ invariablement liée au système d'axes~$\omega \xi \eta$
+et sa développée~$(K)$. La \Ord{2}{e} nappe de la surface focale
+sera engendrée par cette développée~$(K)$ dans le mouvement du
+plan~$P$. C'est une surface moulure. Ainsi \emph{une des nappes de la
+développée d'une surface moulure est une développable, l'autre
+est une surface moulure}.
+
+\MarginNote{Cas
+particuliers.}
+Examinons le cas particulier où la développable enveloppe
+du plan~$P$ est un cylindre ou un cône.
+
+\ParItem{\Primo.} Si \emph{le plan~$P$ enveloppe un cylindre}, les tangentes
+aux trajectoires orthogonales sont parallèles aux plans de
+section droite, les trajectoires sont les développantes des
+sections droites; ce sont des lignes planes; \emph{les \Card{2} systèmes
+de lignes de courbure de la surface sont des courbes planes.
+Le plan~$P$ roule sur le cylindre de façon que son intersection
+avec le plan d'une section droite roule sur cette section
+droite. On peut encore engendrer la surface en considérant
+dans un plan une famille de courbes parallèles (qui sont ici
+les développantes de la section droite du cylindre), et en
+déplaçant chacune de ces courbes d'un mouvement de translation
+perpendiculaire au plan}.
+
+\ParItem{\Secundo.} Si \emph{le plan~$P$ enveloppe un cône} de sommet~$A$, considérons
+une trajectoire orthogonale rencontrant le plan~$P$ en~$M$,
+la tangente en~$M$ est perpendiculaire à~$A M$, donc la trajectoire
+orthogonale est une courbe tracée sur une sphère de centre~$A$\Add{.}
+Coupons alors le cône par une sphère de centre~$A$ et de
+%% -----File: 187.png---Folio 179-------
+rayon~$R$, soit~$(c)$ l'intersection, et considérons dans le plan~$P$
+le cercle~$(S)$ de centre~$A$ et de rayon~$R$. \emph{Le plan~$F$ roule
+sur le cône de façon que le cercle~$(S)$ roule sur la courbe~$(c)$\Add{.}}
+
+\MarginNote{Autres hypothèses.}
+Cherchons maintenant si les deux nappes de la développée
+d'une surface peuvent être des développables. La surface
+est alors surface moulure de \Card{2} manières; les \Card{2} systèmes
+de lignes de courbure sont des courbes planes. Les trajectoires
+orthogonales des plans~$P$, qui enveloppent l'une des nappes
+de la développée, constituant un des systèmes de lignes
+de courbure, doivent être planes. Soit $P'$~le plan de l'une
+d'elles. Les plans~$P$ sont tous normaux à une courbe située
+dans~$P'$; ils sont donc tous perpendiculaires à $P'$. Si donc
+les plans~$P$ ne sont pas parallèles, les plans~$P'$ le sont tous;
+les plans~$P$ enveloppent un cylindre, et les plans~$P'$ sont perpendiculaires
+aux génératrices de ce cylindre, ainsi que les
+normales à la surface; le \DPtypo{profit}{profil} situé dans un plan~$P$ et qui
+engendre la surface moulure est une parallèle aux génératrices
+du cylindre. Les surfaces obtenues sont donc des cylindres;
+la seconde nappe de la développée est une droite rejetée à
+l'infini.
+
+Si les plans~$P$ sont parallèles, on arrive à la même conclusion,
+car les plans~$P'$ enveloppent un cylindre.
+
+Le cas supposé est donc impossible.
+
+Supposons qu'une des nappes de la développée soit une
+développable, l'autre étant une courbe. La surface est une
+surface moulure qui s'obtient par le mouvement d'un profil
+situé dans le plan~$P$ qui enveloppe la développable. La \Ord{2}{e} nappe
+de la développée est engendrée dans ce mouvement par
+%% -----File: 188.png---Folio 180-------
+la développée du profil; pour que ce soit une courbe, il faut
+que la développée du profil soit un point, donc que ce profil
+soit un cercle; imaginons alors la sphère qui a ce profil
+pour grand cercle; elle est inscrite dans la surface; \emph{la surface
+est une enveloppe de sphères de rayon constant}. C'est une
+surface canal.
+
+\emph{Réciproquement toute enveloppe d'une famille de sphères
+égales satisfait à la condition précédente.} Soit la sphère
+\[
+\sum (x - a)^{2} - r^{2} = 0,
+\]
+la caractéristique a pour \Ord{2}{e} équation
+\[
+\sum (x - a)\, da = 0\DPtypo{;}{.}
+\]
+C'est un grand cercle de la sphère; les normales à la surface
+enveloppe sont dans le plan de ce cercle. L'une des nappes de
+la développée sera l'enveloppe des plans de ce cercle. Si
+nous considérons le lieu du centre de la sphère, le plan du
+grand cercle lui est constamment normal; \emph{la surface est engendrée
+par un cercle de rayon constant dont le centre décrit une
+courbe, et dont le plan reste constamment normal à cette
+courbe}.
+
+Enfin comme cas singulier, nous avons encore celui où
+l'une des nappes de la développée est une droite. La surface
+est alors de révolution autour de cette droite.
+
+
+\ExSection{VII}
+
+\begin{Exercises}
+\item[32.] \DPchg{Etudier}{Étudier} la congruence formée des droites tangentes à une
+sphère et normales à une même surface; étudier les surfaces
+normales à ces droites, et leurs lignes de courbure.
+
+\item[33.] \DPchg{Etudier}{Étudier} la congruence formée des droites normales à une surface
+dont une famille de lignes de courbure est située sur
+des sphères concentriques.
+
+\item[34.] Montrer que les surfaces moulures, dans le cas où l'une des
+nappes de la développée est un cylindre ou un cône, peuvent
+être \DPtypo{definies}{définies} par le mouvement d'un profil plan, de forme
+invariable, dont le plan reste constamment normal à un \DPtypo{cylindr}{cylindre}
+ou à un cône. \DPtypo{Prèciser}{Préciser} le mouvement de ce profil. Chercher si
+l'on peut dire quelque chose d'analogue pour les surfaces
+moulures générales.
+
+\item[35.] Montrer que les droites tangentes à deux quadriques homofocales
+constituent une congruence de normales. Si on fait réfléchir
+toutes ces droites, considérées comme des rayons lumineux,
+sur une autre quadrique homofocale aux deux premières,
+quelles seront les multiplicités focales de cette seconde
+congruence?
+
+\item[36.] \DPchg{Etant}{Étant} données deux surfaces homofocales du second \DPtypo{degre}{degré} et un
+plan~$P$, si on mène par les droites du plan~$P$ des plans tangents
+aux deux surfaces, les droites qui joignent les points
+de contact correspondants sont normales à une famille de surfaces
+parallèles. Soit~$(\delta)$ la droite qui contient les pôles
+du plan~$P$ par rapport aux deux quadriques homofocales, et
+$(d')$~la droite du plan~$P$ qui correspond à une droite~$(d)$ de
+la congruence de normales considérée. Le plan mené par~$(\delta)$
+perpendiculairement à~$(d')$ coupe~$(d)$ en un point~$m$. Le lieu
+du point~$m$ est l'une des surfaces cherchées: c'est une cyclide.
+Les développables de la congruence découpent sur les
+surfaces homofocales des réseaux conjugués.
+
+\item[37.] On considère la congruence des droites de l'espace sur lesquelles
+trois plans formant un trièdre trirectangle déterminent
+des segments invariables. Démontrer que c'est une congruence
+de normales et déterminer les surfaces normales aux
+droites de la congruence. Déterminer les points focaux sur
+une quelconque de ces droites. Déterminer les cônes directeurs
+des développables de la congruence.
+
+\item[38.] \DPtypo{Demontrer}{Démontrer} qu'il existe des congruences (isogonales) telles
+que les plans focaux forment un dièdre constant. Quelle est
+la propriété des arêtes de rebroussement des développables de
+la congruence par rapport aux nappes de la surface focale qui
+les contiennent? Chercher l'équation différentielle de ces
+courbes sur la surface focale supposée donnée. Que peut-on
+dire du cas où l'une des nappes de la multiplicité focale est
+une développable, une courbe, une sphère?
+
+\item[39.] Si on considère une famille de sphères dont le lieu des
+centres~$\omega$ est une courbe plane~$C$, et dont les rayons sont
+proportionnels aux distances des centres~$\omega$ à une droite fixe~$\Delta$
+du plan de la courbe~$C$, démontrer que l'enveloppe de ces
+sphères a toutes ses lignes de courbure planes. Que peut-on
+dire des plans de ces lignes de courbure? Réciproquement,
+comment peut-on obtenir toutes les surfaces canaux dont toutes
+les lignes de courbure sont planes?
+\end{Exercises}
+%% -----File: 189.png---Folio 181-------
+
+
+\Chapitre{VIII}{Les Congruences de Droites et les Correspondances Entre
+Deux Surfaces.}
+
+\Section{Nouvelle représentation des congruences.}
+{1\Add{.}}{} Dans ce qui précède, nous avons défini une congruence
+par son support, et en donnant la direction de la
+droite ou des droites~$(D)$ qui passent par chaque point
+du support. On peut plus généralement, et ce sera préférable
+au point de vue projectif, considérer \Card{2} surfaces supports se
+correspondant point par point, les droites de la congruence
+étant celles qui joignent les points homologues des deux surfaces.
+En réalité, les \Card{2} surfaces se correspondront élément
+de contact à élément de contact, et en même temps que la congruence
+des droites joignant les points homologues, on pourra
+considérer celle des intersections des plans tangents homologues.
+
+Il est naturel alors d'employer des coordonnées homogènes\Add{.}
+Soient $M(x\Add{,}y\Add{,}z\Add{,}t)$ et $\DPtypo{M}{M_{1}}(x_{1}\Add{,} y_{1}\Add{,} z_{1}\Add{,} t_{1})$ les points homologues sur les
+\Card{2} surfaces; on pourra définir la congruence par les équations
+\[
+X = x + \rho x_{1}, \qquad
+Y = y + \rho y_{1}, \qquad
+Z = z + \rho z_{1}, \qquad
+T = t + \rho t_{1}.
+\]
+Soient de même $u, v, w, r$ les coordonnées tangentielles d'un plan
+tangent à la \Ord{1}{ère} surface, $u_{1}, v_{1}, w_{1}, r_{1}$ celles du plan tangent
+homologue à la \Ord{2}{e} surface. La congruence pourra être définie
+au point de vue tangentiel par les équations
+\[
+U = u + \rho u_{1}, \qquad
+V = v + \rho v_{1}, \qquad
+W = w + \rho w_{1}, \qquad
+R = r + \rho r_{1}.
+\]
+
+Soient $(S), (S_{1})$ les \Card{2} surfaces supports; les systèmes conjugués
+%% -----File: 190.png---Folio 182-------
+sur ces surfaces étant invariants, d'après leur définition
+même, par toute transformation projective, nous sommes
+conduits à étudier leurs relations. Soient
+\begin{align*}
+\Tag{(S)}
+x &= f(\lambda,\mu), &
+y &= g(\lambda,\mu), &
+z &= h(\lambda,\mu), &
+t &= \DPtypo{h}{k}(\lambda,\mu); \\
+\Tag{(S_{1})}
+x_{1} &= f_{1}(\lambda,\mu), &
+y_{1} &= g_{1}(\lambda,\mu), &
+z_{1} &= h_{1}(\lambda,\mu), &
+t_{1} &= \DPtypo{k}{h}_{1}(\lambda,\mu);
+\end{align*}
+les équations des deux surfaces.
+
+Le choix des paramètres $\lambda\Add{,} \mu$ est fixé par le Théorème suivant:
+\emph{Quand \Card{2} surfaces $(S)\Add{,} (S_{1})$ se correspondent point par point, il
+existe sur~$(S)$ un réseau conjugué qui correspond à un réseau
+conjugué de~$S_{1}$, et en général il n'en existe qu'un.} Soient $d\lambda,
+d\mu$ et~$\delta\lambda, \delta\mu$ les paramètres définissant \Card{2} directions conjuguées
+sur~$(S)$, elles sont conjuguées harmoniques par rapport aux
+directions
+\[
+E'\, d\lambda^{2} + 2F'\, d\lambda · d\mu + G'\, d\mu^{2} = 0.
+\]
+De même sur~$(S_{1})$, \Card{2} directions conjuguées sont conjuguées harmoniques
+par rapport aux directions
+\[
+E'_{1}\, d\lambda^{2} + 2F'_{1}\, d\lambda · d\mu + G'_{1}\, d\mu^{2} = 0.
+\]
+Chercher un système conjugué commun revient donc à chercher un
+couple de points conjugués par rapport à \Card{2} couples donnés par
+\Card{2} équations quadratiques; si les \Card{2} formes quadratiques n'ont
+pas de facteur commun, il y a un couple et un seul répondant
+à la question. Or\Add{,} les \Card{2} équations précédentes définissent les
+lignes asymptotiques des \Card{2} surfaces; si donc \Card{2} surfaces se
+correspondent point par point d'une façon telle qu'il n'y ait
+pas sur~$(S)$ une famille d'asymptotiques correspondant à une
+famille d'asymptotiques de~$(S_{1})$, il existe un système conjugué
+%% -----File: 191.png---Folio 183-------
+de~$(S)$ et un seul qui correspond à un système conjugué de~$(S_{1})$\Add{.}
+Il est défini par l'équation:
+\[
+\begin{vmatrix}
+E'\, d\lambda + F'\, d\mu    & F'\, d\lambda + G'\, d\mu \\
+E_1'\,d\lambda + F_1'\, d\mu & F_1'\, d\lambda + G_1'\, d\mu
+\end{vmatrix} = 0\Add{.}
+\]
+Il y aura impossibilité si les formes ont un facteur commun,
+et indétermination si les \Card{2} facteurs sont communs, c'est-à-dire,
+si les lignes asymptotiques se correspondent sur les
+deux surfaces. \DPchg{Ecartant}{Écartant} ces cas d'exception, nous supposerons
+que les paramètres $\lambda\Add{,} \mu$ correspondent à ce système conjugué
+commun.
+
+\Section{Emploi des coordonnées homogènes.}
+{2.}{} Nous allons reprendre les formules usuelles et
+voir ce qu'elles deviennent en coordonnées homogènes.
+
+%[** TN: Removed several ". ---" start-of-paragraph markers]
+Une \emph{courbe} en coordonnées homogènes est définie
+par \Card{4} équations
+\[
+x = f(\lambda), \qquad
+y = g(\lambda), \qquad
+z = h(\lambda), \qquad
+t = k(\lambda).
+\]
+La tangente au point $M(x,y,z,t)$ joint le point~$M$ au point
+\[%[** TN: Two large expressions are in-line in original]
+M'\left(\dfrac{dx}{d\lambda}, \dfrac{dy}{d\lambda}, \dfrac{dz}{d\lambda}, \dfrac{dt}{d\lambda}\right).
+\]
+Le plan osculateur passe par la droite~$MM'$ et
+par le point
+\[
+M''\left(\dfrac{d^{2}x}{d\lambda^{2}}, \dfrac{d^{2}y}{d\lambda^{2}}, \dfrac{d^{2}z}{d\lambda^{2}}, \dfrac{d^{2}t}{d\lambda^{2}}\right)\Add{.}
+\]
+
+Corrélativement une \emph{développable} sera l'enveloppe du
+plan~$P$
+\[
+u = f(\lambda), \qquad
+v = g(\lambda), \qquad
+w = h(\lambda), \qquad
+r = k(\lambda).
+\]
+La caractéristique (génératrice) sera l'intersection du plan~$P$
+et du plan
+$P'\left(\dfrac{du}{d\lambda},
+         \dfrac{dv}{d\lambda},
+         \dfrac{dw}{d\lambda},
+         \dfrac{dr}{d\lambda}\right)$.
+Le point de contact avec l'arête
+de rebroussement sera en outre dans le plan
+$P''\left(\dfrac{d^{2}u}{d\lambda^{2}},
+          \dfrac{d^{2}v}{d\lambda^{2}},
+          \dfrac{d^{2}w}{d\lambda^{2}},
+          \dfrac{d^{2}r}{d\lambda^{2}}\right)$.
+%% -----File: 192.png---Folio 184-------
+
+Une \emph{surface} quelconque peut se définir au point de
+vue ponctuel par
+\[
+x = f(\lambda, \mu), \qquad
+y = g(\lambda, \mu), \qquad
+z = h(\lambda, \mu), \qquad
+t = k(\lambda, \mu);
+\]
+et au point de vue tangentiel par
+\[
+u = F(\lambda, \mu), \qquad
+v = G(\lambda, \mu), \qquad
+w = H(\lambda, \mu), \qquad
+r = K(\lambda, \mu).
+\]
+On peut définir le \emph{plan tangent} en fonction du point de contact
+$(x, y, z, t)$. Ce plan contient le point, donc
+\[
+\sum ux = 0;
+\]
+il contient les tangentes aux courbes $\lambda = \cte$\DPtypo{.}{,} $\mu = \cte$, donc
+les points
+$\left(\dfrac{\dd x}{\dd \mu}, \dfrac{\dd y}{\dd \mu},
+       \dfrac{\dd z}{\dd \mu}, \dfrac{\dd t}{\dd \mu}\right)$
+et
+$\left(\dfrac{\dd x}{\dd \lambda}, \dfrac{\dd y}{\dd \lambda},
+       \dfrac{\dd z}{\dd \lambda}, \dfrac{\dd t}{\dd \lambda}\right)$.
+\[
+\sum u \frac{\dd x}{\dd \lambda} = 0, \qquad
+\sum u \frac{\dd x}{\dd \mu} = 0;
+\]
+et nous avons ainsi \Card{3} équations définissant des quantités
+proportionnelles à $u, v, w, r$. On peut écrire l'équation ponctuelle
+du plan tangent au point $(x\Add{,}y\Add{,}z\Add{,}t)$
+\[
+\begin{vmatrix}
+X  &   Y  &  Z  &  T \\
+x  &   y  &  z  &  t \\
+\mfrac{\dd x}{\dd \lambda} &
+\mfrac{\dd y}{\dd \lambda} &
+\mfrac{\dd z}{\dd \lambda} &
+\mfrac{\dd t}{\dd \lambda} \\
+%
+\mfrac{\dd x}{\dd \mu} &
+\mfrac{\dd y}{\dd \mu} &
+\mfrac{\dd z}{\dd \mu} &
+\mfrac{\dd t}{\dd \mu}
+\end{vmatrix} = 0.
+\]
+
+\emph{Corrélativement} on définira un \emph{point} de la surface en
+fonction du plan tangent en ce point, au moyen des équations:
+\[
+\sum ux = 0, \qquad
+\sum x \frac{\dd u}{\dd \lambda} = 0, \qquad
+\sum x \frac{\dd u}{\dd \mu} = 0;
+\]
+de sorte qu'en définitive, on peut définir l'un des éléments
+point, plan tangent, en fonction de l'autre au moyen des formules
+\[
+\sum ux = 0, \qquad
+\sum u\, dx = 0, \qquad
+\sum x\, du = 0.
+\]
+
+Proposons-nous maintenant d'\emph{exprimer que \Card{2} directions
+%% -----File: 193.png---Folio 185-------
+$MT(d\lambda, d\mu)$ et $MS(\delta \lambda\DPtypo{.}{,} \delta \mu)$ sont conjuguées}. Ces \Card{2} directions
+sont conjuguées si, le point de contact du plan tangent se déplaçant
+dans la direction $MT, MS$ est la caractéristique de ce
+plan tangent. Or\Add{,} cette caractéristique est
+\[
+\sum uX = 0, \qquad
+\sum du·X = 0;
+\]
+la droite~$MS$ est définie par le point $(x, y, z, t)$ et le point
+$(\delta x, \delta y, \delta z, \delta t)$. Pour exprimer que $MS$~est la caractéristique, il
+faut exprimer que les \Card{2} points précédents sont sur la caractéristique,
+ce qui donne:
+\begin{alignat*}{2}
+&\sum ux = 0, \qquad
+&&\sum du·x = 0; \\
+&\sum u · \delta x = 0, \qquad
+&&\sum du · \delta x = 0;
+\end{alignat*}
+les \Card{3} \Ord{1}{ères} équations sont vérifiées, nous avons donc la condition
+unique
+\[
+\sum du · \delta x = 0,
+\]
+ou la condition symétrique
+\[
+\sum \delta u · dx = 0.
+\]
+En particulier nous trouvons la condition pour qu'une direction
+soit conjuguée d'elle-même, c'est-à-dire soit direction
+asymptotique
+\[
+\sum du · dx = 0.
+\]
+
+Exprimons alors que les directions $\lambda = \cte$, $\mu = \cte$ forment
+un réseau conjugué. Nous avons
+\[
+\Tag{(1)}
+\sum · \frac{\dd u}{\dd \lambda} · \frac{\dd x}{\dd \mu}  = 0.
+\]
+Cette condition peut se transformer: l'équation
+\[
+\sum u \frac{\dd x}{\dd \mu} = 0
+\]
+différentiée par rapport à $\lambda$ donne
+\[
+\sum \frac{\dd u}{\dd \lambda} · \frac{\dd x}{\dd \mu}
+  + \sum u \frac{\dd^{2} x}{\dd \lambda\, \dd \mu} = 0;
+\]
+%% -----File: 194.png---Folio 186-------
+et \Eq{(1)}~s'écrit
+\[
+\Tag{(2)}
+\sum u \frac{\dd^{2} x}{\dd \lambda·\dd \mu} = 0.
+\]
+De même l'équation
+\[
+\sum u \frac{\dd x}{\dd \lambda} = 0
+\]
+différentiée par rapport à~$\mu$ donne
+\[
+\sum \frac{\dd u}{\dd \mu} · \frac{\dd x}{\dd \lambda}
+  + \sum u \frac{\dd^{2} x}{\dd \lambda\, \DPtypo{d}{\dd}\mu} = 0,
+\]
+et \Eq{(1)}~peut s'écrire
+\[
+\Tag{(3)}
+\sum \frac{\dd u}{\dd \mu} · \frac{\dd x}{\dd \lambda} = 0.
+\]
+En partant de l'une des relations
+\[
+\sum x \frac{\dd u}{\dd \lambda} = 0, \qquad
+\sum x \frac{\dd u}{\dd \mu} = 0,
+\]
+on obtiendrait la relation
+\[
+\Tag{(4)}
+\sum x \frac{\dd^{2} u}{\dd \lambda · \dd \mu} = 0.
+\]
+Ces \Card{4} équations \Eq{(1)}\Add{,} \Eq{(2)}\Add{,} \Eq{(3)}\Add{,} \Eq{(4)} dépendent simultanément des éléments
+ponctuel et tangentiel. En exprimant $u, v, w, r$ en fonction
+de~$x, y, z, t$, on obtient la condition en coordonnées ponctuelles:
+\[
+\Tag{(5)}
+\begin{vmatrix}
+x & \mfrac{\dd x}{\dd \lambda}
+  & \mfrac{\dd x}{\dd \mu}
+  & \mfrac{\dd^{2} x }{\dd\lambda\, \dd\mu}
+\end{vmatrix} = 0.
+\]
+Dans cette relation~\Eq{(5)}, le premier membre représente, par
+abréviation, le déterminant dont la première ligne serait la
+ligne écrite entre les deux traits verticaux, et dont les
+trois autres lignes se déduiraient de celle-là en~$y$ remplaçant~$x$
+par~$y, z, t$. Cette notation sera employée couramment dans la
+suite.
+
+Lorsque $t = \cte$, la condition~\Eq{(5)} se réduit à la condition
+connue
+\[
+\begin{vmatrix}
+\mfrac{\dd x}{\dd \lambda} &
+\mfrac{\dd x}{\dd \mu} &
+\mfrac{\dd^{2} x}{\dd \lambda · \dd \mu}
+\end{vmatrix} = F' = 0.
+\]
+
+La condition~\Eq{(5)} peut s'interpréter autrement: il existe
+%% -----File: 195.png---Folio 187-------
+une même relation linéaire et homogène entre les éléments correspondants
+des lignes
+\begin{align*}%[** TN: Filled in last three equations]
+\frac{\dd^2 x}{\dd \lambda · \dd \mu}
+  &= L \frac{\dd x}{\dd \lambda} + M \frac{\dd x}{\dd \mu} + N x\Add{,} \\
+\frac{\dd^2 y}{\dd \lambda · \dd \mu}
+  &= L \frac{\dd y}{\dd \lambda} + M \frac{\dd y}{\dd \mu} + N y\Add{,} \\
+\frac{\dd^2 z}{\dd \lambda · \dd \mu}
+  &= L \frac{\dd z}{\dd \lambda} + M \frac{\dd z}{\dd \mu} + N z\Add{,} \\
+\frac{\dd^2 t}{\dd \lambda · \dd \mu}
+  &= L \frac{\dd t}{\dd \lambda} + M \frac{\dd t}{\dd \mu} + N t\Add{,}
+\end{align*}
+c'est-à-dire: \emph{les \Card{4} coordonnées homogènes~$x,y,z,t$ satisfont à
+une même équation linéaire aux dérivées partielles de la forme}:
+\[
+\frac{\dd^2 f}{\dd \lambda· \dd \mu}
+  = L \frac{\dd f}{\dd \lambda} + M \frac{\dd f}{\dd \mu} + Nf.
+\]
+En \DPtypo{operant}{opérant} au point de vue tangentiel, on verrait de même que
+\emph{la condition cherchée est que $u,v,w,r$~soient intégrales d'une
+même équation}:
+\[
+\frac{\dd^2f}{\dd \lambda\, \dd \mu}
+  = P \frac{\dd f}{\dd \lambda} + Q \frac{\dd f}{\dd \mu} + Rf.
+\]
+On montrerait sans peine que si $x,y,z,t$\DPtypo{,}{} ou $u,v,w,r$ satisfont
+à une équation de la forme précédente, elles ne satisfont qu'à
+une seule.
+
+\Paragraph{Remarque.} En coordonnées cartésiennes, $t = 1$, $r = 1$, et on a
+$R = N = 0$.
+
+Considérons une \emph{surface réglée}; les équations d'une
+génératrice, joignant le point $M(x\Add{,}y\Add{,}z\Add{,}t)$ au point $M_1 (x_1, y_1, z_1, t_1)$
+sont:
+\[
+X = x + \rho x_1, \qquad
+Y = y + \rho y_1, \qquad
+Z = z + \rho z_1, \qquad
+T = t + \rho t_1.
+\]
+Supposons la surface \emph{développable}; les plans tangents aux points
+$(x,y,z,t)$ et $(x_1, y_1, z_1, t_1)$ sont les mêmes. Or\Add{,} le plan tangent en~$M$
+%% -----File: 196.png---Folio 188------
+passant par la génératrice et par la tangente à la courbe
+$\rho = \Err{\cte}{0}$ contient le point $(dx, dy, dz, dt)$. De même le plan tangent
+en~$M_1$, contient le point $(dx_1, dy_1, dz_1, dt_1)$. La condition
+pour que les plans soient confondus est donc
+\[
+\begin{vmatrix}
+x  &  x_1  &  dx  &  dx_1
+\end{vmatrix} = 0.
+\]
+
+Si nous définissions la surface en coordonnées tangentielles,
+nous arriverions de même à la condition
+\[
+\begin{vmatrix}
+u  &  u_1  &  du  &  du_1
+\end{vmatrix} = 0.
+\]
+
+Voyons enfin une \emph{congruence}: nous pouvons encore la
+représenter par les équations
+\[
+X = x + \rho x_1, \qquad
+Y = y + \rho y_1, \qquad
+Z = z + \rho z_1, \qquad
+T = t + \rho t_1;
+\]
+mais ici $x, y, z, t$ et $x_1, y_1, z_1, t_1$, sont fonctions de deux paramètres
+arbitraires $(\lambda, \mu)$. Cherchons les \emph{éléments focaux}. Soit $F$ un
+foyer d'une droite~$D(\lambda, \mu)$. Soit $\rho$ la valeur du paramètre qui
+correspond à ce point. Toutes les surfaces réglées de la congruence
+qui contiennent la droite~$D$ ont en ce point~$F$ même
+plan tangent. Considérons en particulier les surfaces $\lambda = \cte$
+et $\mu = \cte[]$. Les plans tangents à ces surfaces contiennent respectivement
+les points $(x, y, z, t)$\Add{,} $(x_1, y_1, z_1, t_1)$\Add{,}
+$\left(\dfrac{\dd x}{\dd \mu} + \rho \dfrac{\dd x_1}{\dd \mu},\dots\right)$
+et $(x, y, z, t)$\Add{,} $(x_1, y_1, z_1, t_1)$\Add{,}
+$\left(\dfrac{\dd x}{\dd \lambda} + \rho \dfrac{\dd x\Add{_1}}{\dd \lambda},\dots\right)$. La condition pour
+que ces plans \DPtypo{coincident}{coïncident}, c'est-à-dire l'\emph{équation aux points
+focaux}, est donc
+\[
+\begin{vmatrix}
+x  &  x_1  &
+\mfrac{\dd x}{\dd \lambda} + \rho\mfrac{\dd x_1}{\dd \lambda} &
+\mfrac{\dd x}{\dd \mu} + \rho\mfrac{\dd x_1}{\dd \mu}
+\end{vmatrix} = 0;
+\]
+On trouvera de même l'\emph{équation aux plans focaux}:
+\[
+\begin{vmatrix}
+u & u_1 &
+\mfrac{\dd u}{\dd \lambda} + \rho \mfrac{\dd u_1}{\dd \lambda} &
+\mfrac{\dd u}{\dd \mu} + \rho \mfrac{\dd \DPtypo{u}{u_{1}}}{\dd \mu}
+\end{vmatrix} = 0.
+\]
+%% -----File: 197.png---Folio 189-------
+
+\Section{Correspondances spéciales.}
+{3.}{} Nous allons étudier la \emph{correspondance entre \Card{2} points
+$M\Add{,} \Err{M}{M_{1}}$, de \Card{2} surfaces telle que les développables de la congruence
+des droites $M\Err{M}{M_{1}}$ coupent les \Card{2} surfaces suivant les
+\Card{2} réseaux conjugués qui se correspondent}. Nous supposerons que
+les paramètres $\lambda\Add{,} \mu$ qui fixent la position d'un point sur chacune
+des surfaces sont précisément tels quo les courbes conjuguées
+homologues soient $\lambda = \cte$ et $\mu = \cte[]$. Les courbes
+$\lambda = \cte\Add{,} \mu = \cte$ sont conjuguées sur la \Ord{1}{ère} surface~$(S)$ donc
+$x, y, z, t$ satisfont à une même équation différentielle
+\[
+\Tag{(1)}
+\frac{\dd^2 f}{\dd \lambda\, \dd \mu}
+  = P \frac{\dd f}{\dd \lambda} + Q \frac{\dd f}{\dd \mu} + R f;
+\]
+de même les courbes $\lambda = \cte$ et $\mu = \cte$ étant conjuguées sur la
+\Ord{2}{e} surface~$(S_1)$, $x_1, y_1, z_1, t_1$\DPtypo{,}{} satisfont à une même équation différentielle
+\[
+\Tag{(2)}
+\frac{\dd^2 f}{\dd \lambda\, \dd \mu}
+  = P_1 \frac{\dd f}{\dd \lambda} + Q_1 \frac{\dd f}{\dd \mu} + R_1 f.
+\]
+Exprimons maintenant que les développables de la congruence
+correspondent à $\lambda = \cte$ et $\mu = \cte[]$. Si nous représentons la
+congruence par les équations
+\[
+X = x + \rho x, \qquad
+Y = y + \rho y, \qquad
+Z = z + \rho z, \qquad
+T = t + \rho t,
+\]
+les développables sont données par l'équation
+\[
+\begin{vmatrix}
+x  &  x_1  &  dx  &  dx_1
+\end{vmatrix} = 0.
+\]
+Or,
+\begin{alignat*}{5}
+dx &= \frac{\dd x}{\dd \lambda}\, d\lambda
+   &&+ \frac{\dd x}{\dd \mu}\, d\mu,\qquad &
+dy &= \dots\dots,\qquad &
+dz &= \dots\dots,\qquad &
+dt &= \dots\dots, \\
+%
+dx_1 &= \frac{\dd x_1}{\dd\lambda}\, d\lambda
+   &&+ \frac{\dd x_1}{\dd\mu}\, d\mu,\qquad &
+dy_1 &= \dots\dots,\qquad &
+dz_1 &= \dots\dots,\qquad &
+dt_1 &= \dots\dots;
+\end{alignat*}
+\iffalse%%%%%[** TN: Code follows for three sets of equations above]
+dy &= \frac{\dd y}{\dd \lambda}\, d\lambda + \frac{\dd y}{\dd \mu}\, d\mu, &
+dz &= \frac{\dd z}{\dd \lambda}\, d\lambda + \frac{\dd z}{\dd \mu}\, d\mu, &
+dt &= \frac{\dd t}{\dd \lambda}\, d\lambda + \frac{\dd t}{\dd \mu}\, d\mu, \\
+%
+dy_1 &= \frac{\dd y_1}{\dd\lambda}\, d\lambda + \frac{\dd y_1}{\dd\mu}\, d\mu, &
+dz_1 &= \frac{\dd z_1}{\dd\lambda}\, d\lambda + \frac{\dd z_1}{\dd\mu}\, d\mu, &
+dt_1 &= \frac{\dd t_1}{\dd\lambda}\, d\lambda + \frac{\dd t_1}{\dd\mu}\, d\mu,
+\fi %%%% End of code for elided equations
+et l'équation précédente devant être vérifiée pour $d\lambda = 0$,
+$d\mu = 0$, nous avons les conditions
+%% -----File: 198.png---Folio 190-------
+\begin{align}
+\Tag{(3)}
+\begin{vmatrix}
+x  &  x_1  &  \mfrac{\dd x}{\dd \lambda}  &  \mfrac{\dd x_1}{\dd \lambda}
+\end{vmatrix} = 0, \\
+\Tag{(4)}
+\begin{vmatrix}
+x  &  x_1  &  \mfrac{\dd x}{\dd \mu}  &  \mfrac{\dd x_1}{\dd \mu}
+\end{vmatrix} = 0.
+\end{align}
+Il existe une même relation linéaire et homogène entre les
+éléments des colonnes, donc
+\begin{alignat*}{2}
+\Tag{(5)}
+Ax + B \frac{\dd x}{\dd \lambda}
+  &= A_1 x_1 + B_1 \frac{\dd x_1}{\dd \lambda},\qquad
+  && \text{et les analogues}\quad\dots\dots\Add{,}
+ \\
+\Tag{(6)}
+Cx + D \frac{\dd x}{\dd \mu}
+  &= C_1 x_1 + D_1 \frac{\dd x_1}{\dd \mu},\qquad
+  && \text{et les analogues}\quad\dots\dots\Add{.}
+\end{alignat*}
+
+\Paragraph{\DPchg{\Ord{1}{er} Cas}{Premier Cas}.} Voyons d'abord ce qui arrive si l'un des \Card{4} coefficients
+$B\Add{,} B_1\Add{,} D\Add{,} D_1$ est nul. Soit $B_1 = 0$. Alors les équations~\Eq{(5)}
+expriment que le point $M_1(x_1\Add{,} y_1\Add{,} z_1\Add{,} t_1)$ est sur la droite qui
+joint les points $M(x\Add{,}y\Add{,}z\Add{,}t)$ et
+\[%[** TN: Large in-line expression in original]
+M\left(\dfrac{\dd x}{\dd \lambda}, \dfrac{\dd y}{\dd \lambda}, \dfrac{\dd z}{\dd \lambda}, \dfrac{\dd t}{\dd \lambda}\right).
+\]
+La droite~$MM_1$
+est tangente à la courbe~$\mu = \cte$ tracée sur la surface~$(S)$.
+Toutes les droites~$MM_1$ sont ainsi tangentes à la surface~$(S)$
+qui est une des nappes de la surface focale de la congruence.
+Sur la surface~$(S)$ les courbes $\mu = \cte$ correspondent à une
+famille de développables, et par suite les courbes $\lambda = \cte$
+conjuguées des précédentes correspondent à la \Ord{2}{e} famille. Il
+nous faut alors chercher comment on peut définir~$(S_1)$ pour
+que cette surface soit coupée suivant un réseau conjugué par
+les développables de la congruence. Les équations~\Eq{(5)} peuvent
+s'écrire dans le cas considéré
+\[
+x_1 = Ax + B \frac{\dd x}{\dd \lambda}, \quad
+y_1 = Ay + B \frac{\dd y}{\dd \lambda}, \quad
+z_1 = Az + B \frac{\dd z}{\dd \lambda}, \quad
+t_1 = At + B \frac{\dd t}{\dd \lambda}\Add{.}
+\]
+Posons
+\[
+x = \theta X, \qquad
+y = \theta Y, \qquad
+z = \theta Z, \qquad
+t = \theta T;
+\]
+nous avons alors
+\[
+x_1 = A \theta X + B \left(\theta \frac{\dd x}{\dd \lambda}
+  + X \frac{\dd \theta}{\dd \lambda}\right), \quad
+y_1 = \dots\dots, \quad
+z_1 = \dots\dots, \quad
+t_1 = \dots\dots;
+\]
+déterminons la fonction~$\theta$ par la relation
+\[
+A \theta + B \frac{d \theta}{d \lambda} = 0,
+\]
+%% -----File: 199.png---Folio 191-------
+ce qui est toujours possible. Nous avons
+\[
+x_1 = A \frac{\dd x}{\dd \lambda}, \qquad
+y_1 = A \frac{\dd y}{\dd \lambda}, \qquad
+z_1 = A \frac{\dd z}{\dd \lambda}, \qquad
+t_1 = A \frac{\dd t}{\dd \lambda};
+\]
+et comme les coordonnées homogènes ne sont définies qu'à un
+facteur près, nous pouvons écrire
+\[
+\Tag{(7)}
+x_1 = \frac{\dd x}{\dd \lambda}, \qquad
+y_1 = \frac{\dd y}{\dd \lambda}, \qquad
+z_1 = \frac{\dd z}{\dd \lambda}, \qquad
+t_1 = \frac{\dd t}{\dd \lambda}.
+\]
+Alors, d'après ces relations, l'équation différentielle~\Eq{(1)}
+s'écrit
+\[
+\Tag{(8)}
+\frac{\dd x_1}{\dd \mu} = Px_1 + Q \frac{\dd x}{\dd \mu} + Rx,
+\]
+condition de la forme~\Eq{(6)}. Les équations \Eq{(3)}~et~\Eq{(4)} sont alors
+vérifiées. Différentions la relation~\Eq{(8)} par rapport à~$\lambda$
+\[
+\frac{\dd^2 x_1}{\dd \lambda\, \dd \mu}
+  = \frac{\dd P}{\dd \lambda} x_1 + P \frac{\dd x_1}{\dd \lambda}
+  + \frac{\dd Q}{\dd \lambda} · \frac{\dd x}{\dd \mu}
+  + Q \frac{\dd^2 x}{\dd \lambda\, \dd \mu}
+  + \frac{\dd R}{\dd \lambda x} + R \frac{\dd x}{\dd \lambda}.
+\]
+Mais, $\Err{x}{x_{1}}$~vérifiant l'équation~\Eq{(2)}, nous avons
+\[
+\frac{\dd^2 x_1}{\dd \lambda\, \dd \mu}
+  = P_1 \frac{\dd x_1}{\dd \lambda} + Q_1 \frac{\dd x_1}{\dd \mu} + R_1 x_1,
+\]
+et nous obtenons ainsi
+\[
+\Tag{(9)}
+P_1 \frac{\dd x_1}{\dd \lambda} + Q_1 \frac{\dd x_1}{\dd \mu} + R_1 x_1
+  = \frac{\dd P}{\dd \lambda} x_1
+  + P \frac{\dd x_1}{\dd \lambda}
+  + \frac{\dd Q}{\dd \lambda}\Err{}{\, \frac{dx}{d\mu}}
+  + Q \frac{\dd^2 x}{\dd \lambda\, \dd \mu}
+  + \frac{\dd R}{\dd \lambda} · x
+  + R \frac{\dd x}{\dd \lambda};
+\]
+\Eq{(8)}\Add{,}~\Eq{(9)} sont \Card{2} équations en $x$~et~$\dfrac{\dd x}{\dd \mu}$. Si on peut les résoudre, on
+en peut tirer $x$ en particulier, en fonction linéaire de~$x_1$,
+$\dfrac{\dd x_1}{\dd \lambda}$\Add{,}~et~$\dfrac{\dd \Err{x}{x_1}}{\dd \mu}$;
+car~$\dfrac{\dd^2 x}{\dd \lambda\, \dd \mu}\Err{}{=\dfrac{\dd x_1}{\dd \mu}}$\Err{}{ et $\dfrac{\dd x}{\dd\lambda} = x_1$} s'exprime en fonction linéaire de ces \Card{3} quantités;
+donc le point $M(x, y, z, t)$ se trouve dans le plan des \Card{3} points
+$(x_1, y_1, z_1, t_1)$\Add{,}
+%[** TN: Added elided components in next two points]
+$\left(\dfrac{\dd x_{1}}{\dd\lambda}, \dfrac{\dd y_{1}}{\dd\lambda},
+       \dfrac{\dd z_{1}}{\dd\lambda}, \dfrac{\dd t_{1}}{\dd\lambda}\right)$\Add{,}
+$\left(\dfrac{\dd x_{1}}{\dd \mu}, \dfrac{\dd y_{1}}{\dd \mu},
+       \dfrac{\dd z_{1}}{\dd \mu}, \dfrac{\dd t_{1}}{\dd \mu}\right)$,
+c'est-à-dire dans le
+plan tangent en~$M_1$, à la surface~$(S_1)$. La droite~$MM_1$ est donc aussi
+tangente à~$(S_1)$, et $(S_1)$~est la \Ord{2}{e} nappe de la surface focale.
+Nous avons ainsi établi une \emph{correspondance point par point entre
+les \Card{2} nappes de la surface focale d'une congruence}.
+%% -----File: 200.png---Folio 192-------
+
+\emph{\DPchg{Ecartons}{Écartons} ce cas}; il faut alors supposer que les équations
+\Eq{(8)}\Add{,}~\Eq{(9)} ne sont pas résolubles en $x$~et~$\dfrac{\dd x}{\dd \mu}$; ce qui exige que
+l'on ait
+\[
+\begin{vmatrix}
+Q                          & R  \\
+\mfrac{\dd Q}{\dd \lambda} & \mfrac{\dd R}{\dd \lambda}
+\end{vmatrix} = 0,
+\]
+ou
+\[
+Q \frac{\dd R}{\dd \lambda} - R \frac{\dd Q}{\dd \lambda} = 0;
+\]
+ce qui exprime que $\dfrac{Q}{R}$ est fonction de $\mu$ seulement
+\[
+R = Q \psi(\mu).
+\]
+Reprenons alors la relation~\Eq{(8)}, et multiplions les \Card{4} coordonnées
+$x\Add{,}y\Add{,}z\Add{,}t$ par un facteur fonction de~$\mu$ de façon à simplifier
+la relation~\Eq{(8)}, qui s'écrit
+\[
+\frac{\dd x_1}{\dd \mu}
+  = Px_1 + Q \left[ \frac{\dd x}{\dd \mu} + x \psi(\mu)\right].
+\]
+On peut multiplier~$x$ par un facteur~$\omega$ tel que l'expression entre
+crochets se réduise~$\theta\omega \dfrac{\dd x}{\dd \mu}$; comme ce facteur~$\omega$ ne dépend pas de~$\lambda$,
+les équations~\Eq{(7)} subsistent, et nous avons des relations de la forme
+\[
+\frac{\dd x_1}{\dd \mu} = Px_1 + Q \frac{\dd x}{\dd \mu}, \quad
+\frac{\dd y_1}{\dd \mu} = \dots\dots, \quad
+\frac{\dd z_1}{\dd \mu} = \dots\dots, \quad
+\frac{\dd t_1}{\dd \mu} = \dots\dots.
+\]
+Ceci revient à supposer $R = 0$ dans les équations~\Eq{(1)}; ce qui
+donne enfin
+\[
+\Tag{(10)}
+\frac{\dd^2 x}{\dd \lambda\, \dd \mu}
+  = P \frac{\dd x}{\dd \lambda} + Q \frac{\dd x}{\dd \mu}, \quad
+\frac{\dd^2 y}{\dd \lambda\, \dd \mu} = \dots, \quad
+\frac{\dd^2 z}{\dd \lambda\, \dd \mu} = \dots, \quad
+\frac{\dd^2 t}{\dd \lambda\, \dd \mu} = \dots.
+\]
+Il est facile de voir que si $x\Add{,}y\Add{,}z\Add{,}t$ satisfont à~\Eq{(10)}, les conditions
+\Eq{(1)}\Add{,} \Eq{(2)}\Add{,} \Eq{(3)}\Add{,} \Eq{(4)} sont satisfaites. Tout d'abord \Eq{(3)}~et~\Eq{(4)}
+le sont, ainsi que~\Eq{(1)}. Voyons alors~\Eq{(2)}. Les équations~\Eq{(10)}
+peuvent s'écrire
+\[
+\frac{\dd x_1}{\dd \mu} = Px_1 + Q \frac{\dd x}{\dd \mu}\Add{,}
+\]
+d'où
+\[
+\frac{\dd^2 x_1}{\dd \lambda\, \dd \mu}
+  = \frac{\dd P}{\dd \lambda} x_1 + P \frac{\dd x_1}{\dd \lambda}
+  + Q \frac{\dd x_1}{\dd \mu}
+  + \frac{1}{Q} \left( \frac{\dd x_1}{\dd \mu} - Px_1\right)
+      \frac{\dd Q}{\dd \lambda}\Add{,}
+\]
+ce qui est bien une équation de la forme~\Eq{(2)}.
+%% -----File: 201.png---Folio 193-------
+
+\Paragraph{\DPchg{\Ord{2}{ème} Cas}{Deuxième Cas}.} Nous supposons maintenant $B, B_1, D, D_1, \neq 0$. Reprenons
+les équations~\Eq{(5)}\Add{,}~\Eq{(6)}\Add{.} En multipliant $x, y, z, t$ et $x_1, y_1, z_1, t_1$,
+par des facteurs convenables, on peut faire \DPchg{disparaitre}{disparaître} dans~\Err{\Eq{(S)}}{\Eq{(5)}}
+le terme en~$x$ et le terme en~$x_1$, de sorte que nous pouvons
+écrire
+\[
+\Tag{(11)}
+\frac{\dd x_1}{\dd \lambda} = L \frac{\dd x}{\dd \lambda}, \qquad
+\frac{\dd y_1}{\dd \lambda} = L \frac{\dd y}{\dd \lambda}, \qquad
+\frac{\dd z_1}{\dd \lambda} = L \frac{\dd z}{\dd \lambda}, \qquad
+\frac{\dd t_1}{\dd \lambda} = L \frac{\dd t}{\dd \lambda}.
+\]
+L'équation~\Eq{(6)} peut s'écrire
+\[
+\Tag{(12)}
+\frac{\dd x_1}{\dd \mu} = M \frac{\dd x}{\dd \mu} + N x + S x_1;
+\]
+différentions par rapport à~$\lambda$ en tenant compte de~\Eq{(11)}, nous
+avons
+\[
+\frac{\dd}{\dd \mu} \left(L \frac{\dd x}{\dd \lambda}\right)
+  = \frac{\dd}{\dd \lambda} \left(M \frac{\dd x}{\dd \mu}\right)
+  + \frac{\dd}{\dd \lambda} (N x) + \frac{\dd}{\dd \lambda} (S x_1);
+\]
+$\dfrac{\dd^2 x}{\dd \lambda\, \dd \mu}$ peut d'après~\Eq{(1)} s'exprimer en fonction de~$\Err{x_1}{x}$, $\dfrac{\dd x}{\dd \lambda}$ et $\dfrac{\dd x}{\dd \mu}$,
+et la relation précédente s'écrit
+\[
+\frac{\dd}{\dd \lambda} (S x_1)
+  = F \left(x, \frac{\dd x}{\dd \lambda}, \frac{\dd x}{\dd \mu}\right),
+\]
+$F$~étant une fonction linéaire, ce qu'on peut écrire encore
+\[
+\frac{\dd S}{\dd \lambda} x_1 + S L \frac{\dd x}{\dd \lambda}
+  = F \left(x, \frac{\dd x}{\dd \lambda}, \frac{\dd x}{\dd \mu}\right).
+\]
+Si $\dfrac{\dd S}{\dd \lambda} \neq 0$, $x_1$ est fonction linéaire de~$x$, $\dfrac{\dd x}{\dd \lambda}$, $\dfrac{\dd x}{\dd \mu}$. Le point~$M$
+est dans le plan tangent en~$M$ à la surface~$(S)$, qui est alors
+une des nappes de la surface focale, cas qui a été précédemment
+examiné. Il faut donc supposer $\dfrac{\dd S}{\dd \lambda} = 0$, $S$~n'est fonction
+que de~$\mu$. Alors si nous reprenons l'équation~\Eq{(12)}, nous pouvons
+multiplier $x_1, y_1, z_1, t_1$ par une fonction de~$\mu$ telle que le
+terme en~$x_1$ \DPchg{disparaisse}{disparaîsse}, les relations~\Eq{(11)} subsistant. Et
+nous ramènerons~\Eq{(12)} à la forme
+\[
+\frac{\dd x_1}{\dd \mu} = H \frac{\dd x}{\dd \mu} + K x.
+\]
+%% -----File: 202.png---Folio 194-------
+Le même raisonnement montrera que $K$~est indépendant de~$\lambda$ et
+que par suite on peut faire \DPchg{disparaitre}{disparaître} le terme en~$x$; finalement
+on a
+\[
+\Tag{(13)}
+\frac{\dd x_1}{\dd \mu} = M \frac{\dd x}{\dd \mu},  \qquad
+\frac{\dd y_1}{\dd \mu} = M \frac{\dd y}{\dd \mu},  \qquad
+\frac{\dd z_1}{\dd \mu} = M \frac{\dd z}{\dd \mu},  \qquad
+\frac{\dd t\Add{_1}}{\dd \mu} = M \frac{\dd t}{\dd \mu}.
+\]
+Les relations \Eq{(11)}~et~\Eq{(13)} sont d'ailleurs suffisantes, car
+on en conclut
+\begin{align*}
+\frac{\dd^2 x_1}{\dd \lambda\, \dd \mu}
+  &= \frac{\dd}{\dd \mu} \left(L \frac{\dd x}{\dd \lambda}\right), \\
+\frac{\dd^2 x_1}{\dd \lambda\, \dd \mu}
+  &= \frac{\dd}{\dd \lambda} \left(M \frac{\dd x}{\dd \mu}\right); \\
+\intertext{d'où}
+\Tag{(14)}
+\frac{\dd}{\dd \lambda} \left(M \frac{\dd x}{\dd \mu}\right)
+  &= \frac{\dd}{\dd \mu} \left(L \frac{\dd x}{\dd \lambda}\right),
+\end{align*}
+équation de la forme~\Eq{(1)}, où $R = 0$; on obtiendrait de même
+\[
+\Tag{(15)}
+\frac{\dd}{\dd \lambda} \left(\frac{1}{M} · \frac{\dd x_1}{\dd \mu}\right)
+  = \frac{\dd}{\dd \mu} \left(\frac{1}{M}\, \frac{\dd x_1}{\dd \lambda}\right),
+\]
+équation de la forme~\Eq{(2)} où $R_1=0$.
+
+\MarginNote{Conclusions.}
+Dans le \emph{\Ord{1}{er} cas}, nous avons été ramenés à faire \DPchg{disparaitre}{disparaître}
+le terme en~$x$ dans l'équation
+\[
+\Tag{(1)}
+\frac{\dd^2 x}{\dd \lambda\, \dd \mu}
+  = P \frac{\dd x}{\dd \lambda} + Q \frac{\dd x}{\dd \mu} + R x
+\]
+au moyen de la substitution
+\[
+x = \omega X
+\]
+on trouve immédiatement la condition
+\[
+\frac{\dd^2 \omega}{\dd \lambda\, \dd \mu}
+  = P \frac{\dd \omega}{\dd \lambda}
+  + Q \frac{\dd \omega}{\dd \mu} + R \omega\Add{,}
+\]
+et on peut dire alors que \emph{la surface~$(S_1)$ est définie par les
+équations}
+\[
+x_1 = \frac{\dd}{\dd \lambda} \left(\frac{x}{\omega}\right)\Add{,} \qquad
+y_1 = \frac{\dd}{\dd \lambda} \left(\frac{y}{\omega}\right)\Add{,} \qquad
+z_1 = \frac{\dd}{\dd \lambda} \left(\frac{z}{\omega}\right)\Add{,} \qquad
+t_1 = \frac{\dd}{\dd \lambda} \left(\frac{t}{\omega}\right)\Add{,}
+\]
+\emph{$\omega$~étant une solution de l'équation~\Eq{(1)}}.
+%% -----File: 203.png---Folio 195-------
+
+Passons au \emph{\Ord{2}{e} cas}: il faut encore faire \DPtypo{disparaitre}{disparaître} le
+terme en~$x$ de l'équation~\Eq{(1)}, ce qui revient à chercher une
+intégrale de cette équation. L'équation prend alors la forme
+\[
+\Tag{(2)}
+\frac{\dd^2 x}{\dd \lambda\, \dd \mu}
+  = P \frac{\dd x}{\dd \lambda} + Q \frac{\dd x}{\dd \mu}\Add{.}
+\]
+Identifions avec l'équation~\Eq{(14)} précédemment obtenue. Nous
+avons
+\[
+\frac{\dd L}{\dd \mu} = P(M - L), \qquad
+\frac{\dd M}{\dd \lambda} = Q(L - M);
+\]
+posons alors
+\[
+L - M = \theta,
+\]
+et nous aurons
+\[
+\frac{\dd L}{\dd \mu} = -P\theta, \qquad
+\frac{\dd M}{\dd \lambda} = Q\theta;
+\]
+et l'on voit immédiatement que $\theta$~doit être intégrale de l'équation
+\[
+\Tag{(3)}
+\frac{\dd^2 \theta}{\dd \lambda\, \dd \mu}
+  + \frac{\dd (P \theta)}{\dd \lambda}
+  + \frac{\dd (Q \theta)}{\dd \mu} = 0\Add{,}
+\]
+qui est ce qu'on appelle \emph{l'adjointe} de~\Eq{(2)}. Ayant~$\theta$, on détermine
+par quadratures $L$~et~$M$; car
+\[
+L = -\int P\theta · d\mu, \qquad
+M =  \int Q\theta · d\lambda\Add{.}
+\]
+
+\MarginNote{Propriétés de
+la correspondance
+précédente.}
+Soient $M(x, y, z, t)$, $M_1(x_1, y_1, z_1, t_1)$; soit maintenant $P$~le
+\Figure{203a}
+point de coordonnées $\left(\dfrac{\dd x}{\dd \lambda},\dots\right)$ ou
+$\left(\dfrac{\dd x_1}{\dd \lambda},\dots\right)$ et $Q$~le point $\left(\dfrac{\dd x}{\dd \mu},\dots\right)$ ou
+$\left(\dfrac{\dd x_1}{\dd \mu},\dots\right)$, de sorte que la droite~$PM$
+est tangente à la courbe $\mu = \cte$ sur
+la surface~$(S)$ et~$PM_1$ à la courbe $\mu =
+\cte$, sur la surface~$(S_1)$, et de même
+%% -----File: 204.png---Folio 196-------
+la droite~$QM$ est tangente à la courbe~$\lambda = \cte$ sur la surface~$(S)$,
+et~$QM_1$ à la courbe~$\lambda = \cte$ sur la surface~$(S_1)$. Les plans
+tangents aux \Card{2} surfaces $(S), (S_1)$ aux points $M, M_1$ se coupent suivant
+la droite~$PQ$. Considérons la congruence de ces droites~$PQ$.
+On peut la définir par les équations
+\[
+X = \frac{\dd x}{\dd \lambda} + \rho \frac{\dd x}{\dd \mu},  \quad
+Y = \frac{\dd y}{\dd \lambda} + \rho \frac{\dd y}{\dd \mu},  \quad
+Z = \frac{\dd z}{\dd \lambda} + \rho \frac{\dd z}{\dd \mu},  \quad
+T = \frac{\dd t}{\dd \lambda} + \rho \frac{\dd t}{\dd \mu}.
+\]
+Les développables de cette congruence sont définies par l'équation
+\[
+\begin{vmatrix}
+  \mfrac{\dd x}{\dd \lambda} &
+  \mfrac{\dd x}{\dd \mu} &
+   \mfrac{\dd^2 x}{\dd \lambda^2} · d \lambda
+ + \mfrac{\dd^2 x}{\dd \lambda\, \dd \mu} d\mu &
+   \mfrac{\dd^2 x}{\dd \lambda\, \dd \mu} · d\lambda
+ + \mfrac{\dd^2 x}{\dd \mu^2} d\mu
+\end{vmatrix} = 0;
+\]
+mais on a
+\[
+\frac{\dd^2 x}{\dd \lambda\, \dd \mu}
+  = P \frac{\dd x}{\dd \lambda} + Q \frac{\dd x}{\dd \mu},
+\]
+de sorte que l'équation précédente se réduit à
+\[
+d\lambda · d\mu = 0.
+\]
+\emph{Les développables de la congruence des droites~$PQ$ correspondent
+donc aux développables de la congruence des droites~$MM\Add{_1}$,
+c'est-à-dire encore aux systèmes conjugués homologues.}
+
+Cherchons maintenant les points focaux. Ils sont donnés
+par l'équation
+\[
+\begin{vmatrix}
+  \mfrac{\dd x}{\dd \lambda} &
+  \mfrac{\dd x}{\dd \mu} &
+  \mfrac{\dd^2 x}{\dd \lambda^2} + \rho \mfrac{\dd^2 x}{\dd \lambda\, \dd \mu} &
+  \mfrac{\dd^2 x}{\dd \lambda\, \dd \mu} + \mfrac{\rho \dd^2 x}{\dd \mu^2}
+\end{vmatrix} = 0,
+\]
+équation qui, à cause de la même condition que précédemment,
+se réduit à $\rho = 0$; une racine est nulle, l'autre infinie;
+\emph{les points focaux ne sont autres que les points~$P, Q$. Ils sont
+dans les plans focaux de la congruence~$MM_1$}.
+
+Considérons le point~$P$, et supposons que l'on fasse $\lambda =
+\cte[]$. La direction de la tangente à la trajectoire du point~$P$
+est définie par un \Ord{2}{e} point, dont les coordonnées sont
+\[
+\frac{\dd}{\dd \mu} \left(\frac{\dd x}{\dd \lambda}\right)
+  = P \frac{\dd x}{\dd \lambda} + Q \frac{\dd x}{\dd \mu},
+\]
+et les analogues.
+%% -----File: 205.png---Folio 197-------
+C'est un point de~$PQ$. Le point~$P$ décrit une courbe tangente à~$PQ$,
+arête de rebroussement de la développable correspondant à
+$\lambda = \cte[]$. Le point~$Q$ décrira de même l'arête de rebroussement
+de la développable correspondant à~$\mu = \cte[]$.
+
+Les propriétés de la correspondance que nous venons d'étudier
+se transforment en elles-mêmes par dualité. En choisissant
+convenablement les coordonnées tangentielles homogènes,
+on aurait donc
+\begin{align*}
+\frac{\dd u_1}{\dd \lambda} &= H \frac{\dd u}{\dd \lambda},
+  \quad\text{et les analogues;}\quad\dots\dots, \\
+\frac{\dd u_1}{\dd \mu} &= K \frac{\dd u}{\dd \mu},
+  \quad\text{et les analogues.}\quad\dots\dots.
+\end{align*}
+Appelons alors congruence~$(K)$ celle des droites~$MM_1$, congruence~$(K')$
+celle des droites~$PQ$. \emph{Si les développables de la congruence~$(K)$
+coupent les surfaces $(S)\Add{,} (S_1)$ suivant deux réseaux
+conjugués, les développables de la congruence~$(K')$ sont circonscrites
+à ces surfaces suivant les mêmes réseaux, et réciproquement.
+Les points focaux de~$(K')$ sont dans les plans focaux
+de~$(K)$, chaque point focal se trouvant dans le plan focal
+qui ne lui correspond pas.}
+
+\Section{Correspondance par plans tangents parallèles.}
+{4.}{} Soit sur la surface~$(S)$ l'une des courbes~$(c)$ du réseau
+conjugué qui correspond à un réseau conjugué sur~$(S_1)$
+et soit~$(c_1)$ la courbe correspondante sur~$(S_1)$. Supposons
+qu'en deux points homologues les plans tangents aux surfaces
+$(S)\Add{,} (S_1)$ soient parallèles; leurs caractéristiques le sont aussi;
+donc \emph{les directions conjuguées homologues sont parallèles.}
+%% -----File: 206.png---Folio 198-------
+Faisons $t = 1$ et~$t_1 = 1$, nous avons
+\begin{alignat*}{3}
+\Tag{(1)}
+\frac{\dd x_1}{\dd \lambda} &= L \frac{\dd x}{\dd \lambda},  \qquad &
+\frac{\dd y_1}{\dd \lambda} &= L \frac{\dd y}{\dd \lambda},  \qquad &
+\frac{\dd z_1}{\dd \lambda} &= L \frac{\dd z}{\dd \lambda}; \\
+\Tag{(2)}
+\frac{\dd x_1}{\dd \mu} &= M \frac{\dd x}{\dd \mu}  \qquad &
+\frac{\dd y_1}{\dd \mu} &= M \frac{\dd y}{\dd \mu}  \qquad &
+\frac{\dd z_1}{\dd \mu} &= M \frac{\dd z}{\dd \mu}.
+\end{alignat*}
+Nous pouvons donc appliquer les résultats précédemment obtenus\Add{.}
+Les plans tangents en~$M\Add{,} M_1$ étant parallèles, la droite~$PQ$ est à
+l'infini. Les droites de la congruence~$(K')$ sont les droites
+du plan de l'infini. Sur chacune de ces droites, les points~$P\Add{,} Q$
+sont les points où elles sont rencontrées par les tangentes
+conjuguées homologues sur $(S)\Add{,} (S_1)$, et le lieu des points~$P\Add{,}Q$
+est tangent à chaque droite~$PQ$ aux points~$P, Q$.
+
+\MarginNote{Cas particulier.}
+En particulier, supposons que, la surface~$(S)$ étant quelconque,
+la surface~$(S_1)$ soit une sphère. La congruence des
+droites~$MM_1$ a des développables qui découpent sur $(S)\Add{,} (S_1)$ des
+réseaux conjugués, les tangentes homologues étant parallèles.
+Or\Add{,} sur une sphère, un réseau conjugué est un réseau orthogonal;
+donc le réseau conjugué de~$(S)$ est aussi un réseau orthogonal,
+ce sont les \emph{lignes de courbure}, dont la recherche est
+ainsi ramenée à celle des développables d'une congruence. En
+particulier, supposons la surface~$(S)$ du \Ord{2}{e} degré, et considérons
+la congruence des droites~$PQ$ du plan de l'infini. Le
+plan de l'infini coupe $(S),(S_1)$ suivant deux coniques $(\Gamma),(\Gamma_1)$.
+Considérons leurs points d'intersection avec une droite~$PQ$;
+les points d'intersection avec~$(\Gamma)$ correspondent aux directions
+des génératrices de~$(S)$ qui passent par~$M$, et qui sont
+les tangentes asymptotiques; les points~$P\Add{,} Q$ qui correspondent
+aux directions principales sont donc conjugués par rapport à
+ces points d'intersection, c'est-à-dire conjugués par rapport
+%% -----File: 207.png---Folio 199-------
+à la conique~$(\Gamma)$. Ils sont de même conjugués par rapport à~$(\Gamma_1)$.
+Les points~$P,Q$ sont les points doubles de l'involution
+déterminée sur la droite~$PQ$ par le faisceau de coniques ayant
+pour bases $(\Gamma),(\Gamma_1)$. La droite~$PQ$ est tangente en~$P\Add{,} Q$ aux \Card{2} coniques
+de ce faisceau qui lui sont tangentes; de sorte que la
+détermination des développables de la congruence~$(K)$, c'est-à-dire
+des lignes de courbure de la quadrique~$(S)$, revenant à
+celle d'un faisceau de coniques, peut se faire algébriquement.
+Si on prend pour paramètres ceux des génératrices rectilignes
+qui passent par un point de~$(S)$, on obtient ainsi l'intégration
+de l'\emph{équation d'Euler}.
+
+\Paragraph{Remarque.} Au lieu du plan de l'infini, on pourrait considérer
+un plan fixe quelconque~$(\Pi)$. La correspondance serait
+telle que les plans tangents en \Card{2} points homologues de $(S), (S_1)$
+se coupent dans le plan~$\Pi$.\DPnote{** TN: [sic], no ()} Les résultats seraient analogues;
+et de même si, corrélativement, on établissait entre les \Card{2} surfaces
+une correspondance telle que la droite~$MM_1$ passe par
+un point fixe.
+
+Considérons deux surfaces $(S), (S_1)$ qui se correspondent
+par plans tangents parallèles. Prenons dans l'espace un point
+fixe~$O$, et substituons à~$(S_1)$ une de ses homothétiques par
+rapport à~$O,(S'_1)$. A tout réseau conjugué sur~$(S_1)$ correspond
+sur~$(S_1\Add{'})$ un réseau homothétique qui est aussi conjugué, et le
+réseau conjugué de~$(S)$ qui correspond à un réseau conjugué
+sur~$(\Err{S_1}{S'_1})$ correspond aussi à un réseau conjugué sur~$(S'_1)$. Imaginons
+que le rapport d'homothétie croisse indéfiniment le
+point~$M'_1$ homothétique de~$\Err{M}{M_1}$ s'éloigne à l'infini, la droite~$MM_1$
+est la parallèle menée par~$M$ au rayon~$OM$. Donc, \emph{si l'on a
+%% -----File: 208.png---Folio 200-------
+\Card{2} surfaces $(S),(S_1)$ se correspondant par plans tangents parallèles,
+si on prend dans l'espace un point fixe~$O$, et si par
+le point~$M$ de~$(S)$ on mène la parallèle~$MN$ au rayon~$\Err{OM}{OM_1}$, les
+développables de la congruence des droites~$MN$ découpent sur~$(S)$
+le réseau conjugué qui correspond à un réseau conjugué
+sur~$(S_1)$}. Si en particulier nous prenons pour~$(S_1)$ une sphère,
+pour~$O$ son centre, $\Err{OM}{OM_1}$~est perpendiculaire au plan tangent à~$(\Err{S}{S_1})$,
+et par conséquent au plan tangent à~$(S)$; $MN$~qui lui est
+parallèle est la normale à~$(S)$. \emph{La congruence des normales à
+une surface a des développables qui déterminent sur cette
+surface un réseau conjugué orthogonal.} On retrouve donc la
+propriété fondamentale des lignes de courbure de la surface~$(S)$.
+
+Remarquons encore que si le rayon de la sphère~$(S_1)$ est
+égal à~$1, x, y, z,$ sont les cosinus directeurs de la normale,
+et les formules \Eq{(1)}\Add{,}~\Eq{(2)} ne sont autres que les formules
+d'Olinde Rodrigues.
+
+
+\ExSection{VIII}
+
+\begin{Exercises}
+\item[40.] On donne deux courbes $C, C_{1}$. Trouver toutes les surfaces~$S$ sur
+lesquelles les courbes de contact des cônes circonscrits à~$S$,
+ayant leurs sommets sur $C$ et~$C_{1}$, forment un réseau conjugué.
+En définissant $C$~et~$C_{1}$ par les équations
+\begin{alignat*}{4}
+x &= f(\lambda), \qquad &
+y &= g(\lambda), \qquad &
+z &= h(\lambda), \qquad &
+t &= k(\lambda); \\
+%
+x &= \phi(\mu), &
+y &= \psi(\mu), &
+z &= \chi(\mu), &
+t &= \theta(\mu),
+\end{alignat*}
+la surface la plus générale répondant à la question est définie
+par les équations
+\begin{align*}
+x &= \int A(\lambda)f(\lambda)\, d\lambda + \int B(\mu)\phi(\mu)\, d\mu, \\
+y &= \int A(\lambda)g(\lambda)\, d\lambda + \int B(\mu)\psi(\mu)\, d\mu, \\
+z &= \int A(\lambda)h(\lambda)\, d\lambda + \int B(\mu)\chi(\mu)\, d\mu, \\
+t &= \int A(\lambda)k(\lambda)\, d\lambda + \int B(\mu)\theta(\mu)\, d\mu.
+\end{align*}
+
+Interpréter géométriquement les formules obtenues de
+façon à trouver une définition géométrique de ces surfaces.
+Transformer par dualité les divers résultats obtenus.
+
+\item[41.] Soit $\Sigma$ la sphère de centre~$O$ et de rayon égal à un; soit $S$
+une surface quelconque et $S'$~sa polaire \DPtypo{reciproque}{réciproque} par rapport
+à~$\Sigma$. Soit $M$ un point quelconque de~$S$ et~$P$ le plan tangent en
+ce point; soient $M'$~et~$P'$ le point et le plan tangent de~$S'$ qui
+correspondent à~$P$ et~$M$ par polaires réciproques. On considère
+la congruence~$K$ des droites~$MM'$ et la congruence~$K'$ des intersections
+des plans $P$~et~$P'$. Montrer que leurs développables
+se correspondent, et que les développables de~$K$ découpent
+sur~$S$ et~$S'$ des réseaux conjugués. Comment les développables
+de~$K$ coupent-elles~$\Sigma?$ Chercher à déterminer~$S$ de manière
+que $K$~soit une congruence de normales; que peut-on dire alors
+des développables de~$K$ et de la surface~$S$?
+
+\item[42.] \DPchg{Etant}{Étant} \DPtypo{donnee}{donnée} une courbe gauche~$C$, par un point fixe~$O$ on
+mène des segments~$OM$ équipollents aux diverses cordes de~$C$.
+Le lieu des points~$M$ est une surface~$S_{0}$. Par chaque point~$M$
+de cette surface on mène la parallèle~$\Delta$ à l'intersection
+des plans osculateurs de $C$~\DPtypo{menes}{menés} aux points $P$~et~$P_{1}$ de~$C$ tels
+que $PP_{1}$~est équipollent à~$OM$. Soient $S_{1}$~et~$S_{2}$ les deux nappes de
+la surface focale de la congruence des droites~$\Delta$:
+
+%[** TN: Regularized formatting of parts]
+\Primo. déterminer
+$S_{1}$~et~$S_{2}$, leur~$ds^2$, leur~$\sum l\,d^{2}x$. Montrer que les
+asymptotiques se correspondent sur $S_{1}$~et~$S_{2}$. Quelles sont les
+courbes de $S_{0}$ qui leur correspondent?
+
+\Secundo. Condition nécessaire
+et suffisante que doit remplir~$C$ pour que la congruence
+des droites~$\Delta$ soit une congruence de normales. Trouver alors
+l'une des surfaces normales. Montrer que les rayons de courbure
+de~$\Sigma$ sont fonctions l'un de l'autre.
+
+\Tertio. En restant
+dans ce cas, rapporter le~$ds^{2}$, de $S_{1}$~aux géodésiques tangentes
+aux droites~$\Delta$ et à leurs trajectoires orthogonales. En conclure
+que $S_{1}$~est applicable sur un \DPchg{paraboloide}{paraboloïde} de révolution.
+
+\textsc{Nota}. Les deux dernières parties de cet exercice se
+rattachent à la fin du chapitre~XIII\@.
+\end{Exercises}
+%% -----File: 209.png---Folio 201-------
+
+
+\Chapitre{IX}{Complexes de Droites.}
+
+\Section{\DPchg{Eléments}{Éléments} fondamentaux d'un complexe de droites.}
+{1.}{} On appelle \emph{complexe} un système de $\infty^{3}$~droites,
+c'est-à-dire une famille de droites dépendant de \Card{3} paramètres.
+
+Soit $A$ un point de l'espace, toutes les droites~$(D)$ du
+complexe qui passent par ce point sont au nombre de~$\infty^{1}$, et
+constituent le \emph{cône du complexe} attaché au point~$A$: nous
+l'appellerons le cône~$(K)$.
+
+Corrélativement: soit un plan~$P$, toutes les droites~$(D)$
+du complexe situées dans ce plan sont au nombre de~$\infty^{1}$, et enveloppent
+une courbe~$(c)$ qui est la \emph{courbe du complexe} associée
+à~$P$. La tangente en tout point de cette courbe est une
+droite du complexe.
+
+Plus généralement nous appellerons \emph{courbe du complexe}
+une courbe~$(c)$ dont toutes les tangentes appartiennent au complexe.
+Considérons sur une telle courbe un point~$A$, et le
+cône du complexe~$(K)$ associé au point~$A$. Ce cône est tangent
+à la courbe~$(c)$. \emph{Une courbe du complexe est tangente en chacun
+de ses points au cône du complexe associé à ce point.}
+
+\Illustration[2.25in]{210a}%[** TN: Moved to top of paragraph]
+Considérons un plan~$P$, et un point~$A$ de ce plan; cherchons
+les droites du complexe situées dans le plan~$P$ et passant
+par~$A$. Considérons le cône du complexe associé au point~$A$,
+%% -----File: 210.png---Folio 202-------
+les droites cherchées sont les
+génératrices de ce cône situées dans
+le plan~$P$: si nous considérons la
+courbe du complexe associée au plan~$P$,
+les droites cherchées sont aussi
+les tangentes issues de~$A$ à cette
+courbe. Cherchons dans le plan~$P$ le
+lieu des points~$A$ tels que deux des
+droites du complexe situées dans le plan~$P$ et passant par~$A$
+soient confondues; les points~$A$ correspondants seront, d'après
+ce qui précède, tels que le cône du complexe correspondant
+soit tangent au plan~$P$: ils doivent aussi être sur la
+courbe du complexe: les droites du complexe confondues \DPtypo{coincident}{coïncident}
+avec la génératrice de contact du cône du complexe, ou
+avec la tangente à la courbe du complexe. Ainsi l'on peut définir
+la courbe du complexe située dans un plan comme étant
+le lieu des points de ce plan pour lesquels le cône du complexe
+est tangent au plan, et la génératrice de contact n'est
+autre que la tangente en ce point à la courbe. La courbe du
+complexe est ainsi définie par points et par tangentes.
+
+\begin{wrapfigure}[14]{O}{1.625in}
+\Input[1.5in]{210b}
+\end{wrapfigure}
+Considérons alors une droite~$(D)$ du complexe; prenons sur
+cette droite un point~$A$, et considérons le cône~$(K)$ du complexe
+associé au point~$A$; soit $P$ le plan
+tangent à ce cône le long de la génératrice~$(D)$.
+A chaque point~$A$ de la droite
+correspond ainsi un plan~$P$. Considérons
+maintenant la courbe~$(c)$ du complexe située
+dans le plan~$P$, elle est tangente à
+%% -----File: 211.png---Folio 203-------
+la droite~$(D)$ précisément au point~$A$, de sorte qu'à chaque
+plan~$P$ passant par la droite correspond un point de cette
+droite. \emph{Il y a une correspondance homographique entre les
+points et les plans d'une droite du complexe.}
+
+Précisons la nature de cette homographie. Une droite
+quelconque peut être représentée par \Card{2} équations de la forme
+\[
+\Tag{(1)}
+X = a Z + f, \qquad
+Y = b Z + g.
+\]
+Pour qu'elle appartienne à un complexe, il faut et il suffit
+qu'il existe une relation entre les paramètres~$a\Add{,}b\Add{,}f\Add{,}g$:
+\[
+\Tag{(2)}
+\phi (a,b,f,g) = 0.
+\]
+Cherchons alors toutes les droites du complexe infiniment
+voisines de la droite~\Eq{(1)} et rencontrant cette droite. Une
+telle droite peut être représentée par les équations
+\[
+\Tag{(3)}
+X = (a + da) Z + (f + df), \qquad
+Y = (b + db) Z + (g + dg).
+\]
+Exprimons qu'elle rencontre la droite~\Eq{(1)}. Les équations
+\[
+\Tag{(4)}
+Z\, da + df = 0, \qquad
+Z\, db + dg = 0,
+\]
+doivent avoir une solution commune en~$Z$, ce qui donne la condition
+\[
+\Tag{(5)}
+da · dg - db · df = 0.
+\]
+Le point d'intersection~$M$ des \Card{2} droites infiniment voisines
+aura alors pour cote
+\[
+Z = - \frac{df}{da}.
+\]
+Différentions la relation~\Eq{(2)}, nous avons
+\[
+\Tag{(6)}
+\frac{\dd \phi}{\dd a}\, da +
+\frac{\dd \phi}{\dd b}\, db +
+\frac{\dd \phi}{\dd f}\, df +
+\frac{\dd \phi}{\dd g}\, dg = 0.
+\]
+Supposons connu le point~$M$, nous avons les relations~\Eq{(4)} dans
+lesquelles $Z$ est connu, et qui par conséquent déterminent les
+rapports des différentielles. Cherchons alors le plan passant
+%% -----File: 212.png---Folio 204-------
+par les deux droites infiniment voisines. Il suffit de multiplier~\Eq{(3)}
+respectivement par $db$ et~$-da$, et d'ajouter, il
+vient, en tenant compte de~\Eq{(5)}
+\[
+\Tag{(7)}
+(X - aZ - f)\, db - (Y - bZ - g)\, da = 0\Add{.}
+\]
+Telle est l'équation du plan cherché: il ne dépend que du rapport~$\dfrac{db}{da}$.
+Nous en concluons que \emph{toutes les droites du complexe
+infiniment voisines de la droite~$D$ et rencontrant cette droite
+en un point~$M$ donné sont dans un même plan, et inversement
+toutes les droites du complexe infiniment voisines de la droite~$D$
+et situées dans un même plan passant par~$D$ rencontrent $D$
+au même point}. Posons
+\[
+\lambda = \frac{da}{db}\Add{,}
+\]
+l'équation~\Eq{(7)} s'écrit
+\[
+\Tag{(8)}
+X - a Z - f -\lambda (Y - b Z - g) = 0\Err{\lambda}{}\Add{.}
+\]
+Démontrons qu'il y a une relation homographique entre~$\lambda, Z$:
+tirons en effet $df, dg$ des équations~\Eq{(4)} et portons dans~\Eq{(6)},
+nous avons
+\[
+\left(\frac{\dd \phi}{\dd a} - Z \frac{\dd \phi}{\dd f}\right) da +
+\left(\frac{\dd \phi}{\dd b} - Z \frac{\dd \phi}{\dd g}\right) db = 0
+\]
+et la relation d'homographie est
+\[
+\lambda \left(\frac{\dd \phi}{\dd a} - Z \frac{\dd \phi}{\dd f}\right)
+            + \frac{\dd \phi}{\dd b} - Z \frac{\dd \phi}{\dd g} = 0.
+\]
+
+Considérons en particulier le cône du complexe de sommet~$M$;
+la génératrice infiniment voisine est une droite du complexe
+rencontrant $D$ en~$M$: le plan de ces \Card{2} droites est le plan
+tangent au cône du complexe, et nous avons l'homographie
+précédemment définie.
+
+Considérons encore une courbe du complexe quelconque
+%% -----File: 213.png---Folio 205-------
+tangente à la droite~$D$ au point~$A$. Considérons une tangente
+infiniment voisine à cette courbe; à la limite cette tangente
+rencontre $D$ au point~$A$, et le plan de ces \Card{2} droites n'est autre
+que le plan osculateur à la courbe au point~$A$, et ce plan
+osculateur est associé au point~$A$ dans l'homographie précédente.
+Donc \emph{toutes les courbes du complexe tangentes à une
+droite~$D$ en un même point~$A$ ont même plan osculateur en ce
+point: c'est le plan tangent au cône du complexe associé au
+point~$A$}.
+
+Considérons enfin une congruence de droites appartenant
+au complexe; prenons dans cette congruence une droite~$D$, et
+sur cette droite un point focal~$A$; le point~$A$ appartient à
+une des nappes de la surface focale de la congruence; il appartient
+aussi à l'arête de rebroussement d'une des développables
+de la congruence, et cette arête de rebroussement, enveloppe
+de droites~$D$ appartenant au complexe, est une courbe du
+complexe. Son plan osculateur en~$A$ est le \Ord{2}{e} plan focal de la
+congruence; d'après ce qui précède, \emph{toutes les congruences du
+complexe passant par la droite~$D$ et ayant un foyer en~$A$ ont
+même \Ord{2}{e} plan focal relatif à la droite~$D$;} il y a correspondance
+homographique entre ce \Ord{2}{e} plan focal et le point~$A$.
+
+\Section{Surfaces du complexe.}
+{2.}{} Cherchons si dans un complexe il y a des congruences
+ayant une surface focale double. Sur une telle surface~$(\Phi)$
+les arêtes de rebroussement des développables sont des lignes
+asymptotiques; or\Add{,} ce sont des courbes du complexe.
+Il s'agit donc de trouver des surfaces telles qu'une famille
+%% -----File: 214.png---Folio 206-------
+de lignes asymptotiques soit formée de courbes du complexe.
+Considérons une telle asymptotique~$(c)$ et un de ses points~$A$.
+Le plan osculateur à la courbe~$(c)$ en~$A$ est le plan tangent
+au cône~$(K)$ du complexe associé au point~$A$, et ce plan osculateur
+est tangent à la surface~$(\Phi)$. Les surfaces cherchées
+sont donc tangentes en chacun de leurs points au cône du complexe
+associé à ce point. Réciproquement soit~$(\Phi)$ une telle
+surface; considérons en chacun de ses points la génératrice
+de contact~$(D)$ du cône du complexe avec le plan tangent. Nous
+déterminons ainsi sur la surface~$(\Phi)$ une famille de courbes
+tangentes en chaque point aux droites~$(D)$; ces courbes~$(\DPtypo{C}{c})$
+sont des courbes du complexe; leur plan osculateur est le
+plan tangent au cône du complexe le long de la droite~$(D)$,
+c'est le plan tangent à la surface~$(\Phi)$ et les courbes~$(c)$ sont
+des asymptotiques de cette surface. De telles surfaces sont
+appelées \emph{surfaces du complexe}.
+
+Considérons les équations d'une droite du complexe
+\[
+\Tag{(1)}
+x = az + f, \qquad
+y = bz + g,
+\]
+$a\Add{,} b\Add{,} f\Add{,} g$ étant liés par l'équation
+\[
+\Tag{(2)}
+\phi(a, b, f, g) = 0.
+\]
+Transportons l'origine au point~$(x, y, z)$ et appelons $X, Y, Z$ les
+nouvelles coordonnées. $X, Y, Z$~sont alors les coefficients de
+direction d'une droite du complexe
+\[
+a = \frac{X}{Z}, \qquad
+b = \frac{Y}{Z},
+\]
+et l'équation du cône du complexe associé au point $(x\Add{,}y\Add{,}z)$ est
+\[
+\phi\left(\frac{X}{Z},
+          \frac{Y}{Z},
+          x - \frac{X}{Z} z,
+          y - \frac{Y}{Z} z\right) = 0,
+\]
+%% -----File: 215.png---Folio 207-------
+ou, en rendant homogène
+\[
+\Psi(X, Y, Z, xZ - zX, yZ - zY) = 0;
+\]
+les courbes du complexe sont alors définies par l'équation
+différentielle, homogène en~$dx, dy, dz$,
+\[
+\Psi(dx · dy · dz, x\,dz - z\,dx, y\,dz - z\,dy) = 0.
+\]
+Une telle équation s'appelle une \emph{équation de Monge}, et \emph{équation
+de Pfaff} si elle est du \Ord{1}{er} degré.
+
+Prenons maintenant l'équation tangentielle du cône du
+complexe
+\[
+F(x, y, z, U, V, W) = 0;
+\]
+la condition pour qu'une surface $z = G(x, y)$ soit tangente à
+ce cône en chacun de ses points, est que l'équation soit vérifiée
+par $U = \dfrac{\dd G}{\dd x} = p$, $V = \dfrac{\dd G}{\dd y} = q$, $W = - 1$; les surfaces du complexe
+sont donc définies par l'équation aux dérivées partielles
+\[
+F(x, y, z, p, q, -1) = 0,
+\]
+qui est de la forme:
+\[
+\Tag{(3)}
+f(x, y, z, p, q) = 0.
+\]
+Nous obtenons une équation aux dérivées partielles du \Ord{1}{er} ordre.
+Inversement, avec les notations précédentes, toute
+équation aux dérivées partielles du \Ord{1}{er} ordre pouvant se
+mettre sous la forme
+\[
+\Tag{(4)}
+f\left(x, y, z, \Err{}{-}\frac{U}{W}, \Err{}{-}\frac{V}{W}\right) = 0
+\]
+exprime que le plan tangent à une surface intégrale est tangent
+à un certain cône associé au point de contact, mais les
+génératrices de tous ces $\infty^{3}$~cônes remplissent en général tout
+l'espace, et ne forment un complexe qu'exceptionnellement.
+
+Pour pouvoir mieux préciser ce cas d'exception, rappelons
+%% -----File: 216.png---Folio 208-------
+les points essentiels de la théorie des équations générales
+aux dérivées partielles du premier ordre, c'est-à-dire
+de la forme~\Eq{(3)}.
+
+Un \emph{élément de contact intégral} est un élément de contact
+dont les coordonnées $(x, y, z, p, q)$ satisfont à l'équation donnée~\Eq{(3)}.
+
+Le \emph{cône élémentaire} associé au point $(x, y, z)$ est l'enveloppe
+des éléments de contact intégraux appartenant à ce point
+son équation tangentielle est précisément l'équation~\Eq{(4)}.
+Tout élément linéaire formé d'un point et d'une génératrice du
+cône élémentaire associé à ce point s'appelle un \emph{élément linéaire
+intégral}. Si $dx, dy, dz$~sont les coefficients de direction
+d'une telle génératrice, l'équation qui caractérise les
+éléments linéaires intégraux s'obtient en éliminant $p$~et~$q$
+entre les équations:
+\[
+\Tag{(5)}
+f(x, y, z, p, q) = 0, \quad
+dz - p\, dx - q\, dy = 0, \quad
+\frac{\dd f}{\dd p} dy - \frac{\dd f}{\dd q} dx = 0.
+\]
+
+\Illustration[1.75in]{217a}
+\noindent L'équation obtenue est une équation de Monge:
+\[
+\Tag{(6)}
+G(x, y, z, dx, dy, dz) = 0.
+\]
+
+Les \emph{courbes intégrales} sont les courbes dont tous les
+éléments linéaires (points-tangentes) sont intégraux. Elles
+sont définies par l'équation~\Eq{(6)}.
+
+Une \emph{bande intégrale} est un lieu d'éléments de contact
+appartenant à une même courbe (points-plans tangents), et
+qui soient tous des éléments de contact intégraux. C'est donc
+un ensemble de $\infty^{1}$~éléments de contact satisfaisant aux équations
+\[
+\Tag{(7)}
+f(x, y, z, p, q) = 0, \qquad
+dz - p\,dx - q\,dy = 0.
+\]
+%% -----File: 217.png---Folio 209-------
+Si on prend une courbe quelconque et si par chacune de ses
+tangentes on mène un plan tangent au cône élémentaire associé
+au point de contact, on obtient une bande intégrale. Par une
+courbe quelconque passent donc, si l'équation~\Eq{(3)}
+est algébrique en~$p, q$, un nombre
+limité de bandes intégrales. Ce nombre
+se réduit de un dans le cas où la courbe
+est une courbe intégrale.
+
+\Paragraph{Par une bande intégrale passe en général
+une surface intégrale et une seule.}
+Les bandes intégrales qui font exception s'appellent \emph{bandes
+caractéristiques}. Les courbes qui leur servent de supports
+sont des courbes intégrales particulières, qu'on appelle
+\emph{caractéristiques}.
+
+Les bandes caractéristiques sont définies par les équations
+\begin{gather*}
+f(x, y, z, p, q) = 0, \\
+%
+\Tag{(8)}
+\frac{dx}{\dfrac{\dd f}{\dd p}}
+  = \frac{dy}{\dfrac{\dd f}{\dd q}}
+  = \frac{dz}{p \dfrac{\dd f}{\dd p} + q \dfrac{\dd f}{\dd q}}
+  = \frac{dp}{- \dfrac{\dd f}{\dd x} - p \dfrac{\dd f}{\dd z}}
+  = \frac{dq}{- \dfrac{\dd f}{\dd y} - q \dfrac{\dd f}{\dd z}}.
+\end{gather*}
+
+On les obtient donc en intégrant un système d'équations
+différentielles ordinaires. \emph{Par un élément de contact intégral
+passe une bande caractéristique et une seule.}
+
+\emph{La surface intégrale qui passe par une bande intégrale
+non caractéristique donnée est engendrée par les bandes caractéristiques
+passant par les divers éléments de contact intégraux
+de cette bande intégrale.}
+
+Sur une surface intégrale il y a au plus une courbe
+%% -----File: 218.png---Folio 210-------
+intégrale qui ne soit pas une caractéristique.
+
+\Paragraph{Toute courbe intégrale est l'enveloppe d'une famille
+de $\infty^{1}$~courbes caractéristiques.} Ces caractéristiques engendrent
+une surface intégrale.
+
+\emph{Réciproquement}: si une famille de $\infty^{1}$~caractéristiques
+a une enveloppe, cette enveloppe est une courbe intégrale.
+
+L'intégration du système~\Eq{(8)} suffit donc pour l'intégration
+de l'équation~\Eq{(3)} et de l'équation de Monge~\Eq{(6)}, qui lui
+est associée.
+
+Ces trois intégrations s'achèvent enfin immédiatement si
+on a une \emph{intégrale complète}, c'est-à-dire une équation où
+figure deux constantes arbitraires
+\[
+H(x, y, z, a, b) = 0
+\]
+définissant des surfaces intégrales, pour toutes les valeurs
+de ces constantes.
+
+Les \emph{courbes caractéristiques} sont alors définies par les
+équations
+\[
+H = 0,\qquad
+\frac{\dd H}{\dd a} + c \frac{\dd H}{\dd b} = 0,
+\]
+où $c$~est une nouvelle constante arbitraire.
+
+Une \emph{surface intégrale quelconque} s'obtient en prenant
+l'enveloppe de $\infty^{1}$~surfaces, faisant partie de l'intégrale
+complète, c'est-à-dire en éliminant~$a$ entre les équations\DPtypo{.}{}
+\[
+H(x, y, z, a, b) = 0,\qquad
+\frac{\dd H}{\dd a}\, da + \frac{\dd H}{\dd b}\, db = 0,
+\]
+après y avoir remplacé~$b$ par une fonction arbitraire de~$a$.
+
+Les caractéristiques tracées sur une telle surface ont
+nécessairement une enveloppe; par suite on obtient une \emph{courbe
+%% -----File: 219.png---Folio 211-------
+intégrale quelconque}, en éliminant~$a$ entre les équations
+\[
+H = 0,\quad
+\frac{\dd H}{\dd a}\, da + \frac{\dd H}{\dd b}\, db = 0,\quad
+\frac{\dd^2 H}{\dd a^2}\, da^2 + 2 \frac{\dd^2 H}{\dd a\, \dd b}\, da\, db +
+\frac{\dd^2 H}{\dd b^2}\, db^2 + \frac{\dd H}{\dd b}\, d^2 b = 0,
+\]
+après y avoir remplacé~$b$ par une fonction arbitraire de~$a$.
+
+Si nous revenons maintenant au cas particulier où l'équation~\Eq{(3)}
+est celle qui définit les surfaces d'un complexe,
+nous voyons que les courbes intégrales sont les courbes du
+complexe, et que les caractéristiques situées sur une surface
+intégrale constituent la famille de $\infty^{1}$~courbes du complexe
+qui sont les lignes asymptotiques de cette surface. Il en résulte
+que les équations~\Eq{(8)} ont alors pour conséquence
+\[
+dp\, dx + dq\, dy = 0,
+\]
+c'est-à-dire que l'équation~\Eq{(3)} a elle-même pour conséquence
+\[
+\frac{\dd f}{\dd p} \left(\frac{\dd f}{\dd x}
+  + p\, \frac{\dd f}{\dd z}\right) +
+\frac{\dd f}{\dd q} \left(\frac{\dd f}{\dd y}
+  + q\, \frac{\dd f}{\dd z}\right) = 0.
+\]
+On démontre que, réciproquement, les seules équations~\Eq{(3)}
+pour lesquelles les caractéristiques sont les lignes asymptotiques
+des surfaces intégrales, sont, (si on excepte les
+équations linéaires), les équations dont les cônes élémentaires
+sont les cônes des complexes de droites.
+
+\Paragraph{Remarque.} Si le cône du complexe se réduit à un plan, le complexe
+est appelé un \emph{complexe linéaire}. Le cône n'a alors pas
+d'équation tangentielle, et la théorie précédente ne s'applique
+plus.
+
+Le cas des complexes linéaires sera étudié dans le chapitre
+suivant.
+
+
+\Section{Complexes spéciaux.}
+{3.}{} Nous dirons qu'un complexe est \emph{spécial} quand l'homographie
+qui existe entre les points et les plans d'une droite
+%% -----File: 220.png---Folio 212-------
+du complexe est spéciale. A un élément d'un système correspond
+toujours le même élément dans le système associé, sauf
+pour un seul élément du \Ord{1}{er} système, dont le correspondant
+est indéterminé. L'équation de l'homographie étant
+\[
+\lambda \left(\frac{\dd \phi}{\dd a} - z\, \frac{\dd \phi}{\dd f}\right)
+            + \frac{\dd \phi}{\dd b} - z\, \frac{\dd \phi}{\dd b} = 0\Add{,}
+\]
+la condition pour qu'on ait une homographie spéciale est
+\[
+\Tag{(1)}
+\frac{\dd \phi}{\dd a} · \frac{\dd \phi}{\dd g} -
+\frac{\dd \phi}{\dd b} · \frac{\dd \phi}{\dd f} = 0.
+\]
+Considérons le \emph{complexe des droites tangentes à une surface};
+considérons une congruence de ce complexe; \Err{les développables}{ces développables de l'une des familles}
+de la congruence seront circonscrites à la surface, l'un des
+plans focaux sera indépendant de la \Err{développable}{congruence} que l'on
+considère. Même résultat si on considère le \emph{complexe des
+droites rencontrant une courbe donnée}. On obtient donc ainsi
+des complexes spéciaux. Nous allons montrer qu'il n'y en a
+pas d'autres. Prenons l'équation d'un complexe sous la forme
+\[
+\Phi = g - \phi(a, b, f) = 0;
+\]
+\Eq{(1)}~s'écrit
+\[
+%[** TN: Original has a leading +; suspect artifact from typist]
+\Tag{(2)}
+\frac{\dd \phi}{\dd a} + \frac{\dd \phi}{\dd b} · \frac{\dd \phi}{\dd f} = 0.
+\]
+Cette relation ne contient plus~$g$, elle doit être une identité
+par rapport à~$a, b, f$. Considérons une droite~$D$ du complexe, et
+les droites infiniment voisines qui la rencontrent; on a la
+condition
+\[
+da · d\phi - db · df = 0,
+\]
+ou
+\[
+db · df - da \left(
+  \frac{\dd \phi}{\dd a}\, da
++ \frac{\dd \phi}{\dd b}\, db
++ \frac{\dd \phi}{\dd f}\, df\right) = 0;
+\]
+remplaçons $\dfrac{\dd \phi}{\dd a}$ par sa valeur tirée de~\Eq{(2)}, il vient
+\[
+\frac{\dd \phi}{\dd b} · \frac{\dd \phi}{\dd f}\, da^2
+  - \frac{\dd \phi}{\dd b}\, da · db
+  - \frac{\dd \phi}{\dd f}\, da · df + db · df = 0,
+\]
+ou
+\[
+\Tag{(3)}
+\left(\frac{\dd \phi}{\dd b}\, da - df\right)
+\left(\frac{\dd \phi}{\dd f}\, da - db\right) = 0.
+\]
+%% -----File: 221.png---Folio 213-------
+le point de rencontre de la droite~$D$ avec les droites infiniment
+voisines est
+\[
+\Tag{(4)}
+z = - \frac{df}{da} = - \frac{\dd \phi}{\dd b},
+\]
+de sorte qu'à tout plan passant par~$D$ correspond toujours le
+même point~$F$:
+\[
+x = az + f,\qquad
+y = bz + \phi,\qquad
+z = - \frac{\dd \phi}{\dd b}.
+\Tag{(5)}
+\]
+Différentions $x, y$
+\[
+dx = a\, dz + z\, da + df,\qquad
+dy = b\, dz + z\, db + d \phi,
+\]
+d'où, en remplaçant $z$ par sa valeur
+\[
+dx - a\, dz = - \frac{\dd \phi}{\dd b}\, da + df,\qquad
+dy - b\, dz =   \frac{\dd \phi}{\dd a}\, da + \frac{\dd \phi}{\dd f}\, df;
+\]
+d'où la relation
+\[
+\Tag{(6)}
+-\frac{\dd \phi}{\dd f} (dx - a\, dz) + dy - b\, dz = 0.
+\]
+Les différentielles $dx, dy, dz$ sont liées par une relation linéaire
+et homogène; les fonctions~$x, y, z$ sont liées au moins
+par une relation.
+
+Si on n'a qu'une relation, le lieu des points~$F$ est une
+surface, et \Eq{(6)}~exprime que la droite~$D$ est tangente à cette
+surface. Si on a 2~relations, le lieu des points~$F$ est une
+courbe et la droite~$D$ rencontre cette courbe. Tels sont les
+2~seuls cas possibles pour les complexes spéciaux.
+
+\Paragraph{Remarques. \1.} Dans l'équation~\Eq{(3)} nous avons jusqu'à
+présent considéré le seul facteur $\left(\dfrac{\dd \phi}{\dd b}\, da - df\right)$. Annulant l'autre
+facteur
+\[
+\frac{db}{da} = \frac{\dd \phi}{\dd f},
+\]
+nous aurions alors des droites du complexe qui seraient
+%% -----File: 222.png---Folio 214-------
+toutes situées dans un même plan avec~$D$, ce plan serait le
+plan singulier de l'homographie, et précisément le plan tangent
+à la surface lieu des points~$F$. On voit ainsi qu'en prenant
+l'un ou l'autre des facteurs, on définit la même surface
+par points et par plans tangents.
+
+\Paragraph{\2.} Si l'équation du complexe ne contient ni~$f$ ni~$g$, on
+a une relation entre les coefficients de direction de la droite~$D$,
+on a le complexe des droites rencontrant une même courbe
+à l'infini.
+
+\Paragraph{\3.} Le calcul précédent peut s'interpréter dans le cas
+d'un complexe quelconque. L'équation~\Eq{(1)}, qui n'est plus alors
+conséquence de l'équation du complexe, jointe à cette équation
+du complexe, définit une congruence des droites du complexe
+sur lesquelles l'homographie est spéciale. Ce sont les
+\emph{droites singulières} du complexe. Alors \emph{toutes les surfaces
+réglées du complexe passant par une droite singulière ont même
+plan tangent au point~$F$ de cette droite défini précédemment},
+ce plan tangent étant parallèle au plan
+\[
+-\frac{\dd \phi}{\dd f} (x - az) + y - bz = 0.
+\]
+Si le lieu des points singuliers est une surface, \Eq{(6)}~montre
+que cette surface est aussi l'enveloppe des plans singuliers,
+et les droites singulières lui sont tangentes. \emph{La surface des
+singularités est une des nappes de la surface focale de la
+congruence des droites singulières, les points et les plans
+singuliers sont des éléments focaux de cette congruence non
+associés entre eux. Si le lieu des points singuliers est une
+%% -----File: 223.png---Folio 215-------
+courbe, les plans singuliers sont \(d'après~\Eq{(6)}\) tangents à
+cette courbe, qui est une courbe focale de la congruence des
+droites singulières.}
+
+\Paragraph{\4.} Considérons en particulier le cas des \emph{complexes du
+\Ord{2}{e} degré}. En un point quelconque, le plan associé est tangent
+au cône du complexe; il est unique et bien déterminé. Il ne
+peut y avoir indétermination que si le cône du complexe associé
+à ce point se décompose. \emph{La surface des singularités est
+donc le lieu des points où le cône du complexe se décompose;
+c'est aussi l'enveloppe des plans pour lesquels la courbe du
+complexe se décompose}, comme le verrait par un raisonnement
+analogue.
+
+\MarginNote{Surfaces et
+courbes des
+complexes spéciaux.}
+Revenons aux complexes spéciaux: considérons d'abord le
+cas du complexe des tangentes à une surface~$(\Phi)$. Les cônes
+du complexe sont les cônes circonscrits à cette surface.
+Les plans tangents à~$(\Phi)$ constituent une intégrale complète.
+Une intégrale quelconque est donc l'enveloppe de $\infty^{1}$~plans tangents
+à~$(\Phi)$, c'est-à-dire une développable quelconque circonscrite
+à~$(\Phi)$. Les caractéristiques, qui sont en général les
+courbes de contact de la surface intégrale avec les surfaces,
+faisant partie de l'intégrale complète, qu'elle enveloppe,
+sont les génératrices rectilignes de ces développables, c'est-à-dire
+les droites même du complexe. Enfin on obtiendra les
+courbes intégrales en prenant l'enveloppe des caractéristiques
+sur les surfaces intégrales; ce sont précisément les
+arêtes de rebroussement des développables qui sont les courbes
+%% -----File: 224.png---Folio 216-------
+du complexe.
+
+Considérons maintenant le complexe des droites rencontrant
+une courbe; on voit de même que les surfaces du complexe
+sont les développables passant par la courbe, les caractéristiques
+sont les droites du complexe, et les courbes
+du complexe sont les arêtes de rebroussement.
+
+\emph{Dans les complexes spéciaux, l'équation aux dérivées partielles
+du \Ord{1}{e} ordre dont dépend la recherche des surfaces du
+complexe a pour caractéristiques les droites du complexe.
+Réciproquement toute équation aux dérivées partielles du \Ord{1}{er} ordre
+dont les caractéristiques sont des droites est associée
+à un complexe spécial.}
+
+Soit en effet l'équation aux dérivées partielles
+\[
+f(x, y, z, p, q) = 0
+\]
+dont les caractéristiques sont des droites. On obtient les
+surfaces intégrales en prenant une courbe intégrale et en menant
+les caractéristiques tangentes: donc les surfaces intégrales
+sont des développables, et le plan tangent est le même
+le long de chaque caractéristique, c'est-à-dire que $dp = 0$,
+$dq = 0$ doivent être conséquences de l'équation des caractéristiques,
+ce qui revient à dire que $f = 0$ doit \DPtypo{entrainer}{entraîner} comme
+conséquence les équations
+\[
+\frac{\dd f}{\dd x} + p\, \frac{\dd f}{\dd z} = 0,\qquad
+\frac{\dd f}{\dd y} + q\, \frac{\dd f}{\dd z} = 0.
+\]
+Supposons alors que $z$~figure dans l'équation aux dérivées
+partielles et posons
+\[
+f = z - \Phi(x, y, p, q);
+\]
+%% -----File: 225.png---Folio 217-------
+les conditions précédentes s'écriront
+\[
+\frac{\dd \Phi}{\dd x} - p = 0,\qquad
+\frac{\dd \Phi}{\dd y} - q = 0,
+\]
+d'où il résulte que $\Phi$~est de la forme
+\[
+\Phi = px + qy + \Psi(p, q),
+\]
+et l'équation aux dérivées partielles est
+\[
+z - px - qy = \Psi(p, q).
+\]
+Le plan tangent à une quelconque des surfaces intégrales est
+donc
+\[
+pX + qY - Z + \Psi(p\DPtypo{.}{,} q) = 0\Add{.}
+\]
+L'ensemble de tous ces plans a donc une enveloppe, surface ou
+courbe. Le cône élémentaire associé à un point quelconque est
+le cône circonscrit à cette surface ou à cette courbe, et
+l'équation aux dérivées partielles est bien associée à un
+complexe spécial.
+
+\Paragraph{Remarque.} Nous avons dû supposer que $z$~figurait dans
+l'équation aux dérivées partielles; s'il n'en est pas ainsi,
+cette équation s'écrit
+\[
+\Phi(x, y, p, q) = 0
+\]
+et les conditions obtenues plus haut s'écrivent
+\[
+\frac{\dd \Phi}{\dd x} = 0,\qquad
+\frac{\dd \Phi}{\dd y} = 0;
+\]
+$\Phi$ doit être indépendant de~$x\Add{,} y$, et l'équation aux dérivées partielles
+prend la forme
+\[
+\Phi(p, q) = 0.
+\]
+On a alors le complexe des droites rencontrant une courbe à
+l'infini.
+
+Considérons par exemple l'équation
+\[
+1 + p^2 + q^2 = 0
+\]
+%% -----File: 226.png---Folio 218-------
+elle définit le \emph{complexe des droites isotropes}; les courbes
+du complexe sont les courbes minima, et on les obtient sans
+intégration comme arêtes de rebroussement des développables
+isotropes.
+
+\Section{Surfaces normales aux droites du complexe\Add{.}}
+{4.}{} Proposons-nous maintenant de chercher les \emph{surfaces
+dont les normales appartiennent au complexe} défini par l'équation
+\[
+\Phi(a, b, f, g) = 0.
+\]
+Une normale à une surface du complexe est définie par les
+équations
+\[
+\frac{X - x}{p} = \frac{Y - y}{q} = -(Z - z)
+\]
+ou
+\[
+X = -pZ + x + pz,\qquad
+Y = -qZ + y + qz;
+\]
+de sorte que les surfaces cherchées sont définies par l'équation
+aux dérivées partielles
+\[
+\Phi(-p, -q, x + pz, y + qz) = 0.
+\]
+Si une surface répond à la question, il est évident que toutes
+les surfaces parallèles répondent aussi à la question.
+Si le complexe est spécial, le problème revient à la recherche
+d'une congruence de normales, connaissant une des multiplicités
+focales. Pour le cas d'un complexe quelconque, nous
+allons chercher les congruences de normales appartenant au
+complexe: on obtiendra ensuite les surfaces au moyen d'une
+quadrature. Pour que  $\infty^{2}$~droites:
+\[
+\frac{x - f}{a} = \frac{y - g}{b} = \frac{z - 0}{1}
+\]
+%% -----File: 227.png---Folio 219-------
+soient les normales d'une même surface, la condition est, en
+posant
+\[
+\alpha = \frac{a}{\sqrt{a^2 + b^2 + 1}},\qquad
+\beta  = \frac{b}{\sqrt{a^2 + b^2 + 1}},\qquad
+\gamma = \frac{1}{\sqrt{a^2 + b^2 + 1}},
+\]
+que $\alpha\, df + \beta\, dg$ soit une différentielle exacte. Or\Add{,} l'équation
+du complexe, résolue par rapport à~$\beta$ peut s'écrire
+\[
+\beta = \phi(\alpha, f, g)\Add{,}
+\]
+et $\alpha\, df + \phi(\alpha, f, g)\, dg$ doit être une différentielle exacte par
+rapport à deux variables indépendantes. Déterminons~$\alpha$ par
+exemple en fonction de $f, g$, nous aurons la condition
+\[
+\frac{\dd \alpha}{\dd g}
+  = \frac{\dd \phi}{\dd \alpha} · \frac{\dd \alpha}{\dd f}
+  + \frac{\dd \phi}{\dd f}.
+\]
+Cherchons une solution de la forme
+\[
+F(\alpha, f, g) = \cte,
+\]
+nous avons, pour déterminer~$F$,
+\[
+\frac{\dd F}{\dd f} + \frac{\dd \alpha}{\dd f} · \frac{\dd F}{\dd \alpha} = 0,
+\qquad
+\frac{\dd F}{\dd g} + \frac{\dd \alpha}{\dd g} · \frac{\dd F}{\dd \alpha} = 0.
+\]
+On est ramené à l'équation
+\[
+\frac{\dd F}{\dd g}
+  - \frac{\dd \phi}{\dd \alpha} · \frac{\dd F}{\dd f}
+  + \frac{\dd \phi}{\dd f} · \frac{\dd F}{\dd \alpha} = 0,
+\]
+qui se ramène au système d'équations différentielles ordinaires
+\[
+dg = \frac{\ -df\ }{\dfrac{\dd \phi}{\dd \alpha}}
+   = \frac{\ d \alpha\ }{\dfrac{\dd \alpha}{\dd f}}.
+\]
+
+Remarquons encore que \emph{les développées des surfaces cherchées
+sont les surfaces pour lesquelles $\infty^{1}$~géodésiques sont
+des courbes du complexe}. Ce sont les surfaces focales des congruences
+considérées.
+
+
+\ExSection{IX}
+
+\begin{Exercises}
+\item[43.] On considère deux plans rectangulaires, et toutes les droites
+telles que le segment intercepté sur chacune d'elles par les
+plans précédents ait une longueur constante. Trouver les congruences
+de normales du complexe de ces droites.
+
+\item[44.] On considère trois plans formant un trièdre trirectangle et
+les droites telles que le rapport des segments déterminés par
+ces trois plans sur chacune d'elles soit constant. Trouver
+les surfaces dont les normales appartiennent au complexe de
+ces droites. Il y a parmi ces surfaces une infinité de surfaces
+du \Ord{2}{e} ordre admettant les \Card{3} plans donnés comme plans de
+symétrie. Le complexe précédent est celui des normales à une
+famille de quadriques homofocales, ou homothétiques par rapport
+à leur centre.
+\end{Exercises}
+%% -----File: 228.png---Folio 220-------
+
+
+\Chapitre{X}{Complexes Linéaires.}
+
+\Section{Généralités sur les complexes algébriques.}
+{1.}{} Soit une droite
+\[
+\Tag{(1)}
+x = az + f,\qquad
+y = bz + g;
+\]
+un \emph{complexe algébrique} sera défini par une relation algébrique
+entre $a, b, f, g$:
+\[
+\Phi(a, b, f, g) = 0.
+\]
+Si on considère les droites du complexe passant par un point~$A$,
+et situées dans un plan~$P$ passant par ce point, ce sont
+les génératrices d'intersection du plan~$P$ avec le cône du
+complexe associé au point~$A$, ou bien les tangentes issues de~$A$
+à la courbe du complexe située dans le plan~$P$; si le complexe
+est algébrique, le cône et la courbe sont algébriques,
+et on voit que \emph{le degré du cône du complexe est égal à l'ordre
+de la courbe plane du complexe}; leur valeur commune s'appelle
+le \emph{degré du complexe}, c'est le nombre de droites du
+complexe situées dans un plan et passant par un point de ce
+plan.
+
+Si ce nombre est égal à~$1$, on a ce qu'on appelle un
+\emph{complexe linéaire}; le cône du complexe associé au point~$A$ est
+un plan qu'on appelle \emph{plan focal} ou \emph{plan polaire} du point~$A$.
+La courbe du complexe située dans un plan~$P$ se réduit à un
+point, qu'on appelle \emph{foyer} ou \emph{pôle} du plan~$P$; si le plan~$P$ est
+%% -----File: 229.png---Folio 221-------
+le plan polaire du point~$A$, le point~$A$ est le pôle du plan~$P$;
+\emph{il y a réciprocité entre un pôle et son plan polaire}.
+
+\Section{Coordonnées homogènes.}
+{2.}{} Pour l'étude des complexes algébriques il y a avantage
+à remplacer $a, b, f, g$ par les coordonnées homogènes de
+droites.
+
+\Paragraph{Coordonnées de Plücker.} Considérons les équations d'une
+droite en coordonnées cartésiennes
+\[
+\Tag{(2)}
+\frac{X - f}{a} = \frac{Y - g}{b} = \frac{Z - h}{c},
+\]
+équations qui contiennent comme cas particulier les équations~\Eq{(1)}.
+Nous prendrons pour coordonnées pluckériennes de la
+droite les \Card{6} quantités
+\[
+a,\quad b,\quad c,\qquad
+p = gc - hb,\qquad
+q = ha - fc,\qquad
+r = fb - ga\Add{.}
+\]
+Ces \Card{6} coordonnées sont, comme on le voit immédiatement, liées
+par la relation homogène
+\[
+\Tag{(3)}
+pa + qb + rc = 0.
+\]
+Ces \Card{6} paramètres liés par une relation homogène se réduisent
+à~\Card{4} en réalité; $a, b, c$~sont les projections sur les axes d'un
+certain segment porté par la droite; $p, q, r$~sont les moments de
+ce segment par rapport aux axes (en coordonnées rectangulaires).
+
+Voyons ce que devient l'équation du complexe. De~\Eq{(2)} on
+tire
+\[
+X = \frac{a}{c}\, Z - \frac{q}{c},\qquad
+Y = \frac{b}{c}\, Z + \frac{p}{c},
+\]
+et l'équation
+\[
+\Phi(a, b, f, g) = 0
+\]
+devient
+\[
+\Phi \left(\frac{a}{c}, \frac{b}{c}, -\frac{q}{c}, \frac{p}{c}\right) = 0.
+\]
+%% -----File: 230.png---Folio 222-------
+Cette équation peut être rendue homogène, et prend la forme
+\[
+\Psi(a, b, c, p, q) = 0
+\]
+on peut y introduire~$r$ en vertu de l'équation~\Eq{(3)}, et on obtient
+finalement, pour définir le complexe, une équation homogène
+entre les coordonnées pluckériennes:
+\[
+\chi(a, b, c, p, q, r) = 0.
+\]
+Réciproquement, toute équation de la forme précédente peut
+être ramenée à la forme
+\[
+\Psi \left(a, b, c, p, -q, -\frac{pa + qb}{c}\right) = 0,
+\]
+et par suite à la forme primitive de l'équation du complexe.
+
+Cherchons le \emph{cône du complexe} de sommet $(x, y, z)$. Nous
+avons, $X, Y, Z$~étant les coordonnées courantes,
+\begin{alignat*}{3}%[** TN: Set on one line in original]
+a &= X - x, &
+b &= Y - y, &
+c &= Z - z, \\
+\Err{}{\intertext{ou encore}}
+p &= \Err{cY - bZ}{yZ - zY},\qquad &
+q &= \Err{aZ - cX}{zX - xZ},\qquad &
+r &= \Err{bX - aY}{xY - yX};
+\end{alignat*}
+l'équation du cône du complexe s'obtiendra en remplaçant $a, b, c$\Add{,}
+$p, q, r$ par les valeurs précédentes dans l'équation du complexe.
+C'est donc:
+\[
+\chi (X - x, Y - y, Z - z,
+\Err{cY - bZ, aZ - cX, bX - aY}{yZ - zY, zX - xZ, xY - yX}) = 0.
+\]
+
+Si on veut une \emph{courbe du complexe}, on prendra
+\begin{alignat*}{3}%[** TN: Set on one line in original]
+a &= dx, &
+b &= dy, &
+c &= dz, \\
+p &= y\, dz - z\, dy,\qquad &
+q &= z\, dx - x\, dz,\qquad &
+r &= x\, dy - y\, dx,
+\end{alignat*}
+et on a l'équation différentielle des courbes du complexe
+\[
+\chi(dx, dy, dz, y\, dz - z\, dy, z\, dx - x\, dz, x\, dy - y\, dx) = 0.
+\]
+
+La condition pour qu'un complexe soit spécial est
+\[
+\frac{\dd \Phi}{\dd a} · \frac{\dd \Phi}{\dd g}
+  - \frac{\dd \Phi}{\dd b} · \frac{\dd \Phi}{\dd f} = 0;
+\]
+elle devient ici
+\[
+\frac{\dd \chi}{\dd a} · \frac{\dd \chi}{\dd p}
+  + \frac{\dd \chi}{\dd b} · \frac{\dd \chi}{\dd q}
+  + \frac{\dd \chi}{\dd c} · \frac{\dd \chi}{\dd r} = 0;
+\]
+%% -----File: 231.png---Folio 223-------
+dans le cas d'un complexe algébrique quelconque, cette équation,
+jointe à celle du complexe définit \emph{la congruence des
+droites singulières}.
+
+Reprenons l'homographie entre droites et plans d'une
+droite du complexe; les coefficients de cette homographie
+sont $\dfrac{\dd \Phi}{\dd a}$, $\dfrac{\dd \Phi}{\dd b}$, $\dfrac{\dd \Phi}{\dd f}$, $\dfrac{\dd \Phi}{\dd g}$, et par suite en coordonnées homogènes, ce
+sont $\dfrac{\dd \chi}{\dd a}, \dots, \dfrac{\dd \chi}{\dd r}$. Considérons la droite $(a_{0}, b_{0}, c_{0}, p_{0}, q_{0}, r_{0})$.
+L'équation
+\[
+\sum a\, \frac{\dd \chi}{\dd a_{0}} + \sum p\, \frac{\dd \chi}{\dd p_{0}} = 0
+\]
+définit un complexe linéaire contenant la droite considérée,
+et sur cette droite, l'homographie pour ce complexe linéaire
+est précisément la même que pour le complexe primitif. Ce
+complexe linéaire est dit \emph{tangent} au complexe donné.
+
+\Paragraph{Remarques.} Si nous définissons la droite par \Card{2} points
+$(x, y, z)$ et $(x', y', z')$ nous avons
+\begin{alignat*}{3}%[** TN: Set on one line in original]
+a &= x' - x, &
+b &= y' - y, &
+c &= z' - z, \\
+p &= yz' - z\Err{y}{y'},\qquad &
+q &= zx' - x\Err{z}{z'},\qquad &
+r &= xy' - yx';
+\end{alignat*}
+d'où l'équation du cône du complexe
+\[
+\chi (x' - x, y' - y, \DPtypo{z}{z'} - z, yz' - zy', zx' - xz', xy' - yx') = 0;
+\]
+
+Corrélativement, définissons la droite par \Card{2} plans
+$(u, v, w, s)$\Add{,} $(u', v', w', s')$. On trouve facilement
+\begin{alignat*}{3}%[** TN: Set on one line in original]
+a &= vw' - wv', &
+b &= wu' - uw', &
+c &= uv' - vu', \\
+p &= su' - us',\qquad &
+q &= sv' - vs',\qquad &
+r &= sw' - ws';
+\end{alignat*}
+on obtient alors l'équation tangentielle d'une courbe plane
+du complexe
+\[
+\chi (vw' - wv', \dots, su' - us', \dots) = 0,
+\]
+%% -----File: 232.png---Folio 224-------
+et on voit bien ainsi que la classe de cette courbe est égale
+à l'ordre du cône du complexe.
+
+\Paragraph{Coordonnées générales de Grassmann et Klein.} Plus généralement
+prenons un tétraèdre de référence quelconque, et
+soient $x_{1}, x_{2}, x_{3}, x_{4}$ les coordonnées d'un point; $u_{1}, u_{2}, u_{3}, u_{4}$ les
+coordonnées d'un plan. Considérons la droite comme définie
+par \Card{2} points~$(x)\Add{,} (y)$. Nous prendrons comme coordonnées de
+cette droite les quantités
+\[
+p_{ik} = \begin{vmatrix}
+\Err{x}{x_{i}} & x_{k} \\
+\Err{y}{y_{i}} & y_{k}
+\end{vmatrix}
+\qquad
+(i, k = 1, 2, 3, 4);
+\]
+remarquons que l'on a $p_{ii} = 0$ et $p_{ki} = -p_{ik}$, de sorte que
+l'on n'obtient ainsi que \Card{6} coordonnées $p_{12}, p_{13}, p_{14}$, $p_{23}, p_{24}, p_{34}$\Add{.}
+Ce sont les moments par rapport au segment des \Card{2} points~$(x)\Add{,} (y)$
+des segments égaux à~$1$ pris sur les \Card{6} arêtes du tétraèdre,
+ou du moins des quantités proportionnelles à ces moments.
+
+Si on a \Card{2} droites $(p_{ik})$~et~$(p'_{ik})$, leur moment relatif~$M$
+est donné par la formule
+\[
+\rho M = \sum p_{ik} p'_{hl}.
+\]
+Si ce moment est nul, les \Card{2} droites se rencontrent. Or\Add{,} considérons
+le déterminant
+\[
+\Theta = \begin{vmatrix}
+x_{1} & x_{2} & x_{3} & x_{4} \\
+y_{1} & y_{2} & y_{3} & y_{4} \\
+x_{1} & x_{2} & x_{3} & x_{4} \\
+y_{1} & y_{2} & y_{3} & y_{4}
+\end{vmatrix} = 0.
+\]
+Développons d'après la règle de Laplace, nous avons
+\[
+\Theta = 2 (p_{12} p_{34} + p_{13} p_{42} + p_{14} p_{\DPtypo{24}{23}})
+  = 2 \Phi (p_{ik}) = 0;
+\]
+de sorte que la condition de rencontre des \Card{2} droites est
+%% -----File: 233.png---Folio 225-------
+\[
+\sum p'_{ik}\, \frac{\dd \Phi}{\dd p_{ik}} = 0.
+\]
+
+Si nous définissons la droite par \Card{2} plans $(u)\Add{,} (v)$, nous
+prendrons pour coordonnées
+\[
+q_{ik} = \begin{vmatrix}
+u_{i} & u_{k} \\
+v_{i} & v_{k}
+\end{vmatrix}\Add{.}
+\]
+Cherchons les relations entre les~$p, q$. La droite étant l'intersection
+des plans $(u)\Add{,} (v)$, un point de cette droite sera
+l'intersection des plans $(u)\Add{,} (v)\Add{,} (w)$. On aura donc
+\[
+\left\{
+\begin{alignedat}{5}
+&u_{1} x_{1} &&+ u_{2} x_{2} &&+ u_{3} x_{3} &&+ u_{4} x_{4} &&= 0, \\
+&v_{1} x_{1} &&+ v_{2} x_{2} &&+ v_{3} x_{3} &&+ v_{4} x_{4} &&= 0, \\
+&w_{1} x_{1} &&+ w_{2} x_{2} &&+ w_{3} x_{3} &&+ w_{4} x_{4} &&= 0.
+\end{alignedat}
+\right.
+\]
+Considérons le déterminant
+\[
+\Omega = \begin{vmatrix}
+u_{1} & u_{2} & u_{3} & u_{4} \\
+v_{1} & v_{2} & v_{3} & v_{4} \\
+w_{1} & w_{2} & w_{3} & w_{4} \\
+s_{1} & s_{2} & s_{3} & s_{4}
+\end{vmatrix}\Add{;}
+\]
+la coordonnée~$x_{i}$\DPtypo{;}{} est égale au coefficient~$S_{i}$ de~$s_{i}$. Pour avoir
+un \Ord{2}{e} point de la droite, nous le définirons par les \Card{3} plans
+$(u)\Add{,} (v)\Add{,} (s)$, et alors $y_{i} = W_{i}$. Considérons l'adjoint de~$\Omega$
+\[
+\begin{vmatrix}
+U_{1} & U_{2} & U_{3} & U_{4} \\
+V_{1} & V_{2} & V_{3} & V_{4} \\
+W_{1} & W_{2} & W_{3} & W_{4} \\
+S_{1} & S_{2} & S_{3} & S_{4}
+\end{vmatrix},
+\]
+nous avons, en associant à chaque mineur du \Ord{2}{e} ordre de~$\Omega$ le
+mineur complémentaire de l'adjoint
+\[
+p_{ik} = \Omega\, \frac{\dd \Phi}{\dd q_{ik}};
+\]
+on peut prendre~$\Omega$ comme arbitraire, et écrire
+\[
+p_{ik} = \frac{\dd \Phi}{\dd q_{ik}}.
+\]
+%% -----File: 234.png---Folio 226-------
+et de même
+\[
+q_{ik} = \frac{\dd \Phi}{\dd p_{ik}}.
+\]
+
+L'équation du complexe sera alors $F (p_{ik}) = 0$ ou $F (q_{hl})
+= 0$, d'où les équations du cône ou de la courbe du complexe.
+La condition pour que le complexe soit spécial est
+\[
+\frac{\dd F}{\dd p_{12}} · \frac{\dd F}{\dd p_{34}} +
+\frac{\dd F}{\dd p_{13}} · \frac{\dd F}{\dd p_{24}} +
+\frac{\dd F}{\dd p_{14}} · \frac{\dd F}{\dd p_{23}} = 0\Add{.}
+\]
+
+\Section{Complexe linéaire.}
+{3.}{} \DPchg{Etudions}{Étudions} plus spécialement le complexe linéaire. Son
+équation s'écrit
+\[
+\sum A_{hl} p_{ik} = 0;
+\]
+le complexe est spécial s'il satisfait à la relation
+\[
+A_{12} A_{34} + A_{13} A_{42} + A_{14} A_{23} = 0,
+\]
+et cette équation exprime que les~$A$ sont les coordonnées d'une
+droite; l'équation du complexe exprime que toute droite du
+complexe rencontre cette droite. \emph{Un complexe linéaire spécial
+est constitué par les droites rencontrant une droite fixe},
+qu'on appelle \emph{directrice du complexe}.
+
+Si on a une droite du complexe, un point~$A$ de cette droite et
+son plan polaire~$P$, le cône du complexe se réduisant ici au
+plan~$P$, l'homographie du complexe est celle des plans de la
+droite~$D$ associés à leurs pôles.
+
+\Section{Faisceau de complexes.}
+{4.}{} Soient \Card{2} complexes linéaires
+\[
+\sum A_{hl} p_{ik} = 0, \qquad
+\sum B_{hl} p_{ik} = 0;
+\]
+l'équation
+\[
+\sum (A_{hl} + \lambda B_{hl}) p_{ik} = 0
+\]
+représentera un \emph{faisceau de complexes}. Cherchons dans ce
+faisceau les complexes spéciaux. Ils sont définis par
+%% -----File: 235.png---Folio 227-------
+l'équation
+\begin{align*}%[** TN: Rebroken]
+     &(A_{14} + \lambda B_{14})(A_{23} + \lambda B_{23}) \\
+{}+{}&(A_{12} + \lambda B_{12})(A_{34} + \lambda B_{34}) \\
+{}+{}&(A_{13} + \lambda B_{13})(A_{24} + \lambda B_{24}) = 0,
+\end{align*}
+équation du \Ord{2}{e} degré. \emph{Dans un faisceau de complexes linéaires
+il y a donc \Card{2} complexes \DPtypo{speciaux}{spéciaux}.} Cherchons a quelles conditions
+il y a racine double. Supposons que $\lambda = 0$ soit racine,
+on a
+\[
+\sum A_{12} A_{34} = 0,
+\]
+et l'équation précédente se réduit à
+\[
+A_{12} B_{34} + A_{34} B_{12} + \dots + \lambda (B_{12} B_{34} + \dots ) = 0.
+\]
+Nous appellerons \emph{invariant du complexe} la quantité
+\[
+\Delta_{A} = A_{12} A_{34} + A_{13} A_{\DPtypo{42}{24}} + A_{14} A_{23},
+\]
+et \emph{invariant relatif} la quantité
+\[
+\Delta_{AB} = \sum  B_{ik}\, \frac{\dd \Delta_{A}}{\dd A_{ik}};
+\]
+l'équation devient alors
+\[
+\Delta_{AB} + \lambda \Delta_{B} = 0;
+\]
+pour que $\lambda = 0$ soit racine double, il faut que $\Delta_{AB} = 0$. Or\Add{,} les
+$A_{ik}$ sont des coordonnées de droite, la condition $\Delta_{AB} = 0$ exprime
+que cette droite appartient au \Ord{2}{e} complexe qui définit le
+faisceau. Elle appartient évidemment au \Ord{1}{er}. Donc \emph{pour que
+l'un des complexes spéciaux soit double, il faut et il suffit
+que sa directrice appartienne à tous les complexes du faisceau}\Add{.}
+Pour que l'équation se réduise à une identité, c'est-à-dire
+pour que tous les complexes du faisceau soient spéciaux, il
+faut encore que $\Delta_{B} = 0$; il faut donc que les \Card{2} complexes
+soient spéciaux, et que leurs directrices se rencontrent.
+
+Nous appellerons \emph{congruence linéaire} l'ensemble des
+droites communes à \Card{2} complexes linéaires. Par tout point de
+%% -----File: 236.png---Folio 228-------
+l'espace passe une droite de cette congruence, et dans tout
+plan il y a une droite. Considérons le faisceau déterminé par
+les \Card{2} complexes qui définissent la congruence. Si ce faisceau
+a \Card{2} complexes spéciaux distincts, toutes les droites de la
+congruence appartiennent à ces complexes spéciaux, et par
+suite rencontrent \Card{2} directrices fixes. \emph{Une congruence linéaire
+est formée en général des droites rencontrant \Card{2} directrices
+fixes.} Si les complexes spéciaux sont confondus, soit $\Delta$ leur
+directrice commune; considérons un complexe quelconque~$(c)$ du
+faisceau. $\Delta$~est une droite du complexe~$(c)$; à chaque point~$A$
+de~$\Delta$ correspond son plan polaire par rapport au complexe~$(c)$;
+les droites de la congruence passant par~$A$ et appartenant au
+complexe~$(c)$ sont dans ce plan polaire. Or\Add{,} les points de~$\Delta$
+ont même plan polaire par rapport à tous les complexes du
+faisceau. Les droites de la congruence rencontrent la droite~$\Delta$,
+et pour chaque point de cette droite sont situées dans le
+plan polaire correspondant.
+
+\Section{Complexes en involution.}
+{5.}{} Reprenons le faisceau de complexes précédent. Les \Card{2} complexes
+de base sont dits \emph{en involution} si on a $\Delta_{AB} = 0$.
+Considérons une droite~$D$ commune aux \Card{2} complexes. A un point~$A$
+de cette droite correspond son plan polaire dans chacun des
+complexes, soient $P, Q$ ces plans; il en résulte une correspondance
+homographique entre les plans $P, Q$ de la droite.
+De même, en partant d'un plan de la droite, on verrait qu'il
+existe une homographie entre les points de la droite. Cherchons
+les plans doubles de cette homographie. Considérons une des
+directrices~$\Delta$ de la congruence linéaire définie par les \Card{2} complexes,
+%% -----File: 237.png---Folio 229-------
+et le plan~$D \Delta$; le pôle de ce plan est l'intersection~$A'$
+de~$D$ avec la \Ord{2}{e} directrice~$\Delta'$,
+car toutes les droites passant par~$A'$
+et rencontrant $\Delta$ appartiennent à la
+congruence, et par suite aux \Card{2} complexes.
+Ainsi $A'$~est foyer du plan~$D \Delta$;
+il est aussi évidemment foyer
+du plan~$D \Delta'$; et l'on voit facilement
+que ces \Card{2} plans sont les plans doubles cherchés. Maintenant
+pour que l'homographie entre les plans~$P, Q$ soit une involution,
+il faut que les plans $P\Add{,}Q$ soient conjugués par rapport à
+ces plans doubles. L'équation du plan polaire d'un point par
+rapport à un complexe quelconque du faisceau est
+\[
+\sum (A_{hl} + \lambda B_{hl})
+\begin{vmatrix}
+X_{i}   &   X_{k}  \\
+x_{i}   &   x_{k}
+\end{vmatrix} = 0,
+\]
+
+\Illustration[1.25in]{237a}
+\noindent équation de la forme
+\[
+P + \lambda Q = 0.
+\]
+Considérons alors \Card{4} complexes quelconques du faisceau, le
+rapport anharmonique des \Card{4} plans polaires d'un même point
+dans ces \Card{4} complexes est égal au rapport anharmonique des \Card{4} quantités~$\lambda$
+correspondantes. Or\Add{,} prenons en particulier les \Card{2} complexes
+de base et les complexes spéciaux. Les valeurs de
+correspondantes sont~$0, \infty$, et les racines de l'équation
+\[
+\sum (A_{14} + \lambda B_{14}) (A_{23} + \lambda B_{23}) = 0;
+\]
+et la condition pour que les \Card{2} \Ord{1}{ères} soient conjuguées harmoniques
+par rapport aux \Card{2} autres est
+\[
+\lambda_{1} + \lambda_{2} = 0
+\]
+ou $\Delta_{AB} = 0$. \emph{Ainsi donc si \Card{2} complexes sont en involution, les
+%% -----File: 238.png---Folio 230-------
+plans polaires d'un point dans ces \Card{2} complexes sont conjugués
+harmoniques par rapport aux plans passant par ce point et
+par les directrices de la congruence commune aux \Card{2} complexes.
+Et réciproquement.}
+
+\Paragraph{Application.} On peut généraliser encore les coordonnées
+de droites. Reprenons la relation fondamentale
+\[
+ap + bq + cr = 0;
+\]
+elle est homogène et du \Ord{2}{e} degré. Or\Add{,} il existe un type remarquable
+d'équations du \Ord{2}{e} degré, celui où ne figurent que les
+carrés. Posons
+\begin{alignat*}{3}
+a + ip &= t_{1},        & b + iq &= t_{3},        & c + ir &= t_{5}, \\
+a - ip &= it_{2},\qquad & b - iq &= it_{4},\qquad & c - ir &= it_{6};
+\end{alignat*}
+la condition précédente devient
+\[
+t_{1}^{2} + t_{3}^{2} + t_{5}^{2} + t_{2}^{2} + t_{4}^{2} + t_{6}^{2} = 0.
+\]
+On introduit comme coordonnées homogènes les $t$, qui sont des
+fonctions linéaires homogènes des coordonnées pluckériennes.
+En égalant ces \Card{6} coordonnées à~$0$, on a les équations de \Card{6} complexes
+qui sont \Card{2} à \Card{2} en involution, car on voit facilement
+que la condition pour que les \Card{2} complexes
+\[
+\sum A_{i} t_{i} = 0, \qquad
+\sum B_{i} t_{i} = 0,
+\]
+soient en involution est
+\[
+\sum A_{i} B_{i} = 0.
+\]
+
+\Section{Droites conjuguées.}
+{6.}{} Considérons un complexe~$(c)$ et une droite~$\Delta$ n'appartenant
+pas à ce complexe\DPtypo{;}{.} Considérons la congruence commune
+à~$(c)$ et au complexe spécial de directrice~$\Delta$\DPtypo{;}{.} Cette
+congruence a une \Ord{2}{e} directrice $\Delta'$ qui est dite la \emph{droite conjuguée} %[**TN: Partial hyphenated word in original]
+%% -----File: 239.png---Folio 231-------
+de~$\Delta$. Il y a évidemment réciprocité entre ces \Card{2} droites.
+\emph{Toutes les droites du complexe~$(c)$ qui rencontrent
+la droite~$\Delta$ rencontrent sa conjuguée~$\Delta'$}, puisque ce sont des
+droites de la congruence, et inversement \emph{toute droite rencontrant
+à la fois les \Card{2} droites conjuguées~$\Delta\Add{,} \Delta'$ appartient à la
+congruence et par suite au complexe}. Si on considère un point~$A$
+de~$\Delta$, son plan polaire passe par~$\Delta'$, puisque toutes les
+droites passant par~$A$ et rencontrant~$\Delta'$ appartiennent au complexe.
+\emph{$\Delta'$~est donc l'enveloppe des plans polaires des points
+de sa conjuguée~$\Delta$}. On voit de même \emph{que $\Delta'$~est le lieu des pôles
+des plans passant par sa conjuguée~$\Delta$}. Si la droite~$\Delta$ appartient
+au complexe~$(c)$, la congruence précédente a ses \Card{2} directrices
+confondues. \emph{Les droites du complexe sont à elles-mêmes
+leurs conjuguées.}
+
+Supposons l'équation du complexe
+\[
+F (a, b, c, p, q, r) = Pa + Qb + Rc + Ap + Bq + Cr = 0\DPtypo{;}{.}
+\]
+Cherchons les coordonnées $(a', b', c', p', q', r')$ de la conjuguée
+d'une droite $(a, b, c, p, q, r)$. Il suffit d'exprimer que le complexe
+donné, et les complexes spéciaux ayant pour directrices
+les droites $(a, b, c, p, q, r)$\Add{,} $(a', b', c', p', q', r')$ appartiennent à un
+même faisceau, ce qui donne
+\[
+P + \lambda p + \lambda' p' = 0, \quad\text{et les analogues}\dots.
+\]
+Multiplions respectivement par $a, b, c, p, q, r$ et ajoutons membre
+à membre, le coefficient de~$\lambda$ \DPchg{disparait}{disparaît} et nous avons
+\[
+F(a, b, c, p, q, r) + \lambda' \sum (ap' + pa') = 0;
+\]
+posons pour abréger
+\[
+\sum (ap' + pa') = \sigma,
+\]
+%% -----File: 240.png---Folio 232-------
+nous avons
+\[
+\Tag{(1)}
+F(a, b, c, p, q, r) + \lambda' \sigma = 0.
+\]
+Si nous multiplions par $a', b', c', p', q', r'$ et si nous ajoutons, c'est
+le coefficient de~$\lambda'$ qui \DPchg{disparaitra}{disparaîtra} et nous aurons
+\[
+\Tag{(2)}
+F(a', b', c', p', q', r') + \lambda \sigma = 0.
+\]
+Enfin si nous multiplions par $A, B, C, P, Q, R$, nous obtenons, en
+posant
+\begin{gather*}
+\Delta = AP + BQ + \Err{c}{C}R, \\
+2\Delta + \lambda F (a, b, c, p, q, r)
+  + \lambda' F(a', b', c', p', q', r') = 0\Add{,}
+\end{gather*}
+ce qui peut s'écrire, en tenant compte de~\Eq{(1)}\Add{,}~\Eq{(2)}
+\[
+\Delta = \lambda \lambda' \sigma,
+\]
+d'où
+\[
+\lambda = \frac{\Delta}{\lambda' \sigma}
+      = - \frac{\Delta}{F(a, b,c, p, q, r)};
+\]
+et nous pouvons prendre pour coordonnées de la droite conjuguée
+\[
+a = A - \frac{\Delta}{F(a\Add{,} \dots)}\, a, \quad\text{et les analogues}, \dots
+\]
+ou
+\[
+a' = AF(a, b, c, p, q, r) - \Delta a, \quad\text{et les analogues}, \dots.
+\]
+
+Supposons qu'on prenne \Card{2} droites conjuguées pour arêtes
+opposées du tétraèdre de référence. Si nous appelons $x, y, z, t$
+les coordonnées tétraédriques, nous aurons
+\begin{alignat*}{3}
+a &= xt' - tx', &
+b &= yt' - ty', &
+c &= zt'- tz', \\
+p &= yz' - zy', \qquad &
+q &= zx' - xz', \qquad &
+r &= xy' - yx'.
+\end{alignat*}
+Supposons qu'on prenne pour droites conjuguées les droites
+$(x = 0, y = 0)$ et $(z = 0, t = 0)$. Leurs coordonnées sont
+\begin{align*}
+&a = 0,     && b = 0,    && c,         && p = 0, && q = 0, && r = 0; \\
+&a' = 0, && b' = 0, && c' = 0, && p' = 0, && q' = 0, && r'.
+\end{align*}
+Exprimons que ces droites sont conjuguées. D'après les conditions
+%% -----File: 241.png---Folio 233-------
+trouvées précédemment, nous avons
+\begin{alignat*}{3}
+0 &= AF (a\Add{,} \dots),\qquad &
+0 &= BF (a\Add{,} \dots),\qquad &
+0 &= CF - \Delta c, \\
+%
+0 &= PF, &
+0 &= QF, &
+r' &= RF.
+\end{alignat*}
+Or,
+\[
+F(a, b, c, p, q, r) = F(0, 0, c, 0, 0, 0) = Rc;
+\]
+il en résulte que $A = 0, B = 0, P = 0, Q = 0$. Alors
+\[
+\Delta = RC,
+\]
+et l'équation du complexe devient
+\[
+Cr + Rc = 0,
+\]
+ou
+\[
+r = kc.
+\]
+
+En particulier cherchons à effectuer cette réduction en
+axes cartésiens. Nous supposerons que $\Delta$ soit l'axe~$Oz$ et que
+$\Delta'$ soit rejetée à l'infini dans le plan des~$x\Add{,}y$. Il faut
+d'abord montrer qu'il y a des droites dont la conjuguée peut
+être rejetée à l'infini. Pour qu'une droite $(a, b, c, p, q, r)$
+soit à l'infini, il faut que $a = 0$, $b = 0$, $c = 0$; et d'après
+les formules précédemment trouvées, nous avons pour les conjuguées
+de ces droites
+\[
+\frac{a'}{A} = \frac{b'}{B} = \frac{c'}{C}
+  = \frac{F(0, 0, 0, p, q, r)}{\Delta};
+\]
+$a', b', c'$ sont donc proportionnels à des quantités fixes. \emph{Les
+conjuguées des droites de l'infini sont parallèles à une
+même direction. Ces droites sont les lieux des pôles des plans
+parallèles à un plan fixe.} On les appelle \emph{diamètres}. En rapportant
+donc un complexe à un diamètre et au plan conjugué,
+on peut mettre l'équation du complexe sous la forme
+\[
+r = kc.
+\]
+
+On peut obtenir cette réduction en axes rectangulaires.
+Il existe en effet une infinité de droites perpendiculaires
+à leurs conjuguées. Elles sont définies par la relation
+%% -----File: 242.png---Folio 234-------
+\[
+aa' + bb' + cc' = 0,
+\]
+ou
+\[
+(Aa + Bb + Cc) F(a, b, c, p, q, r) - \Delta (a^{2} + b^{2} + c^{2}) = 0.
+\]
+Ces droites constituent donc un complexe du \Ord{2}{e} degré. Prenons
+un diamètre quelconque $(a, b, c, p, q, r)$. Le plan conjugué, passant
+par l'origine a pour équation
+\[
+p'X + q'Y + r'Z = 0;
+\]
+la condition pour qu'il soit perpendiculaire au diamètre est
+\[
+\frac{a}{p'} = \frac{b}{q'} = \frac{c}{r'},
+\]
+ou
+\[
+\frac{a}{PF - \Delta p} = \frac{b}{QF - \Delta q} = \frac{c}{RF - \Delta r};
+\]
+la droite conjuguée du diamètre étant à l'infini, on peut remplacer
+$a, b, c$ par $A, B, C$, ce qui donne
+\[
+\frac{A}{PF - \Delta p} = \frac{B}{QF - \Delta q} = \frac{C}{RF - \Delta r}.
+\]
+On a
+\[
+ap + bq + cr = 0,
+\]
+donc ici
+\[
+Ap + Bq + Cr = 0;
+\]
+et
+\[
+F(a, b, c, p, q, r) = Pa + Qb + Rc\Add{.}
+\]
+Multiplions alors les \Card{2} termes des rapports précédents respectivement
+par $A, B, C$ et ajoutons, nous obtenons le rapport égal~$\dfrac{\sum A^{2}}{\Delta F}$;
+nous pouvons alors prendre  $a = A$, $b = B$, $c = C$ et $F =\Delta$,
+et enfin
+\[
+\frac{A}{P \Delta - p \Delta} = \frac{\sum A^{2}}{\Delta^{2}},
+  \quad\text{et les analogues\Add{,}}
+\]
+d'où
+\[
+p = P - \frac{A \Delta}{\sum A^{2}}, \qquad
+q = Q - \frac{B \Delta}{\sum B^{2}}, \qquad
+r = R - \frac{C \Delta}{\sum C^{2}}.
+\]
+Nous obtenons ainsi un diamètre perpendiculaire au plan conjugué,
+c'est \emph{l'axe du complexe} et on a l'équation réduite en
+coordonnées rectangulaires
+\[
+r - mc = 0.
+\]
+Le complexe ne dépend que d'un seul paramètre~$m$ par rapport
+%% -----File: 243.png---Folio 235-------
+au groupe des mouvements.
+
+Si $r = 0$, $c = 0$, l'équation est satisfaite; or\Add{,} $r = 0$,
+$c = 0$ sont les coordonnées des droites rencontrant $Oz$ et perpendiculaire
+à~$Oz$. \emph{Le complexe contient toutes les droites
+rencontrant l'axe et perpendiculaires à l'axe}; $c, r$~sont des
+coordonnées qui ne changent pas si on fait tourner la droite
+autour de~$Oz$; de même si on la déplace parallèlement à~$Oz$.
+Autrement dit \emph{un mouvement \DPchg{hélicoidal}{hélicoïdal} d'axe~$Oz$ laisse le complexe
+inaltéré. Il en résulte que si on a $\infty^{1}$~droites appartenant
+au complexe et ne dérivant pas les unes des autres par
+un mouvement \DPchg{hélicoidal}{hélicoïdal}, on obtiendra toutes les droites du
+complexe en faisant subir à ce système de droites les translations
+et rotations précédentes}. Considérons les droites dont
+les coordonnées~$a, p$ sont nulles, et cherchons parmi ces
+droites celles qui appartiennent au complexe; nous trouvons
+les droites
+\[
+bx = mc, \qquad
+cy - bz = 0,
+\]
+qui constituent une famille de génératrices du \DPchg{paraboloide}{paraboloïde}
+\[
+xy - mz = 0.
+\]
+Par conséquent, \emph{pour obtenir toutes les droites d'un complexe,
+il suffit de prendre un système de génératrices d'un \DPchg{paraboloide}{paraboloïde}
+et de faire subir à chacune d'elles un des mouvements
+précédents}.
+
+\Section{Réseau de complexes.}
+{7.}{}
+$\Phi = 0$, $\Phi' = 0$, $\Phi'' = 0$ étant les équations de \Card{3} complexes
+linéaires, un \emph{réseau de complexes} sera défini par l'équation
+%% -----File: 244.png---Folio 236-------
+\[
+\lambda \Phi + \lambda' \Phi' + \lambda'' \Phi'' \DPtypo{+}{} = 0.
+\]
+Considérons les droites communes à tous les complexes du réseau,
+c'est-à-dire communes aux \Card{3} complexes $\Phi = 0$, $\Phi' = 0$,
+$\Phi'' = 0$; il y en a $\infty^{1}$; elles appartiennent aux complexes spéciaux
+du réseau, on peut les définir au moyen de \Card{3} de ces complexes
+spéciaux. Or\Add{,} un complexe spécial est formé de toutes
+les droites rencontrant sa directrice; les droites précédentes
+rencontrent donc \Card{3} droites fixes, elles constituent un
+système de génératrices d'une quadrique, le \Ord{2}{e} système de génératrices
+comprenant les directrices des complexes spéciaux
+du réseau.
+
+\Paragraph{Application. On peut définir un complexe par \Card{5} droites
+n'appartenant pas à une même congruence linéaire.} Soient en
+effet les droites $1, 2, 3, 4, 5$;  donnons-nous un point~$P$ et cherchons-en
+le plan polaire; considérons les droites $1, 2, 3, 4$;
+il existe deux droites $\Delta\Add{,} \Delta'$ qui rencontrent ces \Card{4} droites, ces
+droites sont conjuguées par rapport au complexe, et alors la
+droite passant par~$P$ et s'appuyant sur~$\Delta\Add{,} \Delta'$ appartient au complexe.
+De même en considérant les droites $2, 3, 4, 5$, nous aurons
+une \Ord{2}{e} droite passant par~$P$ et appartenant au complexe;
+le plan polaire de~$P$ est alors déterminé par ces \Card{2} droites.
+
+\Section{Courbes du complexe.}
+{8.}{} Proposons-nous de déterminer les courbes du complexe
+\[
+r = kc.
+\]
+Considérons une droite passant par un point~$(x, y, z)$ et de
+coefficients directeurs~$a, b, c$; pour qu'elle appartienne au
+complexe, il faut que l'on ait
+%% -----File: 245.png---Folio 237-------
+\[
+bx - ay = kc\Add{,}
+\]
+et l'équation différentielle des courbes du complexe est alors
+\[
+\Tag{(1)}
+x\, dy - y\, dx = k · dz.
+\]
+Cette équation s'écrit
+\[
+x^{2}\, d\left(\frac{y}{x}\right) = d(kz)\Add{,}
+\]
+posons
+\[
+\Tag{(2)}
+kz = Y, \qquad \frac{y}{x} = X, \qquad x^{2} = P;
+\]
+l'équation précédente s'écrit
+\[
+dY - P\, dX = 0\Add{,}
+\]
+elle montre que $P$~est la dérivée de~$Y$ par rapport à~$X$. On a
+donc la solution générale de~\Eq{(1)}
+\[
+\Tag{(3)}
+X = \phi(t), \qquad
+Y = \Psi(t), \qquad
+P = \frac{d\Psi}{d\phi};
+\]
+d'où $x\Add{,}y\Add{,}z$ exprimées en fonction d'une \DPtypo{varable}{variable} arbitraire~$t$ au
+moyen de \Card{2} fonctions arbitraires. Si on prend pour variable
+indépendante~$X$, on aura
+\[
+Y = f(X)\Add{,} \qquad
+P = f'(X)\Add{,}
+\]
+d'où les équations de la courbe
+\[
+\Tag{(4)}
+kz = f \left(\frac{y}{x}\right), \qquad
+x^{2} = f' \left(\frac{y}{x}\right)\DPtypo{;}{.}
+\]
+On pourra poser
+\[
+\frac{y}{x} = u\Add{,}
+\]
+d'où les expressions de $x\Add{,}y\Add{,}z$ en fonction de~$u$
+\[
+\Tag{(5)}
+x =   \sqrt{f'(u)}, \qquad
+y = u \sqrt{f'(u)}, \qquad
+z = \frac{1}{k}\, f(u).
+\]
+
+Il est facile, en particularisant la forme de la fonction~$f$,
+d'obtenir des courbes remarquables du complexe.
+
+\ParItem{\Primo.} \DPtypo{on}{On} obtiendra toutes les courbes algébriques du complexe
+en prenant pour~$f$ une fonction algébrique de~$u$. Posons
+en particulier
+\[
+f(u) = \frac{u^{3}}{3}\Add{,}
+\]
+%% -----File: 246.png---Folio 238-------
+alors
+\[
+f'(u) = u^2\Add{,}
+\]
+et nous avons
+\[
+\Tag{(6)}
+x = u, \qquad
+y = u^2, \qquad
+z = \frac{u^3}{3k};
+\]
+ces équations sont celles d'une cubique gauche osculatrice au
+plan de l'infini dans le direction $x = 0$, $y = 0$. Réciproquement
+on peut par une transformation projective ramener les équations
+de toute cubique gauche à la forme précédente, d'où il
+résulte que \emph{les tangentes à toute cubique gauche appartiennent
+à un complexe linéaire}.
+
+\ParItem{\Secundo.} Les formules générales~\Eq{(5)} contiennent un radical,
+provenant de ce qu'on a posé $x^2 = P$. On fera \DPchg{disparaitre}{disparaître} le
+radical en choisissant le paramètre de façon que $P$~soit carré
+parfait. Pour cela considérons la courbe plane~$(X\Add{,}Y)$ considérée
+comme enveloppe de la droite
+\[
+Y - u^2X + 2 \phi(u) = 0,
+\]
+car on a bien alors
+\[
+\frac{dY}{dX} = u^2;
+\]
+l'enveloppe est définie par l'équation de la droite et par
+\[
+-uX + \phi'(u) = 0\Add{,}
+\]
+d'où l'on tire
+\[
+X = \frac{\phi'(u)}{u}, \qquad
+Y = u\phi'(u) - 2\phi(u);
+\]
+d'où
+\[
+\Tag{(7)}
+x = u, \qquad
+y = \phi' (u), \qquad
+z = \frac{1}{k} \bigl[u\phi'(u) - 2\phi(u)\bigr];
+\]
+et ces formules permettent de trouver toutes les courbes unicursales
+du complexe; il n'y a qu'à prendre pour~$u$ une fonction
+rationnelle d'un paramètre arbitraire, et pour~$\phi$ une
+%% -----File: 247.png---Folio 239-------
+fonction rationnelle de~$u$.
+
+\ParItem{\Tertio.} L'équation différentielle~\Eq{(1)} peut encore s'écrire
+\[
+(x^2 + y^2)\, d \left(\arctg \frac{y}{x}\right) = k\, dz;
+\]
+posons
+\[
+kz = Y, \qquad
+\arctg \frac{y}{x} = X, \qquad
+x^2 + y^2 = P = \frac{dY}{dX}.
+\]
+En prenant $X$ comme variable indépendante, on aura la solution
+générale
+\[
+\arctg \frac{y}{x} = \omega, \qquad
+kz = f(\omega), \qquad
+x^2 + y^2 = f'(\omega);
+\]
+qu'on peut encore écrire
+\[
+\Tag{(8)}
+x = \sqrt{f'(\omega)} · \cos\omega, \qquad
+y = \sqrt{f'(\omega)} \Add{·} \sin\omega, \qquad
+z = \frac{1}{k} f(\omega).
+\]
+On obtient des courbes particulières en prenant
+\[
+f(\omega) = R^2 \omega + C;
+\]
+d'où
+\[
+\Tag{(9)}
+x = R \cos\omega, \qquad
+y = R \sin\omega, \qquad
+z = \frac{R^2}{k} \omega + a;
+\]
+\DPtypo{Ce}{ce} sont des hélices tracées sur des cylindres de révolution
+autour de l'axe du complexe. Le pas de ces hélices~$\dfrac{2 \pi R^2}{k}$ est
+uniquement fonction de~$R$, donc \emph{toutes les hélices du complexe
+tracées sur un même cylindre ayant l'axe du complexe pour axe
+ont même pas}.
+
+\MarginNote{Propriétés
+générales des
+courbes du
+complexe.}
+Il résulte immédiatement de la définition des courbes
+d'un complexe que, \emph{dans un complexe linéaire, le plan polaire
+d'un point d'une courbe du complexe est le plan osculateur à
+la courbe en ce point}. Considérons alors les plan osculateurs
+à une courbe du complexe issus d'un point~$P$. Soit~$A$ l'un des
+points de contact; le plan osculateur en~$A$ étant le plan
+%% -----File: 248.png---Folio 240-------
+polaire de~$A$, la droite~$PA$ appartient au complexe, et par suite
+est dans le plan polaire de~$P$. Il en résulte que \emph{les points
+de contact des plans osculateurs issus d'un point à une courbe
+d'un complexe linéaire sont dans un même plan}. En particulier
+\emph{les points de contact des plans osculateurs issus d'un
+point à une cubique gauche sont dans un même plan passant par
+le point donné}.
+
+Prenons les formules~\Eq{(7)}. Nous trouvons
+\begin{align*}
+A &= y' z''- z' y'' = \frac{1}{k} \phi' \phi'' = \frac{y}{k} \phi''', \\
+B &= z' x'' - x' z'' = - \frac{u}{k} \phi''' = - \frac{x}{k} \phi''', \\
+C &= x' y'' - y' x'' = \phi''' = \phi''';
+\end{align*}
+et
+\[
+\begin{vmatrix}
+x'    &  y'    &  z'   \\
+x''   &  y''   &  z''  \\
+x'''  &  y'''  &  z'''
+\end{vmatrix} = \frac{1}{k}\, \phi'''{}^2.
+\]
+On voit alors que la torsion au point $(x,y,z)$ est donnée par
+\[
+T = -\frac{x^2 + y^2 + z^2}{k};
+\]
+Elle ne dépend que du point, et pas de la courbe. Donc \emph{toutes
+les courbes du complexe linéaire passant par un point ont
+même torsion en ce point \(Sophus Lie\)}.
+
+\Section{Surfaces normales du complexe.}
+{9.}{} Il n'y a pas lieu de rechercher les surfaces d'un
+complexe linéaire. Soit en effet le complexe linéaire
+\[
+ay - bx + kc = 0;
+\]
+le plan polaire du point $(x,y,z)$ est parallèle au plan
+\[
+Xy - Yx + kZ = 0,
+\]
+%% -----File: 249.png---Folio 241-------
+et pour qu'une surface soit tangente à ce plan, il faudrait
+que l'on eût
+\[
+\frac{p}{y} = \frac{q}{-x} = \frac{-1}{k},
+\]
+ou
+\[
+p = -\frac{y}{k}, \qquad
+q =  \frac{x}{k};
+\]
+et la condition d'intégrabilité
+\[
+\frac{\dd p}{\dd y} = \frac{\dd q}{\dd x}
+\]
+n'est pas réalisée. Le problème est impossible.
+
+Nous nous proposerons alors de chercher les surfaces
+dont les normales sont des droites du complexe. Nous aurons à
+intégrer l'équation aux dérivées partielles
+\[
+py - qx - k = 0,
+\]
+ce qui revient à l'intégration du système
+\[
+\frac{dx}{y} = \frac{dy}{-x} = \frac{dz}{k} = - dt,
+\]
+qui est précisément le système auquel on arrive lorsqu'on recherche
+les courbes normales aux plans polaires de leurs
+points\Add{.} Nous pouvons écrire
+\[
+dx = -y·dt, \qquad
+dy =  x·dt, \qquad
+dz = -k·dt;
+\]
+système qui s'intègre immédiatement et donne
+\[
+x = R \cos t, \qquad
+y = R \sin t, \qquad
+z = - kt + h.
+\]
+Ces trajectoires orthogonales dépendent de \Card{2} constantes arbitraires.
+Ce sont des hélices circulaires ayant toutes même
+pas, trajectoires d'un mouvement \DPchg{hélicoidal}{hélicoïdal} uniforme de pas~$-2k$.
+D'où l'interprétation cinématique du complexe linéaire:
+considérons un mouvement \DPchg{hélicoidal}{hélicoïdal} uniforme; à chaque point~$M$
+correspond la vitesse de ce point, et le plan polaire du
+point~$M$ dans le complexe est le plan perpendiculaire à cette
+%% -----File: 250.png---Folio 242-------
+vitesse. \emph{Le complexe linéaire est constitué par les normales
+aux vitesses du mouvement instantané d'un corps solide.}
+
+Les surfaces normales du complexe sont définies par les
+équations
+\[
+x = v \cos u, \qquad
+y = v \sin u, \qquad
+z = - ku + \phi(v);
+\]
+car elles sont évidemment engendrées par les hélices précédentes.
+Ce sont les \DPchg{hélicoides}{hélicoïdes} engendrés par un profil quelconque
+dans le mouvement précédent. Les \DPtypo{equations}{équations} précédentes
+\DPtypo{representent}{représentent} d'ailleurs \DPchg{l'hélicoide}{l'hélicoïde} le plus général. Il en
+résulte que \emph{les normales issues d'un point à un \DPchg{hélicoide}{hélicoïde}
+sont dans un même plan} (plan polaire de ce point).
+
+\Paragraph{Remarque.} Les hélices trajectoires orthogonales des
+plans polaires s'obtiennent en faisant $v = \cte$, et leurs
+trajectoires orthogonales sont les courbes du complexe situées
+sur les surfaces précédentes. Cherchons-les. Formons l'élément
+linéaire sur ces surfaces:
+\begin{align*}%[** TN: Rebroken]
+ds^2 = dx^2 + dy^2 + dz^2
+  &= (\cos u · dv - v · \sin u · du)^2 \\
+  &+ (\sin u · dv + v · \cos u · du)^2 + (- m · du + \phi' · dv)^2,
+\end{align*}
+\[
+ds^2 = (v^2 + m^2)\Add{·} du^2 - 2 m \phi' · du\,dv + (1 + \phi'{}^2) · dv^2;
+\]
+et les trajectoires orthogonales des hélices $v = \cte$, $dv = 0$,
+sont définies par l'équation
+\[
+(v^2 + m^2)\Add{·} du - m \phi' · dv = 0\Add{,}
+\]
+d'où
+\[
+u = \int \frac{m \phi'}{v^2 + m^2}\, dv\Add{.}
+\]
+Leur détermination dépend d'une quadrature.
+%% -----File: 251.png---Folio 243-------
+
+\Section{Surfaces réglées du complexe.}
+{10.}{} Considérons une surface réglée dont les génératrices
+appartiennent au complexe; soit~$G$ une de ses génératrices;
+elle appartient au complexe, donc à chacun de
+ses points~$M$ correspond un plan~$P$ qui en est le plan \Err{focal}{polaire};
+d'autre part au point~$M$ correspond aussi homographiquement le
+plan tangent à la surface en ce point; il en résulte qu'\emph{il y
+a correspondance homographique entre le plan polaire d'un
+point de la génératrice et le plan tangent à la surface en ce
+point}; dans cette homographie il y a \Card{2} éléments doubles, donc
+\emph{sur chaque génératrice de la surface il existe \Card{2} points $A\Add{,} B$
+tels que les plans polaires de ces points soient tangents à la
+surface}. Considérons le lieu des points~$A$ sur la surface; en
+chacun de ses points le plan tangent à la surface est le plan
+polaire de~$A$; la tangente à la courbe, qui est dans le plan
+tangent à la surface, est donc dans le plan polaire; donc \emph{le
+lieu des points~$A$, et aussi le lieu des points~$B$, qui peuvent
+d'ailleurs se confondre algébriquement, sont des courbes du
+complexe}. Le plan osculateur en chaque point est le plan polaire,
+donc il est tangent à la surface; \emph{ces courbes sont
+donc des asymptotiques de la surface réglée}; les asymptotiques
+se déterminent au moyen d'une seule quadrature.
+
+Il peut arriver que les génératrices de la surface appartiennent
+à une congruence linéaire; elles appartiennent alors
+à une infinité de complexes linéaires, et pour chaque complexe,
+on aura \Card{2} lignes asymptotiques courbes de ce complexe.
+On obtiendra ainsi toutes les asymptotiques sans aucune intégration.
+%% -----File: 252.png---Folio 244-------
+\emph{Les génératrices de la surface précédente s'appuient
+alors sur \Card{2} directrices fixes}. C'est le cas des \DPtypo{conoides}{conoïdes} à
+plan directeur et des surfaces réglées du \Ord{3}{e} ordre. Inversement
+on verrait facilement qu'une courbe quelconque du complexe
+est asymptotique d'une infinité de surfaces réglées du
+complexe; on peut donc au moyen de ces surfaces réglées trouver
+une courbe quelconque du complexe.
+
+Si les génératrices de la surface appartiennent à un
+complexe linéaire spécial, les courbes du complexe sont des
+courbes planes dont les plans contiennent la directrice du
+complexe; \emph{les surfaces normales du complexe sont de révolution
+autour de la directrice; les surfaces réglées du complexe sont
+des surfaces dont les génératrices rencontrent une droite
+fixe}; cette directrice est une asymptotique de la surface,
+et les autres asymptotiques se déterminent par \Card{2} quadratures.
+
+
+\ExSection{X}
+
+\begin{Exercises}
+\item[45.] \DPchg{Etudier}{Étudier} les asymptotiques des surfaces réglées du \Ord{3}{e} ordre.
+Montrer que ce sont des unicursales du \Ord{4}{e} ordre, et que chaque
+génératrice rencontre une asymptotique en deux points conjugués
+harmoniques par rapport aux points où la génératrice
+s'appuie sur la droite double et sur la droite singulière.
+
+\item[46.] Déterminer les asymptotiques de la surface de Steiner. Par
+quelles courbes sont-elles représentées dans la représentation
+paramétrique de la surface? \DPchg{Etudier}{Étudier} les cas de dégénérescence.
+
+\item[47.] Déterminer la surface canal la plus générale dont toutes les
+lignes de courbure soient sphériques; montrer que ces lignes
+de courbure se déterminent sans intégration.
+
+\item[48.] Que peut-on dire de la détermination des lignes de courbure
+d'une surface canal, enveloppe de $\infty^{1}$~sphères coupant une
+sphère fixe sous un angle constant?
+
+\item[49.] Déterminer les surfaces réglées d'un complexe linéaire qui
+admettent pour ligne asymptotique une courbe donnée. Montrer
+que toutes leurs asymptotiques se \DPtypo{déterminant}{déterminent} sans intégration,
+et qu'elles sont algébriques si la courbe donnée est
+algébrique.
+\end{Exercises}
+%% -----File: 253.png---Folio 245-------
+
+
+\Chapitre{XI}
+{Transformations Dualistiques. Transformation de Sophus Lie.}
+
+\Section{\DPchg{Eléments}{Éléments} et multiplicités de contact.}
+{1.}{} On appelle \emph{élément de contact} l'ensemble d'un point~$M$
+et d'un plan~$P$ passant par ce point. Un tel élément sera
+défini par ses \emph{coordonnées}, coordonnées $x, y, z$ du point, et
+coefficients de direction $p, q, -1$ de la normale au plan. Un
+élément de contact est ainsi défini par \Card{5} coordonnées.
+
+Considérons un point~$A$, les éléments de contact de ce
+point sont formés par ce point et tous les plans passant par
+ce point; les coordonnées $x, y, z$ sont fixes, et $p, q$ arbitraires.
+Un point possède $\infty^{2}$~éléments de contact.
+
+Considérons une courbe; un de ses éléments de contact
+est formé d'un point de la courbe et d'un plan tangent à la
+courbe en ce point; les coordonnées sont $x, y, z$ fonctions d'un
+paramètre arbitraire, et $p, q$ \DPtypo{lies}{liés} par la relation
+\[
+px' + qy'- z' = 0\Add{,}
+\]
+il y a donc \Card{2} paramètres arbitraires. Une courbe possède $\infty^{2}$
+éléments de contact.
+
+Considérons maintenant une surface; un de ses éléments de
+contact est formé par un point et le plan tangent en ce point;
+%% -----File: 254.png---Folio 246-------
+ses coordonnées sont $x, y, z = f (x, y)$, $p=\dfrac{\dd f}{\dd x}$, $q=\dfrac{\dd f}{\dd y}$. Il y a \Card{2} paramètres
+arbitraires, donc une surface possède $\infty^{2}$~éléments
+de contact. Remarquons que $p, q$ peuvent ne dépendre que d'un
+seul paramètre; c'est le cas des surfaces développables, qui
+possèdent ainsi $\infty^{2}$~points et $\infty^{1}$~plans tangents, et correspondent
+par dualité aux courbes, qui possèdent $\infty^{1}$~points et $\infty^{2}$~plans
+tangents.
+
+Les points, courbes et surfaces, qui sont engendrées par
+$\infty^{2}$~éléments de contact, sont appelés \emph{multiplicités~$M_2$}. Plus
+généralement on appellera \emph{multiplicité} toute famille d'éléments
+de contact dont les coordonnées vérifient la relation
+\[
+\Tag{(1)}
+dz - p\, dx - q\, dy = 0.
+\]
+Si ces coordonnées ne dépendent que d'un paramètre arbitraire,
+on aura les multiplicités~$M_1$; si elles dépendent de \Card{2} paramètres
+arbitraires, on aura les \emph{multiplicités~$M_2$}.
+
+Cherchons à déterminer toutes les multiplicités~$M_2$:
+$x, y, z$, $p, q$ sont fonctions de \Card{2} paramètres arbitraires
+\[
+x = f (u, v), \quad
+y = g (u, v), \quad
+z = h (u, v), \quad
+p = k (u, v), \quad
+q = l (u, v).
+\]
+Considérons les \Card{3} \Ord{1}{ères} relations; entre elles on peut éliminer
+$u, v$, et par suite de cette élimination on peut obtenir
+\Card[f]{1},~ou \Card{2}, ou \Card{3} relations.
+
+Supposons d'abord qu'on obtienne une relation
+\[
+F(x, y, z) = 0\Add{,}
+\]
+on peut considérer $z$ comme fonction de $x, y$; et si on écrit
+que la relation~\Eq{(1)} est satisfaite quels que soient, $x, y$, on a
+\[
+p = \frac{\dd z}{\dd x}, \qquad
+q = \frac{\dd z}{\dd y},
+\]
+%% -----File: 255.png---Folio 247-------
+et on a les éléments de contact d'une surface.
+
+Supposons qu'on obtienne \Card{2} relations
+\[
+F(x, y, z) = 0, \qquad
+G(x, y, z) = 0;
+\]
+on peut considérer $x, y$ comme fonctions de~$z$
+\[
+x = \phi(z), \qquad
+%[** TN: Appears to be \Psi in original, but using \psi for consistency]
+y = \psi(z),
+\]
+et l'équation~\Eq{(1)} devient
+\[
+dz - p \phi'(z)\, dz - q \psi'(z)\, dz = 0,
+\]
+ou
+\[
+p \phi'(z) + q \psi'(z) - 1 = 0;
+\]
+le plan de l'élément de contact est tangent à la courbe
+$x = \phi(z)$, $y = \psi (z)$, on a les éléments de contact d'une courbe.
+
+Enfin si on obtient \Card{3} relations, c'est que $x, y, z$ sont des
+constantes; l'équation~\Eq{(1)} est alors vérifiée quels que soient
+$p\Add{,} q$, qui sont alors les paramètres arbitraires, et on a les
+éléments de contact d'un point.
+
+Cherchons maintenant les multiplicités~$M$; nous avons
+\[
+x = f (t), \qquad
+y = g (t), \qquad
+z = h (t), \qquad
+p = k (t), \qquad
+q = l (t).
+\]
+Considérons les \Card{3} \Ord{1}{ères} équations, et entre elles éliminons~$t$.
+Il y a \Card{2} ou \Card{3} relations.
+
+S'il y a \Card{2} relations, le lieu des points de la multiplicité,
+qu'on appelle aussi \emph{support de la multiplicité}, est une
+courbe, et les plans ne dépendant que d'un paramètre, pour
+chaque point de la courbe il y a un plan tangent déterminé;
+on a une \emph{bande d'éléments de contact}.
+
+S'il y a \Card{3} relations, $x, y, z$ sont des constantes, le support
+est un point; on a alors une famille de plans dépendant
+d'un paramètre et passant par un point fixe; c'est ce qu'on
+appelle un \emph{cône élémentaire}.
+%% -----File: 256.png---Folio 248-------
+
+Considérons \Card{2} multiplicités~$M_2$; elles peuvent avoir en
+commun, \Card{0}\DPtypo{,}{}~ou \Card{1} élément de contact, ou une infinité.
+
+Considérons le cas d'\emph{un élément de contact commun}; si les
+multiplicités sont \Card{2} points $A\Add{,} A'$, il ne peut y avoir un élément
+de contact commun que si les \Card{2} points sont confondus, et
+alors il y a $\infty^{2}$~éléments de contact communs. Si on a un point
+et une courbe, le point est sur la courbe, et tous les plans
+tangents à la courbe en ce point appartiennent à des éléments
+de contact communs, qui sont ainsi au nombre de~$\infty^{1}$. Si on a un
+point et une surface, le point sera sur la surface, et l'élément
+de contact commun sera constitué par le point et le plan
+tangent à la surface en ce point. Considérons \Card{2} courbes; si
+elles ont un élément de contact commun, elles se rencontrent
+en un point, et si elles n'y sont pas tangentes, il n'y a
+qu'un élément de contact commun. Considérons une courbe et
+une surface; il y aura un élément de contact commun si la
+courbe est tangente à la surface. Enfin \Card{2} surfaces ont un
+élément de contact commun si elles sont tangentes en un point.
+
+Il y aura \emph{$\infty^{1}$~éléments de contact communs} pour un point
+sur une courbe, \Card{2} courbes tangentes en un point, une courbe
+sur une surface, \Card{2} surfaces circonscrites le long d'une courbe\Add{.}
+
+Considérons un \emph{point qui décrit une courbe}; on a une
+famille de $\infty^{1}$~points dont chacun donne à la courbe $\infty^{1}$~éléments
+de contact. Considérons une \emph{surface engendrée par une courbe};
+nous avons $\infty^{1}$~courbes~$(c)$ dont chacune a en commun avec la
+surface une bande, et par suite donne à la surface $\infty^{1}$~éléments
+de contact. Considérons $\infty^{1}$~surfaces; leur \emph{enveloppe} a avec
+%% -----File: 257.png---Folio 249-------
+chacune d'elles une bande commune; nous avons encore $\infty^{1}$~éléments
+générateurs d'une multiplicité~$M_2$, donnant chacun à la
+multiplicité $\infty^{1}$~éléments de contact.
+
+On pourrait considérer le cas où chaque élément générateur
+ne donne qu'un élément de contact à la multiplicité:
+$\infty^{2}$~points engendrant une surface; $\infty^{2}$~courbes formant une congruence
+de courbes (dans ce cas, comme dans celui des congruences
+de droites, il y a en général une surface focale,
+tangente à chacune de ces courbes, et ayant avec chacune un
+élément de contact commun); enfin si on considère $\infty^{2}$~surfaces,
+leur enveloppe a en commun avec chacune d'elles un élément de
+contact.
+
+\Paragraph{Remarque.} Dans les trois cas précédents, quand nous disons
+que chaque élément générateur donne un élément de contact
+à la multiplicité, il faut entendre que cette multiplicité
+peut se décomposer en nappes, et que cela s'applique alors à
+chacune des nappes séparément.
+
+Il y a un cas exceptionnel, celui de $\infty^{1}$~courbes ayant
+une enveloppe; on a alors $\infty^{1}$~courbes cédant chacune à la multiplicité
+$\infty^{1}$~éléments de contact.
+
+\Section{Transformations de contact.}
+{2.}{} On appelle \emph{transformation de contact} toute transformation
+des éléments de contact qui change une multiplicité~$M_2$
+en une multiplicité~$M_2$. On a \Card{5} équations de
+transformation
+\begin{gather*}
+x' = f (x, y, z, p, q), \quad
+y' = g (x, y, z, p, q), \quad
+z' = h (x, y, z, p, q), \\
+p' = k (x, y, z, p, q), \qquad
+q' = l (x, y, z, p, q).
+\end{gather*}
+%% -----File: 258.png---Folio 250-------
+Si l'élément de contact~$(x, y, z, p, q)$ appartient à un multiplicité,
+on a
+\[
+\Tag{(1)}
+dz - p\, dx - q\, dy = 0,
+\]
+et pour que l'élément transformé $(x', y', z', p', q')$ appartienne
+aussi à une multiplicité, il faut que l'on ait
+\[
+dz' - p'\, dx'- q'\, dy'= 0.
+\]
+Une transformation de contact laisse invariante l'équation~\Eq{(1)}\Add{.}
+Une telle transformation change \Card{2} multiplicités ayant un élément
+de contact commun en \Card{2} multiplicités ayant un élément de
+contact commun, et de même \Card{2} multiplicités ayant $\infty^{1}$~éléments
+de contact communs en \Card{2} multiplicités ayant $\infty^{1}$~éléments de
+contact communs. Une transformation de contact change les
+points, courbes et surfaces en points, courbes, ou surfaces
+indistinctement.
+
+Reprenons les équations de la transformation, et entre
+elles éliminons~$p\Add{,} q$, nous obtenons \Card[f]{1},~ou \Card{2}, ou \Card{3} relations entre
+$x, y, z$,~$x', y', z'$.
+
+Si on obtient \Card{3} relations,
+\[
+\Tag{(2)}
+x' = f(x, y, z), \qquad
+y' = g(x, y, z), \qquad
+z' = h(x, y, z)\Add{,}
+\]
+dans la transformation de contact est contenue une transformation
+ponctuelle. Une telle transformation change un point en
+point, une courbe en courbe, une surface en surface; \Card{2} courbes
+qui se rencontrent se transforment en \Card{2} courbes qui se
+rencontrent, deux surfaces tangentes en \Card{2} surfaces tangentes.
+A un élément de contact commun à \Card{2} multiplicité correspond un
+élément de contact commun aux \Card{2} multiplicités transformées.
+On obtiendra $p', q'$ en fonction de~$p, q$ en considérant $z'$ comme
+%% -----File: 259.png---Folio 251-------
+fonction de $x', y'$. On a
+\begin{alignat*}{3}%[** TN: Added elided equations]
+dx' &= \frac{\dd f}{\dd x}\, dx
+     &&+ \frac{\dd f}{\dd y}\, dy
+     &&+ \frac{\dd f}{\dd z}\, (p\, dx + q\, dy), \\
+dy' &= \frac{\dd g}{\dd x}\, dx
+     &&+ \frac{\dd g}{\dd y}\, dy
+     &&+ \frac{\dd g}{\dd z}\, (p\, dx + q\, dy), \\
+dz' &= \frac{\dd h}{\dd x}\, dx
+     &&+ \frac{\dd h}{\dd y}\, dy
+     &&+ \frac{\dd h}{\dd z}\, (p\, dx + q\, dy);
+\end{alignat*}
+éliminant $dx, dy$ entre ces \Card{3} relations, on a
+\[
+dz' = k(x,y,z,p,q)\, dx' + l(x,y,z,p,q)\, dy',
+\]
+d'où
+\[
+p' = k(x,y,z,p,q), \qquad
+q' = l(x,y,z,p,q).
+\]
+
+Supposons ensuite que l'on obtienne \Card[f]{1}~relation d'élimination
+\[
+\Tag{(3)}
+\Omega (x,y,z, x',y',z') = 0\Add{.}
+\]
+Considérons un point $A(x\Add{,}y\Add{,}z)$ du \Ord{1}{er} espace; cherchons la multiplicité
+qui lui correspond dans le \Ord{2}{e} espace; elle est engendrée
+par des éléments de contact dont les points sont liés
+au point~$A$ par l'équation~\Eq{(3)} qui représente une surface~$S_A'$.
+La multiplicité correspondant à un point est une surface. Si
+on a une courbe lieu de points~$A$, il lui correspond une famille
+de $\infty^{1}$~surfaces, et la multiplicité engendrée par ces
+surfaces, c'est-à-dire leur enveloppe, sera la transformée de
+la courbe. Enfin si on a une surface lieu de $\infty^{2}$~points~$A$, il
+leur correspondra $\infty^{2}$~surfaces dont l'enveloppe correspondra à
+la surface donnée.
+
+\MarginNote{Transformations
+dualistiques.}
+Supposons la relation~\Eq{(3)} bilinéaire en $x,y,z$,
+$x',y',z'$. A chaque point du \Ord{1}{er} espace correspond un plan
+du \Ord{2}{e} espace et réciproquement. A $\infty^{3}$~points du \Ord{1}{er} espace
+correspondent $\infty^{3}$~plans distincts. \DPchg{Ecrivons}{Écrivons}
+\[
+\Omega = Ax' + By' + Cz' + D
+\]
+%% -----File: 260.png---Folio 252-------
+ou
+\[
+A = ux + vy + wz + h, \quad
+B = u'  x + \ldots, \quad
+C = u'' x + \ldots, \quad
+D = u'''x + \ldots;
+\]
+pour avoir la transformée d'une surface
+\[
+f(x',y',z') = 0
+\]
+il faut prendre l'enveloppe des plans $\Omega = 0$, $x'\Add{,} y'\Add{,} z'$~étant liés
+par la relation précédente, ce qui donne
+\[
+\frac{A}{\ \dfrac{\dd f}{\dd x'}\ } =
+\frac{B}{\ \dfrac{\dd f}{\dd y'}\ } =
+\frac{C}{\ \dfrac{\dd f}{\dd z'}\ } =
+\frac{D}{\ \dfrac{\dd f}{\dd t'}\ }\Add{.}
+\]
+Telles sont les équations de la transformation. Il faudra que
+l'on en puisse tirer $x, y, z$: donc que les formes $A, B, C, D$, soient
+indépendantes, et alors l'ensemble des plans $\Omega = 0$ constitue
+bien l'ensemble de tous les plans de l'espace. La transformation
+précédente est une \emph{transformation dualistique}. L'ensemble
+des transformations de contact forme évidemment un groupe;
+une transformation de contact peut se décomposer en transformations
+de contact plus simples.
+
+Prenons pour nouvelles variables
+\[
+X = A, \qquad
+Y = B, \qquad
+Z = C, \qquad
+T = D;
+\]
+alors
+\[
+\Omega = Xx' + Yy' + Zz' + 1 = 0\Add{,}
+\]
+la transformation est une transformation par polaires réciproques
+par rapport à la sphère
+\[
+x^2 + y^2 + z^2 + 1 = 0\Add{,}
+\]
+et toute transformation dualistique se ramène à la transformation
+précédente suivie d'une transformation projective.
+
+Considérons celles de ces transformations qui sont \emph{symétriques}
+ou \emph{involutives}, telles que le plan homologue d'un
+%% -----File: 261.png---Folio 253-------
+point soit le même, qu'on considère le point comme appartenant
+à l'un ou à l'autre espace. Les équations
+\[
+\Omega(x,y,z,x',y',z') = 0, \qquad
+\Omega(x',y',z',x,y,z) = 0,
+\]
+doivent être équivalentes; on doit donc avoir, $k$~étant un
+facteur constant
+\[
+\Omega(x,y,z,x',y',z') = k \Omega(x',y',z',x,y,z);
+\]
+faisons $x' = x$, $y' = y$, $z' = z$,
+\[
+\Omega(x,y,z,x,y,z) = k \Omega(x,y,z,x,y,z);
+\]
+alors ou bien $\Omega(x,y,z,x,y,z) = 0$, ou bien $k = 1$.
+
+Si $\Omega = 0$, le plan correspondant à un point passe par ce
+point. On a
+\begin{multline*}%[** TN: Filled in missing terms, added break]
+x(ux + vy + wz + h) + y(u'x + v'y + w'z + h') \\
+  + z (u''x + v''y + w''z + h'')
+  + u'''x + v'''y + w'''z + h''' = 0,
+\end{multline*}
+ce qui revient à écrire que le déterminant
+\[
+\begin{vmatrix}
+u  &  v  &  w  &   h \\
+u'  & v'  &  w' &   h' \\
+u'' & v''  & w'' &   h'' \\
+u''' & v''' & w''' &  h'''
+\end{vmatrix}
+\]
+est un déterminant, symétrique gauche, donc de la forme
+\[
+\left\lvert
+\begin{array}{@{}rrrr@{}}
+  0   &    C  &    -B  &   \phantom{-}P \\
+  -C   &    0   &    A  &  Q \\
+  B   &  -A   &     0 &    R  \\
+  -P    & -Q   &   -R &    0
+\end{array}
+\right\rvert
+\]
+et l'équation devient
+\[
+\Omega = x'(\DPtypo{c}{C}y - Bz + P)
+       + y'(-Cx + Az + Q)
+       + z'(Bx - Ay + R)
+       - Px - \DPtypo{A}{Q}y - Rz = 0\Add{,}
+\]
+ou
+\begin{multline*}%[** TN: Added break]
+A(yz' - zy') + B(zx' - xz') + C(xy' - yx') \\
+  + P(x' - x) + Q (y' - y) + R(z' - z) = 0\Add{,}
+\end{multline*}
+%% -----File: 262.png---Folio 254-------
+équation d'un complexe linéaire. Le lieu des points $(x',y',z')$
+associés au point $(x,y,z)$ est le plan polaire du point $(x,y,z)$
+par rapport au complexe. Le plan polaire d'un point est la
+multiplicité transformée de ce point et réciproquement. La
+transformée d'une droite est sa conjuguée, une droite du complexe
+est à elle-même sa transformée. \Card{2} multiplicités transformées
+sont les \Card{2} multiplicités focales d'une congruence de
+droites du complexe et réciproquement: à une courbe correspond
+en général une développable; à une courbe du complexe
+correspond la développable de ses tangentes.
+
+Si nous prenons maintenant la solution $k = 1$, nous avons
+\[
+x'(ux + vy + wz + h) + \ldots = x(ux' + vy' + wz' + h) + \ldots\Add{,}
+\]
+la forme $\Omega$ est symétrique en $x\Add{,}y\Add{,}z$, $x'\Add{,}y'\Add{,}z'$, et on a
+\begin{multline*}%[** TN: Added break]
+\Omega = axx' +  byy' + czz' \\
+  + m(yz' + zy') + n(zx' + xz') + p(xy' + yx') \\
+  + r(x + x') + s(y + y') + t(z + z') + u\Add{.}
+\end{multline*}
+\DPtypo{les}{Les} \Card{2} points $(x,y,z)$\Add{,} $(x',y,' z')$ sont conjugués par rapport à la
+quadrique
+\[
+ax^2 + by^2 + cz^2 + 2myz + 2nzx + 2pxy + 2rx + 2sy + 2tz + u = 0\Add{.}
+\]
+Nous avons la transformation par polaires réciproques.
+
+D'une façon générale, pour avoir les équations d'une
+transformation de contact définie par une seule relation
+$\Omega = 0$, on écrira que la relation
+\[
+dz' - p'\, dx' - q'\, dy' = 0\Add{,}
+\]
+est conséquence des relations
+\[
+dz - p\, dx - q\, dy = 0, \qquad
+d\Omega = 0;
+\]
+ce qui donne
+\[
+dz' - p'\, dx' - q'\, dy'
+  = \lambda (dz - p\, dx - q\, dy) + \mu\, d\Omega\Add{.}
+\]
+%% -----File: 263.png---Folio 255-------
+En identifiant, on a \Card{6} équations; si entre elles on élimine~$\lambda\Add{,} \mu$,
+on a \Card{4} équations qui jointes à $\Omega = 0$ donnent $x',y',z',p',q'$
+en fonction de~$x,y,z,p,q$.
+
+Passons enfin au cas où on a \Card{2} relations d'élimination
+\[
+\Tag{(4)}
+\Omega(x,y,z,x',y',z') = 0, \qquad
+\Theta(x,y,z,x',y',z') = 0.
+\]
+A un point~$M$ du \Ord{1}{er} espace correspond dans le \Ord{2}{e} espace une
+courbe~$(c')$. A une courbe lieu de $\infty^{1}$~points correspond une
+surface engendrée par $\infty^{1}$~courbes; à une surface~$(S)$ lieu de
+$\infty^{2}$~points correspond une congruence de courbes; une telle
+congruence a en général une surface focale, tangente à toutes
+ces courbes, et qui sera la transformée de la surface~$(S)$.
+
+Pour avoir les équations d'une telle transformation, on
+écrira que la relation
+\[
+dz'- p'\, dx' - q'\, dy' = 0
+\]
+est conséquence des relations
+\[
+dz - p\, dx - q\, dy = 0, \qquad
+d\Omega = 0, \qquad
+d\Theta = 0;
+\]
+ce qui donne
+\[
+dz' - p'\, dx' - q'\, dy'
+  = \lambda (dz - p\, dx - q\, dy) + \mu d\Omega +  \nu d\Theta.
+\]
+En identifiant on a \Card{6} équations; si entre elles on \DPtypo{elimine}{élimine}
+$\lambda, \mu, \nu$ on a \Card{3} équations qui jointes à $\Omega = 0$, $\Theta = 0$, donnent
+les formules de transformation.
+
+\Section{Transformation de Sophus Lie.}
+{3.}{} Supposons les équations~\Eq{(4)} bilinéaires. A un point~$M(x\Add{,}y\Add{,}z)$
+correspond une droite~$D'$. Aux $\infty^{3}$~points~$M$ correspond un
+complexe de droites~$D'$, soit~$(K')$. De même à tous les points
+du \Ord{2}{e} espace correspond dans le \Ord{1}{er} espace un complexe~$(K)$.
+Considérons une seule des équations~\Eq{(4)}; à chaque point~$M$
+%% -----File: 264.png---Folio 256-------
+correspond un plan~$P'$; l'autre équation au même point~$M$ fait
+correspondre un plan~$Q'$ et la droite~$D'$ est l'intersection des
+plans $P'\Add{,} Q'$ qui correspondent au point~$M$ dans les \Card{2} homographies
+les plans~$P'$ correspondent homographiquement aux plans~$Q'$; le
+complexe~$(K')$ est le complexe des droites intersections des
+plans qui se correspondent dans \Card{2} homographies. (C'est le complexe
+de Reye, ou \emph{complexe tétraédral}; les droites sont coupées
+par un tétraèdre en \Card{4} points dont le rapport anharmonique
+est constant. Le rapport anharmonique des \Card{4} plans menés par
+une droite du complexe et par les \Card{4} sommets du tétraèdre est
+constant (Von~Staudt). Le complexe~$(K')$ est du \Ord{2}{e} degré,
+la surface des singularités est constituée par les \Card{4} faces
+du tétraèdre). A une courbe~$(c)$ correspond une surface
+réglée du complexe~$(K')$. A une surface~$(S)$ correspond une
+congruence de droites appartenant au complexe~$(K')$; cette
+congruence admet \Card{2} multiplicités focales. A un élément de
+contact du \Ord{1}{er} espace correspondent \Card{2} éléments de contact de
+l'autre.
+
+Cherchons les équations des \Card{2} complexes. Soient
+\[
+\Omega = Ax' + By' + Cz' + D, \qquad
+\Theta = Lx' + My' + Nz' + P\Add{.}
+\]
+Soit $M'(x',y'z')$ un point du \Ord{2}{e} espace. Soit $D$ la droite correspondante;
+si $(x,y,z)$ et $(x_0,y_0,z_0)$ sont \Card{2} points de cette
+droite, on a
+\begin{alignat*}{2}
+&\Omega (x,y,z,x',y',z') = 0, &
+&\Theta (x,y,z,x'\DPtypo{.}{,}y',z') = 0\Add{,} \\
+&\Omega (x_0,y_0,z_0, x',y',z') = 0, \qquad &
+&\Theta (x_0,y_0,z_0,x',y',z') = 0.
+\end{alignat*}
+\DPchg{Eliminons}{Éliminons} $x',y,'z'$ entre ces \Card{4} équations, nous avons
+%% -----File: 265.png---Folio 257-------
+\[
+\begin{vmatrix}
+A    &  B    &  C    &  D   \\
+A_0  &  B_0  &  C_0  &  D_0 \\
+L    &  M    &  N    &  P   \\
+L_0  &  M_0  &  N_0  &  P_0
+\end{vmatrix} = 0;
+\]
+c'est l'équation du complexe. En développant par la règle de
+Laplace, on trouvera une équation du \Ord{2}{e} degré par rapport aux
+coordonnées de la droite. Le complexe~$(K)$, et de même le complexe~$(K')$,
+est en général du \Ord{2}{e} degré.
+
+A une courbe~$(c)$ correspond une surface réglée engendrée
+par la droite~$D'$. Cherchons si cette surface réglée peut être
+développable. Les droites~$D'$ ont pour équations
+\[
+Ax' + By'+ Cz' + D = 0, \qquad
+Lx' + My'+ Nz' + P = 0;
+\]
+$x,y,z$, et par suite $A,B,C,D$, étant fonctions d'un paramètre~$t$.
+Exprimons que cette droite rencontre la droite infiniment
+voisine: nous adjoignons a ses équations les équations:
+\[
+x'\, dA + y'\, dB + z'\, dC + dD = 0, \qquad
+x'\, dL + y'\, dM + z'\, dN + dP = 0;
+\]
+d'où la condition
+\[
+\begin{vmatrix}
+A   &  B   &  C   &  D  \\
+L   &  M   &  N   &  P  \\
+dA  &  dB  &  dC  &  dD \\
+dL  &  dM  &  dN  &  dP
+\end{vmatrix} = 0.
+\]
+Or\Add{,} l'équation du complexe~$(K)$ peut s'écrire
+\[
+\begin{vmatrix}
+A         &  B         &  C         &  D        \\
+L         &  M         &  N         &  P        \\
+\Delta A  &  \Delta B  &  \Delta C  &  \Delta D \\
+\Delta L  &  \Delta M  &  \Delta N  &  \Delta P
+\end{vmatrix} = 0,
+\]
+$A,B,C,D$ étant des fonctions linéaires, les accroissements~$\Delta$
+sont proportionnels aux différentielles~$d$. La courbe~$(c)$ est
+%% -----File: 266.png---Folio 258-------
+donc telle que sa tangente appartient au complexe~$(K)$. Aux
+courbes du \Ord{1}{er} complexe correspondent des développables engendrées
+par les tangentes aux courbes du \Ord{2}{e} complexe. Au
+point~$M$ d'une courbe~$(c)$ du \Ord{1}{er} complexe correspond une génératrice~$T'$
+d'une développable, soit $M'$ son point de contact avec
+l'arête de rebroussement; si on considère un élément linéaire
+formé d'un point~$M$ et d'une droite~$D$ du \Ord{1}{er} complexe passant
+par ce point, il lui correspondra un élément linéaire déterminé
+du \Ord{2}{e} complexe. Les courbes des \Card{2} complexes se correspondent
+ainsi par points et par tangentes.
+
+\Illustration{266a}
+Soit une surface~$(S)$, et supposons le complexe~$(K)$ effectivement
+du \Ord{2}{e} degré. Considérons un point~$M$ de la surface et
+le plan tangent~$P$. Le cône du complexe~$(K)$ de sommet~$M$ est
+coupé par le plan~$P$ suivant \Card{2} droites
+$D, D_1$ qui appartiennent au complexe~$(K)$\Add{.}
+Par chaque point de~$(S)$ passent ainsi
+\Card{2} droites du complexe~$(K)$ tangentes à
+la surface. Par tout point de la
+surface~$(S)$ passent donc \Card{2} courbes
+$(\gamma)\Add{,} (\gamma_1)$ du complexe~$(K)$ situées sur cette surface. Au point~$M$
+correspond une droite~$D'$ du complexe~$(K')$. A la droite~$D$ du complexe~$(K)$
+correspond un point~$M'$ de~$D'$; et de même à la droite~$D$,
+correspond un point~$M_1'$ de~$D'$. Aux courbes $(\gamma)\Add{,} (\gamma_1)$ du complexe~$(K)$
+correspondent \Card{2} courbes $(\gamma')\Add{,} (\gamma_1')$ du complexe~$(K')$
+tangentes en $M'\Add{,} M_1'$ à la droite~$D'$. Si le point~$M$ décrit la
+courbe~$(\gamma)$, les droites~$D'$ ont pour enveloppe la courbe~$(\gamma_1')$,
+%% -----File: 267.png---Folio 259-------
+et si $M$~décrit~$(\gamma_1)$, $D'$~enveloppe~$(\gamma_1')$. Si on considère la
+congruence des droites~$D'$ correspondant aux points~$M$ de la surface~$(S)$,
+les courbes~$(\gamma_t')$ sont les arêtes de rebroussement
+d'une des familles de développables de cette congruence et les
+courbes~$(\gamma_1')$ sont les arêtes de rebroussement de l'autre famille.
+Les courbes~$(\gamma')$ engendrent une des nappes de la surface
+focale, les courbes~$(\gamma_1')$ engendrent l'autre nappe. Le
+plan tangent en~$M'$ à la multiplicité focale est le plan osculateur
+à~$(\gamma_1')$, et par suite le plan tangent au cône du complexe~$(K')$
+de sommet~$M_1'$. Un élément de contact correspondant
+à l'élément~$(M,P)$ est formé du point~$M'$ et du plan tangent au
+cône du complexe~$(K')$ qui a pour sommet~$M_1'$. L'autre élément
+correspondant à~$(M,P)$ est formé du point~$M_1'$ et du plan tangent
+au cône du complexe~$(K')$ qui a pour sommet~$M'$.
+
+Si la surface~$(S)$ est une surface du complexe~$(K)$, tangente
+en chacun de ses points au cône du complexe, les droites
+$D,~D_1$ sont confondues; alors les \Card{2} éléments de contact correspondant
+à l'élément~$(M\Add{,}P)$ sont confondus, et la surface~$(S')$
+définie par ces éléments est une surface du complexe~$(K')$.
+
+\Paragraph{Remarques.} Les seuls cas possibles sont les suivants:
+
+\ParItem{\Primo.} Les complexes~$(K)\Add{,} (K')$ sont effectivement du \Ord{2}{e} degré.
+On démontre alors, comme nous l'avons dit précédemment,
+qu'ils sont tous \Card{2} tétraédraux.
+
+\ParItem{\Secundo.} Un des complexes est linéaire. On démontre que
+l'autre est constitué par les droites qui rencontrent une
+conique.
+
+\ParItem{\Tertio.} Les \Card{2} complexes sont linéaires. On démontre qu'ils
+%% -----File: 268.png---Folio 260-------
+sont tous \Card{2} spéciaux. Ce cas donne la \emph{transformation d'Ampère},
+définie par les équations
+\[
+x' = p, \qquad
+y' = -y, \qquad
+z' = -z - px\DPtypo{.}{,}\qquad
+p' = x, \qquad
+q' = -q\Add{.}
+\]
+
+\Section{Transformation des droites en sphères.}
+{4.}{} Prenons en particulier:
+\[
+\Omega = x + iy + x'z + z' = 0, \qquad
+\Theta = x' (x - iy) - z - y' = 0
+\]
+L'équation du \Ord{1}{er} complexe est:
+\[
+\left\lvert
+\begin{array}{@{}crrc@{}}
+z - z_0 &  0 &  \phantom{-}0 & x + iy - (x_0 + iy_0) \\
+z_0     &  0 &             1 & x_0 + iy_0 \\
+x - iy  & -1 &             0 & -z \\
+x_0 -iy_0 - (x-iy) & 0 &   0 & z - z_0
+\end{array}
+\right\rvert
+= 0,
+\]
+ce qui devient:
+\[
+(x - x_0)^2 + (y - y_0)^2 + (z - z_0)^ 2 = 0,
+\]
+c'est-à-dire:
+\[
+\Tag{(K)}
+a^2 + b^2 + c^ 2 = 0.
+\]
+Le complexe~$(K)$ est le complexe des droites minima. Cherchons
+le \Ord{2}{e} complexe. Il suffit de \DPtypo{considerer}{considérer} \Card{2} points~$(x',y',z')$
+correspondant au même point~$(x,y,z)$. Nous avons:
+\[
+\begin{vmatrix}
+0      &         0         &  x' - x'_0  &  z' - z'_0 \\
+1      &          i        &       x'_0  &       z'_0 \\
+x'- x'_0  &  -i(x' - x'_0) &   0          & -(y' - y'_0) \\
+x'_0     &      -ix'_0     &  \llap{$-$} 1 &     y'_0
+\end{vmatrix} = 0,
+\]
+ce qui devient:
+\[
+(x' - x'_0) (x'y'_0 - y'x'_0) - (z' - z'_0) (x' - x'_0) = 0,
+\]
+c'est-à-dire:
+\[
+a(r + c) = 0\DPtypo{.}{;}
+\]
+la solution $a = 0$ est singulière, et on a pour le complexe~$(K')$
+\[
+\Tag{(K')}
+r + c =0.
+\]
+Nous avons ainsi une \emph{correspondance entre un complexe spécial
+%% -----File: 269.png---Folio 261-------
+du \Ord{2}{e} degré et un complexe linéaire}. Le cône du complexe~$(K)$
+sera le cône isotrope. A chaque élément de contact du \Ord{1}{er} espace
+correspondent \Card{2} éléments de contact du \Ord{2}{e} espace conjugués
+par rapport au complexe~$(K')$, car d'une façon générale
+les points~$M'\Add{,} M_1'$, sont sur une droite~$D'$ de~$(K')$ et le plan associé
+à~$M'$ est ici le plan polaire de~$M_1'$ et inversement.
+
+Partons d'une sphère; prenons \Card{2} génératrices d'un système;
+ce sont des droites minima~$D, D_1$. Le \Ord{2}{e} système de génératrices
+est entièrement défini, car chacune d'elles doit
+rencontrer $D, D_1$, et le cercle imaginaire à l'infini. Aux \Card{2} droites~$D, D_1$
+correspondent \Card{2} points~$M', M_1'$. Considérons une génératrice
+isotrope~$\Delta$ rencontrant~$D, D_1$, il lui correspond un
+point~$\mu'$; $\Delta$~rencontrant la droite~$D$, la droite~$M' \mu'$ est une
+droite du complexe linéaire, et de même~$M_1' \mu'$; donc $\mu_1$~est le
+pôle d'un plan passant par~$M_1'M'$. Lorsque $\Delta$~décrit la sphère,
+le plan~$\mu' M' M_1'$ tourne autour de~$M'M_1'$, et le lieu de~$\mu'$ est la
+droite conjuguée de~$M_1' M'$. A la sphère correspond une droite.
+En partant du \Ord{2}{e} système de génératrices, au système~$D$,~$D_1$, correspondrait
+la droite~$M_1'M'$. \emph{A une sphère correspondent en
+réalité \Card{2} droites conjuguées par rapport au complexe linéaire.}
+
+Ceci peut se voir par le calcul. Prenons la droite~$(\Delta')$
+\[
+\Tag{(\Delta')}
+x' = az' - q, \qquad
+y' = bz' + p;
+\]
+$c = 1$ et~$r = - ap - bq$. La surface réglée correspondante est
+engendrée par les droites
+\[
+x + iy + z(az' - q) + z' = 0, \qquad
+(az'- q)(x - iy) - z - bz' - p = 0\Add{,}
+\]
+ou
+\[
+x + iy - qz + z'(az + 1) = 0, \qquad
+-q(x - iy) - z - p + z' \bigl[a(x - iy) - b\bigr] = 0\Add{.}
+\]
+%% -----File: 270.png---Folio 262-------
+\DPchg{Eliminant}{Éliminant} $z'$ on a la surface
+\[
+(x + iy - qz) \bigl[a(x - iy) - b\bigr]
+   + (az + 1) \bigl[q(x - iy) + z + p\bigr] = 0\Add{,}
+\]
+ou
+\[
+a(x^2 + y^2+ z^2) - b(x + iy) + q(x - iy) + (c - r)z + p = 0\Add{,}
+\]
+c'est l'équation d'une sphère et il est facile de voir que ce
+peut être une sphère quelconque, en choisissant~$(\Delta')$ convenablement.
+
+Cherchons la conjuguée de~$(\Delta')$ par rapport à~$(K')$. Nous
+avons à exprimer que le complexe~$(K')$ et les complexes spéciaux
+$(\Delta), (\Delta')$ appartiennent à un même faisceau. Ce qui donne
+\begin{alignat*}{3}%[** TN: Added break, rearranged terms]
+\lambda a + \lambda' a' &= 0, &
+\lambda b + \lambda' b' &= 0, &
+\lambda c + \lambda' c' + 1 &= 0, \\
+\lambda p + \lambda' p' &= 0,\qquad &
+\lambda q + \lambda' q' &= 0,\qquad &
+\lambda r + \lambda' r' + 1 &= 0;
+\end{alignat*}
+prenons $\lambda' = -\lambda$, nous avons:
+\begin{alignat*}{3}%[** TN: Added break, rearranged terms]
+a' &= a, & b' &= b, & c' &= c + \frac{1}{\lambda}, \\
+p' &= p, \qquad & q' &= q, \qquad & r' &= r + \frac{1}{\lambda};
+\end{alignat*}
+prenons $\lambda = -\dfrac{1}{c+r}$, alors $c' = -r$ et $r'= -c$, et l'on voit
+que l'on retrouve la même sphère.
+
+Les formules de la transformation se calculent par la
+méthode générale. On trouve:
+\begin{gather*}
+x = -\frac{z'}{2}
+  + \frac{1}{2}\, \frac{x' (x' p + y' q) + y' - p'}{q' + x'}, \qquad
+y =  \frac{i z'}{2}
+  - \frac{i}{2}\, \frac{x' (x' p + y' q) - y' + p'}{q' + x'}, \\
+z =  -\frac{x'p' + y'q'}{q' + x'}, \qquad
+p =   \frac{x'q' + 1}{q' - x'}, \qquad
+q = -i\frac{x'q' - 1}{q' - x'}.
+\end{gather*}
+Cette transformation de Sophus Lie, changeant des droites qui
+se rencontrent en sphères tangentes, c'est-à-dire des droites
+qui ont un élément de contact commun en sphères ayant un élément
+de contact commun, réalise par suite la correspondance
+%% -----File: 271.png---Folio 263-------
+signalée dans les chapitres précédents entre les droites et
+les sphères.
+
+Par exemple elle transforme une surface réglée en surface
+canal; une quadrique en cyclide de Dupin; une surface
+développable en surface canal isotrope; une bande asymptotique
+d'une surface en une bande de courbure de la transformée;
+de sorte qu'on peut dire qu'elle transforme les lignes asymptotiques
+en lignes de courbure.
+
+On vérifiera facilement qu'elle transforme un complexe
+linéaire de droites en une famille de $\infty^{2}$~sphères coupant une
+sphère fixe sous un angle constant; et que cet angle constant
+est droit, lorsque le complexe linéaire est en involution avec
+le complexe~$(K')$.
+
+\Section{Transformation des lignes asymptotiques.}
+{5.}{} Proposons-nous de trouver toutes les transformations
+de contact qui changent les lignes asymptotiques d'une
+surface quelconque en les lignes asymptotiques de la transformée
+de cette surface; c'est-à-dire qui changent toute
+bande asymptotique en une bande asymptotique. Remarquons à
+cet effet qu'une telle transformation changera toute multiplicité~$M_2$
+sur laquelle les bandes asymptotiques ne dépendent
+pas seulement de constantes arbitraires, mais dépendent de
+fonctions arbitraires, en une multiplicité~$M_2$ de même nature.
+Or\Add{,} les bandes asymptotiques (ou de rebroussement) étant définies
+par les équations
+\[
+dz - p\, dx - q\, dy = 0, \qquad
+dp\, dx + dq\, dy = 0,
+\]
+on devra considérer comme bande asymptotique, dans la question
+%% -----File: 272.png---Folio 264-------
+actuelle, $\infty^{1}$~éléments de contact ayant le même point\Add{,}
+\[%[** TN: Inline parenthetical remark in original]
+dx = dy = dz = 0,\qquad  p = f(q)\Add{,}
+\]
+c'est-à-dire un cône élémentaire.
+
+Et, dès lors, les $M_2$~particulières en question sont les plans,
+les droites et les points. Les transformations cherchées
+échangent donc entre elles les figures qui sont des droites,
+des points ou des plans. De là plusieurs cas à examiner:
+
+\ParItem{\Primo.} Si la transformation est ponctuelle, elle échange les
+points en points, les plans en plans, et les droites en droites.
+C'est par suite une \emph{transformation homographique}.
+
+\ParItem{\Secundo.} Si la transformation est une transformation de contact
+de la première espèce, c'est-à-dire fait correspondre à chaque
+point du premier espace~$(E)$ une surface du second espace~$(E')$,
+elle change les points de~$(E)$ en les plans de~$(E')$; et comme
+elle fait alors correspondre aussi à chaque point de~$(E')$ une
+surface de~$(E)$, elle change les points de~$(E')$ en les plans de~$(E)$;
+et dès lors elle change les points en plans, les plans
+en points, et les droites en droites. Si donc on la compose
+avec une transformation par polaires réciproques, on obtient
+une transformation homographique; et, par suite, elle s'obtient
+en composant une transformation homographique avec une
+transformation par polaires réciproques. C'est donc une
+\emph{transformation dualistique}.
+
+\ParItem{\Tertio.} Si la transformation est une transformation de contact
+de la deuxième espèce, c'est-à-dire si à tout point de l'un
+des espaces correspond dans l'autre une courbe, à tout point
+de l'un des espaces correspondra dans l'autre une droite. Or,
+prenons dans l'espace~$(E)$ quatre points $P_1,P_2,P_3,P_4$ non situés
+%% -----File: 273.png---Folio 265-------
+dans un même plan, et soient $D_1,D_2,D_3,D_4$ les droites qui leur
+correspondent dans l'espace~$(E')$. Il existe au moins une
+droite~$\Delta$ ayant avec chacune des quatre droites $D_1,D_2,D_3,D_4$
+un élément de contact commun; et à~$\Delta$ devrait correspondre dans~$(E)$
+un point, un plan ou une droite ayant un élément de contact
+commun avec chacun des quatre points $P_1,P_2,P_3,P_4$. Or\Add{,} il
+n'en existe pas. Donc \emph{ce troisième cas est impossible}.
+
+Les seules transformations pouvant répondre à la question
+sont donc homographiques ou dualistiques. Mais toute
+transformation de contact changeant les droites en droites
+répond à la question, car elle changera une famille de génératrices
+d'une développable, dont chacune a un élément de
+contact commun avec la génératrice infiniment voisine, en les
+génératrices d'une autre développable; et, par suite, la bande
+de rebroussement de la première développable ou la bande de
+rebroussement de la seconde.
+
+Nous pouvons donc conclure:
+
+\ParItem{\Primo.} \emph{Les transformations homographiques et les transformations
+dualistiques changent les lignes asymptotiques en lignes
+asymptotiques; et ce sont les seules transformations de
+contact possédant cette propriété.}
+
+\ParItem{\Secundo.} \emph{Ces transformations sont aussi les seules transformations
+de contact changeant toute droite en une droite.}
+
+\Section{Transformations des lignes de courbure.}
+{6.}{} La transformation de contact des droites en sphères,
+de Lie, permet de déduire immédiatement des résultats
+précédents toutes les transformations de contact qui
+%% -----File: 274.png---Folio 266-------
+changent les lignes de courbure d'une surface quelconque en
+les lignes de courbure de sa transformée.
+
+On voit de plus que ce sont aussi celles qui changent
+toute sphère en une sphère. On aurait pu du reste refaire un
+raisonnement direct analogue à celui du \Numero~précédent, en partant
+des multiplicités~$M_2$ pour lesquelles les bandes de courbure
+dépendent de fonctions arbitraires.
+
+Cherchons, plus spécialement, celles des transformations
+considérées qui sont des transformations ponctuelles. Dans
+la transformation de Lie, les points de l'espace~$(E)$ correspondent
+aux droites d'un complexe linéaire~$(K')$. Les transformations
+cherchées proviennent donc des transformations
+projectives ou dualistiques qui laissent invariant ce complexe\Add{.}
+On les obtient en composant avec la transformation par polaires
+réciproques définie par ce complexe~$(K')$ l'une quelconque
+des transformations projectives qui laissent le complexe
+invariant.
+
+Ainsi se trouve établie une correspondance entre le
+groupe projectif d'un complexe linéaire et le groupe des
+transformations ponctuelles qui changent toute sphère en
+sphère. Ce dernier est ce qu'on appelle le groupe conforme;
+on sait que ses transformations s'obtiennent en combinant des
+inversions, des homothéties et des déplacements.
+
+Parmi les transformations de contact qui changent les
+lignes de courbure en lignes de courbure figurent les \emph{dilatations},
+dans lesquelles chaque élément de contact subit une
+translation perpendiculaire à son plan et d'amplitude donnée,
+c'est-à-dire dans lesquelles chaque surface est remplacée par
+%% -----File: 275.png---Folio 267-------
+une surface parallèle.
+
+Parmi ces transformations figurent aussi les transformations
+de \DPtypo{Ribeaucour}{Ribaucour} qui seront définies au chapitre~XIII\@.
+
+
+\ExSection{XI}
+
+\begin{Exercises}
+\item[50.] \DPchg{Etudier}{Étudier} la congruence des droites définies par les équations
+\[
+A\lambda + B\mu + C = 0, \qquad
+A_{1}\lambda + B_{1}\mu + C_{1} = 0,
+\]
+où $A,B,C$, $A_{1},B_{1},C_{1}$ sont des fonctions linéaires des coordonnées
+et $\lambda, \mu$ des paramètres arbitraires. Discuter en particulier la
+question des droites passant par un point, des droites rencontrant
+une droite fixe, des droites situées dans un plan, des
+multiplicités focales.
+
+\item[51.] Démontrer les résultats énoncés à la fin du~\No3 de ce chapitre.
+
+\item[52.] Démontrer par le calcul les propriétés de la transformation
+de Lie énoncées à la fin du~\No4 de ce chapitre.
+\end{Exercises}
+%% -----File: 276.png---Folio 268-------
+
+
+\Chapitre{XII}{\DPchg{Systemes}{Systèmes} Triples Orthogonaux.}
+
+
+\Section{Théorème de Dupin.}
+{1.}{} L'emploi des coordonnées rectangulaires revient à
+celui d'un système de \Card{3} plans orthogonaux. On peut généraliser
+et employer comme surfaces coordonnées un système triple
+quelconque:
+\[
+\phi(x,y,z) = u, \qquad
+\psi(x,y,z) = v, \qquad
+\chi(x,y,z) = w;
+\]
+ces formules transforment les coordonnées $u,v,w$ en coordonnées
+$x,y,z$\DPtypo{,}{.} Si nous résolvons les équations précédentes en $x,y,z$, ce
+que nous supposons possible, nous aurons
+\[
+\Tag{(1)}
+x = f(u,v,w), \qquad
+y = g(u,v,w), \qquad
+z = h(u,v,w).
+\]
+On emploie en général un \emph{système triple orthogonal}. Cherchons
+donc à exprimer que les équations~\Eq{(1)} définissent un système
+triple orthogonal. Les intersections des surfaces \Card{2} à \Card{2} doivent
+être orthogonales. Les surfaces des \Card{3} familles s'obtiendront
+en faisant successivement $u = \cte$, $v = \cte$, $w = \cte[]$.
+Les intersections des surfaces \Card{2} à \Card{2} sont respectivement
+$(v = \cte, w = \cte)$\Add{,} $(w = \cte, u = \cte)$\Add{,} $(u = \cte, v = \cte)$, et les
+directions des tangentes sont respectivement $\dfrac{\dd f}{\dd u}, \dfrac{\dd g}{\dd u}, \dfrac{\dd h}{\dd u}$;
+$\dfrac{\dd f}{\dd v}, \dfrac{\dd g}{\dd v}, \dfrac{\dd h}{\dd v}$; et
+$\dfrac{\dd f}{\dd w}, \dfrac{\dd g}{\dd w}, \dfrac{\dd h}{\dd w}$. La condition d'orthogonalité est
+\[
+\Tag{(2)}
+\sum \frac{\dd f}{\dd v} \Add{·} \frac{\dd f}{\dd w} = 0, \qquad
+\sum \frac{\dd f}{\dd w} · \frac{\dd f}{\dd u} = 0, \qquad
+\sum \frac{\dd f}{\dd u} \Add{·} \frac{\dd f}{\dd v} = 0.
+\]
+Interprétons ces conditions. Prenons la surface $w = \cte[]$. La
+\Ord{3}{e} condition exprime que sur cette surface les lignes $u = \cte$,
+%% -----File: 277.png---Folio 269-------
+$v = \cte$ sont orthogonales, et les \Card{2} premières expriment que
+$\dfrac{\dd f}{\dd w}, \dfrac{\dd g}{\dd w}, \dfrac{\dd h}{\dd w}$ est une direction perpendiculaire aux tangentes à
+ces \Card{2} courbes, et par suite, que c'est la direction de la
+normale; soient $l,m,n$ ces \Card{3} coefficients de direction. \DPtypo{Differentions}{Différentions}
+la \Ord{3}{e} relation par rapport à~$w$ nous avons
+\[
+\sum \frac{\dd f}{\dd u}\, \frac{\dd^2 f}{\dd v\, \dd w} +
+\sum \frac{\dd f}{\dd v}\, \frac{\dd^2 f}{\dd u\, \dd w} = 0,
+\]
+ou
+\[
+\sum \frac{\dd f}{\dd u}\, \frac{\dd l}{\dd v} +
+\sum \frac{\dd f}{\dd v}\, \frac{\dd l}{\dd u} = 0.
+\]
+Or\Add{,} on a
+\[
+\sum l\, \frac{\dd f}{\dd u} = 0, \qquad
+\sum l\, \frac{\dd f}{\dd v} = 0,
+\]
+d'où
+\[
+\sum l\, \frac{\dd^2 f}{\dd u\, \dd v}
+  = -\sum \frac{\dd l}{\dd v}\, \frac{\dd f}{\dd u}, \qquad
+\sum l\, \frac{\dd^2 f}{\dd u\, \dd v}
+  = -\sum \frac{\dd l}{\dd u}\, \frac{\dd f}{\dd v},
+\]
+et la condition précédente s'écrit
+\[
+\sum l\, \frac{\dd^2 f}{\dd u\, \dd v}  = 0,
+\]
+ou $F' = 0$, ce qui exprime que les lignes $u = \cte$, $v = \cte$,
+c'est-à-dire les intersections de la surface $w = \cte$ avec les
+surfaces $u = \cte$ et $v = \cte$ sont conjuguées sur cette surface,
+et comme elles sont orthogonales, ce sont des lignes de courbure.
+D'où le \emph{Théorème de Dupin: sur chaque surface d'un
+système triple orthogonal, les intersections avec les autres
+surfaces de ce système sont des lignes de courbure}.
+
+\Section{\DPchg{Equation}{Équation} aux dérivées partielles de Darboux.}
+{2.}{} Proposons-nous de rechercher les systèmes triples
+orthogonaux. Prenons une famille de surfaces:
+\[
+\Tag{(1)}
+\phi(x,y,z) = u
+\]
+et cherchons à déterminer \Card{2} autres familles constituant avec
+celle-ci un système triple orthogonal. Prenons dans l'espace
+%% -----File: 278.png---Folio 270-------
+un point~$M$; par ce point~$M$ passe une surface~$u$; prenons les
+tangentes~$MT, MT'$ en $M$~à ses lignes de courbure. Ces droites
+sont parfaitement déterminées; si $p, q, -1$ sont les coefficients
+de direction de~$MT$, ce sont des fonctions connues de~$x, y, z$.
+De même pour~$MT'$. Il faudra alors qu'en chaque point~$M$,
+une surface d'une autre famille, soit
+\[
+%[** TN: Looks like \Psi in original, but using \psi for consistency]
+%[** TN: No equation number (2) in original]
+\Tag{(2)}
+\psi (x,y,z) = v,
+\]
+soit normale à~$MT$; il faudra donc que $p, q$ soient les dérivées
+partielles de~$z$ par rapport à $x, y$, ($z$~étant défini par l'équation
+précédente). On aura donc
+\[
+\frac{\dd \psi}{\dd x} + p \frac{\dd \psi}{\dd z} = 0, \qquad
+\frac{\dd \psi}{\dd y} + \Err{}{q} \frac{\dd \psi}{\dd z} = 0.
+\]
+Ces équations ne sont pas compatibles en général: pour
+qu'elles le soient, il faut et il suffit, d'après la théorie
+des systèmes complets, que l'on ait:
+\[
+\Tag{(3)}
+\frac{\dd q}{\dd x} + p \frac{\dd q}{\dd z} =
+\frac{\dd p}{\dd y} + q \frac{\dd p}{\dd z}\Add{,}
+\]
+équation aux dérivées partielles du \Ord{3}{e} ordre, puisque $p, q$
+s'expriment en fonction des dérivées \Ord{1}{ères} et \Ord{2}{mes} de~$\phi$ par
+rapport à $x, y, z$. Ainsi donc \emph{une famille de surfaces données ne
+peut en général faire partie d'un système triple orthogonal}.
+Si la condition~\Eq{(3)} est réalisée, la fonction~$\psi(x,y,z)$ est déterminée
+à une fonction arbitraire près, et nous avons une
+\Ord{2}{e} famille de surfaces dont chacune coupe à angle droit chacune
+des surfaces~$(S)$ de la famille~$\phi(x,y,z) = \const.$\ suivant
+une ligne de courbure de cette surface~$(S)$. Et, d'après le
+théorème de Joachimsthal, l'intersection de chaque surface~$(S_1)$
+de cette seconde famille avec chaque surface~$(S)$ de la première
+est aussi ligne de courbure sur~$(S_1)$.
+%% -----File: 279.png---Folio 271-------
+
+En résumé nous avons deux familles de surfaces
+\begin{align*}
+\Tag{(S)}
+\phi(x,y,z) &= \const. \\
+\Tag{(S_1)}
+\psi(x,y,z) &= \const.
+\end{align*}
+qui se coupent orthogonalement suivant des courbes qui sont
+lignes de courbure à la fois pour les deux surfaces qui s'y
+croisent. Et il reste à étudier si l'on peut déterminer une
+troisième famille de surfaces
+\[
+\Tag{(S_2)}
+\chi(x,y,z) = \const.
+\
+\]
+qui constitue avec les deux premières un système triple orthogonal,
+c'est-à-dire à étudier le système d'équations linéaires
+aux dérivées partielles dont dépend la fonction inconnue~$\chi$:
+\[
+\Tag{(4)}
+\left\{
+\begin{aligned}
+\frac{\dd \phi}{\dd x} · \frac{\dd \chi}{\dd x} +
+\frac{\dd \phi}{\dd y} · \frac{\dd \chi}{\dd y} +
+\frac{\dd \phi}{\dd z} · \frac{\dd \chi}{\dd z} &= 0, \\
+%
+\frac{\dd \psi}{\dd x} · \frac{\dd \chi}{\dd x} +
+\frac{\dd \psi}{\dd y} · \frac{\dd \chi}{\dd y} +
+\frac{\dd \psi}{\dd z} · \frac{\dd \chi}{\dd z} &= 0.
+\end{aligned}
+\right.
+\]
+Posons, pour abréger,
+\begin{align*}
+Af &= \frac{\dd \phi}{\dd x} · \frac{\dd f}{\dd x}
+    + \frac{\dd \phi}{\dd y} · \frac{\dd f}{\dd y}
+    + \frac{\dd \phi}{\dd z} · \frac{\dd f}{\dd z}, \\
+Bf &= \frac{\dd \phi}{\dd x} · \frac{\dd f}{\dd x}
+    + \frac{\dd \phi}{\dd y} · \frac{\dd f}{\dd y}
+    + \frac{\dd \phi}{\dd z} · \frac{\dd f}{\dd z};
+\end{align*}
+et, d'après la théorie des systèmes complets d'équations linéaires,
+la condition nécessaire et suffisante pour l'\Err{intégralité}{intégrabilité}
+du système~\Eq{(4)} est que l'équation:
+\[
+\left(A \frac{\dd\psi}{\dd x} - B \frac{\dd\phi}{\dd x}\right)
+  \Del{·} \frac{\dd\chi}{\dd x} +
+\left(A \frac{\dd\psi}{\dd y} - B \frac{\dd\phi}{\dd y}\right)
+  \Del{·} \frac{\dd\chi}{\dd y} +
+\left(A \frac{\dd\psi}{\dd z} - B \frac{\dd\phi}{\dd z}\right)
+  \Del{·} \frac{\dd\chi}{\dd z} = 0
+\]
+soit une conséquence algébrique des équations~\Eq{(4)}, c'est-à-dire
+que l'on ait:
+\[
+\Tag{(5)}
+\begin{vmatrix}
+A \mfrac{\dd\psi}{\dd x} - B \mfrac{\dd\phi}{\dd x} &
+  \mfrac{\dd\phi}{\dd x} & \mfrac{\dd\psi}{\dd x} \\
+A \mfrac{\dd\psi}{\dd y} - B \mfrac{\dd\phi}{\dd y} &
+  \mfrac{\dd\phi}{\dd y} & \mfrac{\dd\psi}{\dd y} \\
+A \mfrac{\dd\psi}{\dd z} - B \mfrac{\dd\phi}{\dd z} &
+  \mfrac{\dd\phi}{\dd z} & \mfrac{\dd\psi}{\dd z}
+\end{vmatrix}
+= 0.
+\]
+%% -----File: 280.png---Folio 272-------
+Cette condition se simplifie. Remarquons en effet que
+\begin{align*}
+A \frac{\dd\psi}{\dd x} + B \frac{\dd\phi}{\dd x}
+  &= \frac{\dd\phi}{\dd x}\, \frac{\dd^2\psi}{\dd x^2}
+   + \frac{\dd\phi}{\dd y}\, \frac{\dd^2\psi}{\dd y\, \dd x}
+   + \frac{\dd\phi}{\dd z}\, \frac{\dd^2\psi}{\dd z\, \dd x} \\
+%
+  &+ \frac{\dd\psi}{\dd x}\, \frac{\dd^2\phi}{\dd x^2}\,
+   + \frac{\dd\psi}{\dd y}\, \frac{\dd^2\phi}{\dd y\, \dd x}
+   + \frac{\dd\psi}{\dd z} · \frac{\dd^2\phi}{\dd z\, \dd x} \\
+%
+  &= \frac{\dd}{\dd x} \left\{
+       \frac{\dd\phi}{\dd x}\, \frac{\dd\psi}{\dd x}
+     + \frac{\dd\phi}{\dd y}\, \frac{\dd\psi}{\dd y}
+     + \frac{\dd\phi}{\dd z}\, \frac{\dd\psi}{\dd z}
+  \right\},
+\end{align*}
+c'est-à-dire que l'on a, à cause de l'\DPtypo{orthogonalite}{orthogonalité} des surfaces $(S)$~et~$(S_1)$,
+\[
+A \frac{\dd\psi}{\dd x} + B \frac{\dd\phi}{\dd x} = 0.
+\]
+On voit de même que l'on a aussi
+\[
+A \frac{\dd\psi}{\dd y} + B\frac{\dd\phi}{\dd y} = 0, \qquad
+A \frac{\dd\psi}{\dd z} + B\frac{\dd\phi}{\dd z} = 0.
+\]
+Par suite, la condition~\Eq{(4)} peut s'écrire simplement
+\[
+\Tag{(6)}
+\begin{vmatrix}
+A \mfrac{\dd\psi}{\dd x} & \mfrac{\dd\phi}{\dd x} & \mfrac{\dd\psi}{\dd x} \\
+A \mfrac{\dd\psi}{\dd y} & \mfrac{\dd\phi}{\dd y} & \mfrac{\dd\psi}{\dd y} \\
+A \mfrac{\dd\psi}{\dd z} & \mfrac{\dd\phi}{\dd z} & \mfrac{\dd\psi}{\dd z}
+\end{vmatrix}
+= 0.
+\]
+Or, pour une valeur quelconque de $x,y,z$, les dérivées~$\dfrac{\dd \psi}{\dd x},\dfrac{\dd \psi}{\dd y},\dfrac{\dd \psi}{\dd z}$ sont les coefficients de direction~$l,m,n$ de la normale
+à la surface~$(S_1)$ qui passe par le point de coordonnées~$x,y,z$;
+et~$\dfrac{\dd\phi}{\dd x},\dfrac{\dd\phi}{\dd y},\dfrac{\dd\phi}{\dd z}$ sont les coefficients de direction de la
+normale à la surface~$(S)$ qui passe par ce même point, c'est-à-dire
+de la tangente à une ligne de courbure de~$(S_1)$; en désignant
+par~$dx,dy,dz$ un déplacement effectué suivant la direction
+de cette tangente, on aura donc
+\[
+\frac{\ \dfrac{\dd\phi}{\dd x}\ }{dz} =
+\frac{\ \dfrac{\dd\phi}{\dd y}\ }{dy} =
+\frac{\ \dfrac{\dd\phi}{\dd z}\ }{dx} =
+\frac{A \dfrac{\dd\psi}{\dd x}}{dl} =
+\frac{A \dfrac{\dd\psi}{\dd y}}{dm} =
+\frac{A \dfrac{\dd\psi}{\dd z}}{dn},
+\]
+%% -----File: 281.png---Folio 273-------
+et la condition~\Eq{(6)} deviendra:
+\[
+\begin{vmatrix}
+dl & dx & l \\
+dm & dy & m \\
+dn & dz & n
+\end{vmatrix}
+= 0\DPtypo{.}{,}
+\]
+ce qui est une identité puisque la différentiation~$d$ considérée
+a lieu suivait une ligne de courbure.
+
+La condition d'intégrabilité du système~\Eq{(4)} est donc remplie,
+et la troisième famille~$(S_2)$ existe toujours et est
+entièrement déterminée. On peut donc énoncer les résultats
+suivants:
+
+\ParItem{\Primo.} \emph{Il existe une équation aux dérivées partielles du troisième
+ordre $\bigl($l'équation \Eq{(3)}$\bigr)$ qui exprime la condition nécessaire
+et suffisante pour qu'une fonction $\phi(x,y,z)$ fournisse
+une famille de surfaces~$(S)$ faisant partie d'un système triple
+orthogonal. Si la famille~$(S)$ est donnée, les deux autres familles $(S_1)$~et~$(S_2)$ sont entièrement déterminées}.
+
+\ParItem{\Secundo.} \emph{Pour que deux familles de surfaces, $(S)$~et~$(S_1)$, fassent
+partie d'un système triple orthogonal, il faut et il suffit
+qu'elles se coupent à angle droit, et que les intersections
+soient lignes de courbure sur les surfaces~$(S)$, ou sur les
+surfaces~$(S_1)$.}
+
+On peut remarquer enfin, que si l'on \DPchg{connait}{connaît} les lignes
+de courbure~$(C_1)$ des surfaces~$(S_1)$ par exemple, qui ne sont
+pas les intersections des surfaces~$(S_1)$ et des surfaces~$(S)$,
+et les lignes de courbure~$(C)$ d'une seule surface~$(S)$, chaque
+%% -----File: 282.png---Folio 274-------
+surface~$(S_2)$ sera engendrée par les courbes~$(C_1)$ qui s'appuient
+sur une même courbe~$(C)$.
+
+\Section{Systèmes triples orthogonaux contenant une surface.}
+{3.}{} Cherchons maintenant si une surface donnée peut
+faire partie d'un système triple orthogonal. Traçons
+sur cette surface les lignes de courbure, et menons les
+normales à la surface en tous les points de ces lignes, elles
+engendrent deux familles de développables orthogonales à la
+surface donnée. En adjoignant à cette surface les surfaces
+parallèles, on a un système triple orthogonal.
+
+\Paragraph{Remarque \1.} Les surfaces parallèles à une surface~$(S)$
+en dérivent par une transformation de contact définie par
+l'équation
+\[
+(X - x)^2 + (Y - y)^2 + (Z - z)^2 - r^2 = 0\Add{,}
+\]
+où $r$~est une constante arbitraire; en effet la surface parallèle
+est l'enveloppe d'une famille de sphères de rayon
+constant ayant leurs centres sur la surface~$(S)$. Cette
+transformation de contact s'appelle dilatation.
+
+%[** TN: Roman numeral in original]
+\Paragraph{Remarque \2.} Lorsqu'on sait qu'une famille de surfaces
+appartient à un système triple orthogonal, on peut sur
+chacune de ces surfaces déterminer les lignes de courbure,
+et les autres familles du système sont engendrées par les
+trajectoires orthogonales des surfaces qui s'appuient sur
+les lignes de courbure. Dans le cas particulier d'une famille
+de surfaces parallèles, les trajectoires orthogonales
+sont les normales à ces surfaces.
+%% -----File: 283.png---Folio 275-------
+
+\Section{Systèmes triples orthogonaux contenant une famille de plans.}
+{4.}{} Considérons une famille de plans~$P$; les trajectoires
+orthogonales s'obtiennent, comme on l'a vu à propos
+des surfaces moulures, en faisant rouler un plan mobile
+sur la développable enveloppe des plans~$P$. Prenons
+dans le plan deux systèmes de courbes orthogonales, ce qui
+est toujours possible, car si nous nous donnons l'un des
+systèmes
+\[
+\phi(x,y) = a,
+\]
+l'autre se détermine par l'intégration de l'équation
+\[
+\dfrac{dx}{\ \dfrac{\dd\phi}{\dd x}\ } =
+\dfrac{dy}{\ \dfrac{\dd\phi}{\dd y}\ }.
+\]
+On peut engendrer les autres familles du système triple
+orthogonal au moyen de ces courbes des plans~$P$, assujetties
+à rencontrer les trajectoires orthogonales; ces familles
+sont ainsi constituées de surfaces moulures. On peut ainsi,
+au moyen du Théorème de Dupin, retrouver leurs lignes de
+courbure.
+
+\Section{Systèmes triples orthogonaux contenant une famille de sphères.}
+{5.}{} Le fait que toute famille de plans fait partie
+d'un système triple orthogonal tient au fond à ce que
+toute courbe d'un plan est ligne de courbure du plan; de
+sorte qu'une famille de surfaces orthogonales aux plans
+donnés satisfera à la condition \DPtypo{necessaire}{nécessaire} et suffisante
+pour qu'il existe une troisième famille complétant le système
+triple orthogonal.
+
+Le même fait sera donc vrai aussi pour une famille de
+sphères. Et pour construire un système triple orthogonal
+%% -----File: 284.png---Folio 276-------
+quelconque contenant la famille de sphères~$(S)$ donnée, il
+suffira: \Primo.~de prendre sur une des sphères deux familles
+de courbes~$(C), (C_1)$ orthogonales; \Secundo.~de déterminer les trajectoires
+orthogonales~$(T)$ des sphères~$(S)$. Car alors les
+courbes~$(T)$ qui s'appuient sur les courbes~$(C)$, et les courbes~$(T)$
+qui s'appuient sur les courbes~$(C_1)$ engendreront les
+surfaces des deux familles $(S_1)$~et~$(S_2)$ formant avec les sphères~$(S)$
+le système triple cherché.
+
+Tout revient donc à résoudre les deux problèmes suivants:
+\Primo.~déterminer sur une sphère un système orthogonal quelconque;
+\Secundo.~déterminer les trajectoires orthogonales d'une famille de sphères.
+
+Le premier problème se ramène immédiatement au problème
+analogue dans le plan au moyen d'une projection stéréographique.
+
+\DPchg{Etudions}{Étudions} donc le second:
+
+Si nous considérons \Card{2} sphères de la famille, les trajectoires
+orthogonales établissent entre elles une correspondance
+point par point, et cette correspondance d'après ce qui précède,
+sera telle qu'à un système orthogonal sur l'une des
+sphères corresponde un système orthogonal sur l'autre. Or,
+deux directions rectangulaires sont conjuguées harmoniques
+par rapport aux directions isotropes, et dans une transformation
+ponctuelle quelconque, le rapport anharmonique des tangentes
+est un invariant, donc les directions isotropes se
+transforment en directions isotropes, les génératrices rectilignes
+de l'une des sphères se transforment en génératrices
+%% -----File: 285.png---Folio 277-------
+rectilignes de l'autre; et le rapport anharmonique de \Card{2} directions
+quelconques avec les directions isotropes restant
+constant, les angles se conservent; la transformation est
+conforme.
+
+Soit alors
+\[
+\sum (x - \Err{x}{x_0})^2 - R^2 = 0\Add{,}
+\]
+l'équation générale des sphères considérées, dépendant d'un
+paramètre~$t$\DPtypo{,}{.}
+
+Les considérations précédentes nous conduisent à introduire
+les génératrices rectilignes. On posera donc:
+\begin{align*}
+x - x_0 + i(y - y_0) &= \lambda \bigl[(z - z_0) + R\bigr], \\
+x - x_0 - i(y - y_0) &= \frac{-1}{\lambda} \bigl[(z - z_0) - R\bigr]
+  = \mu \bigl[z - z_0 + R\bigr];
+\end{align*}
+d'où:
+\begin{gather*}%[** TN: Re-arranged]
+(z - z_0)\left(\mu + \frac{1}{\lambda}\right)
+  = R \left(\frac{1}{\lambda} - \mu\right),\qquad
+z - z_0 = R\, \frac{1 - \lambda \mu}{1 + \lambda \mu}, \\
+\begin{aligned}
+x - x_0 + i(y - y_0) &= \frac{2R \lambda}{1 + \lambda \mu}, \\
+x - x_0 - i(y - y_0) &= \frac{2R \mu}{1 + \lambda \mu}.
+\end{aligned}
+\end{gather*}
+Les équations \DPtypo{différentièllés}{différentielles} à des trajectoires orthogonales
+sont:
+\[
+\frac{dx}{x - x_0} =
+\frac{dy}{y - y_0} =
+\frac{dz}{z - z_0} =
+\frac{d(x + iy)}{x - x_0 + i(y - y_0)} =
+\frac{d(x - iy)}{x - x_0 - i(y - y_0)}\Add{.}
+\]
+\DPtypo{en}{En} égalant le \Ord{3}{e} rapport successivement aux \Card{2} derniers, et
+posant:
+\[
+dA = \frac{d(x_0 + i y_0)}{2R}, \qquad
+dB = \frac{d(x_0 - i y_0)}{2R}, \qquad
+dC = \frac{dz_0}{2\Err{R_0}{R}},
+\]
+\DPtypo{On}{on} obtient \Card{2} équations de Riccati
+\[
+\frac{d\lambda}{dt}
+  = \lambda^2 \frac{dB}{dt} - 2\lambda \frac{dC}{dt} - \frac{dA}{dt},
+\qquad
+\frac{d \mu}{dt}
+  = \mu^2 \frac{dA}{dt} - 2\mu \frac{dC}{dt} - \frac{dB}{dt}.
+\]
+Si on \DPchg{connait}{connaît} une trajectoire orthogonale, on a une solution
+%% -----File: 286.png---Folio 278-------
+de chaque équation, et la résolution du problème est ramenée
+à \Card{2} quadratures. Pour \Card{2} trajectoires orthogonales, on n'a
+plus qu'une seule quadrature, et pour \Card{3} trajectoires orthogonales,
+le problème s'achève sans quadrature. On a alors,
+comme intégrale générale de la première équation:
+\[
+\frac{\lambda - \lambda_1}{\lambda - \lambda_2} :
+\frac{\lambda_3 - \lambda_1}{\lambda_3 - \lambda_2} =
+\frac{\lambda^0 - \lambda^0_1}{\lambda^0 - \lambda^0_2} :
+\frac{\lambda^0_3 - \lambda^0_1}{\lambda^0_3 - \lambda^0_\DPtypo{1}{2}},
+\]
+en désignant par l'indice zéro les valeurs qui correspondent
+à~$t = t_0$. C'est donc une relation de la forme:
+\[
+\lambda = \frac{M \lambda^0 + N}{P \lambda^0 + Q}.
+\]
+On aura de même, pour la seconde équation de Riccati, une
+intégrale de la forme
+\[
+\mu = \frac{R \mu^0 + S}{T \mu^0 + U};
+\]
+et ces deux formules définissent la correspondance établie par
+les trajectoires orthogonales entre la sphère qui correspond
+à la valeur~$t_0$ du paramètre, et la sphère qui correspond à la
+valeur~$t$ du paramètre.
+
+On \DPchg{reconnait}{reconnaît} alors que cette transformation change les
+cercles d'une des sphères en cercles de l'autre; et par projection
+stéréographique elle deviendrait une des transformations
+planes du groupe des \Err{}{des transformations par }rayons vecteurs réciproques.
+
+\Section{Systèmes triples orthogonaux particuliers.}
+{6.}{} Rappelons comme systèmes triples orthogonaux
+particuliers, le système des quadriques homofocales
+\[
+\frac{x^2}{a - \lambda} +
+\frac{y^2}{b - \lambda} +
+\frac{z^2}{c - \lambda} - 1 = 0;
+\]
+et le système des cyclides du quatrième degré homofocales
+\[
+\frac{x^2}{a - \lambda} +
+\frac{y^2}{b - \lambda} +
+\frac{z^2}{c - \lambda} +
+\frac{(x^2 + y^2 + z^2 - R^2)^2}{4R^2 (d - \lambda)} -
+\frac{(x^2 + y^2 + z^2 + R^2)\Err{}{^2}}{4R^2 (e - \lambda)} = 0.
+\]
+%% -----File: 287.png---Folio 279-------
+On vérifie qu'on obtient un autre système, formé de cyclides
+de Dupin du troisième degré, en considérant les surfaces
+lieux des points de contact des plans tangents menés, par un
+point d'un des axes, à une famille de quadriques homofocales.
+
+
+\ExSection{XII}
+
+\begin{Exercises}
+\item[53.] On considère une famille de $\infty^{1}$~\DPchg{paraboloides}{paraboloïdes}~$P$ ayant mêmes
+plans principaux. Comment faut-il choisir ces \DPchg{paraboloides}{paraboloïdes}
+pour que la congruence des génératrices rectilignes d'un même
+système de tous ces \DPchg{paraboloides}{paraboloïdes} soit une congruence de normales?
+Montrer qu'alors les \DPchg{paraboloides}{paraboloïdes}~$P$ constituent l'une
+des trois familles d'un système triple orthogonal et trouver
+les deux autres familles. Montrer qu'on peut choisir les \DPchg{paraboloides}{paraboloïdes}~$P$
+plus particulièrement de manière que l'une de
+ces autres familles soit encore formée de \DPchg{paraboloides}{paraboloïdes}; et
+donner, dans ce cas, la signification géométrique des deux
+familles de \DPchg{paraboloides}{paraboloïdes}.
+\end{Exercises}
+%% -----File: 288.png---Folio 280-------
+
+
+\Chapitre{XIII}{Congruences de \DPchg{Spheres}{Sphères} et \DPchg{Systemes}{Systèmes} Cycliques.}
+
+\Section{Généralités.}
+{1.}{} Nous appellerons \emph{congruence de sphères} une famille
+de $\infty^{2}$~sphères~$(\Sigma)$
+\[
+\Tag{(1)}
+\sum (x - f)^2 - r^2 = 0,
+\]
+$f,g,h,r$ étant fonctions de \Card{2} paramètres~$u,v$. Le lieu des centres
+de ces sphères est une surface~$(S)$
+\[
+x = f(u,v), \qquad
+y = g(u,v), \qquad
+r = h(u,v).
+\]
+Cherchons l'enveloppe de ces sphères. A l'équation~\Eq{(1)} nous
+devrons adjoindre les \Card{2} équations
+\[
+\Tag{(2)}
+\sum (x - f) \frac{\dd f}{\dd u} + r \frac{\dd r}{\dd u} = 0, \qquad
+\sum (x - f) \frac{\dd f}{\dd v} + r \frac{\dd r}{\dd v} = 0.
+\]
+Ces équations~\Eq{(2)} représentent une droite, donc l'enveloppe
+des sphères~$(\Sigma)$ touche chacune de ces sphères en \Card{2} points,
+que l'on appelle \emph{points focaux}; l'enveloppe, que l'on appellera
+\emph{surface focale}, se \DPtypo{decompose}{décompose} donc en \Card{2} nappes~$(F_1), (F_2)$.
+
+Considérons dans la congruence~\Eq{(1)} une famille de $\infty^{1}$~sphères~$(\Sigma)$;
+il suffit de définir~$u, v$ en fonction d'un paramètre~$t$;
+ces sphères admettent une enveloppe, qui touche chacune
+d'elles le long d'un cercle \DPtypo{caracteristique}{caractéristique} ayant pour plan
+\[
+\Tag{(3)}
+\sum (x - f)\, df + r\, dr = 0;
+\]
+lorsque les expressions de~$u, v$ en fonction de~$t$ varient, tous
+ces cercles caractéristiques passent par \Card{2} points fixes, qui
+sont les points focaux de la sphère considérée. Les enveloppes
+%% -----File: 289.png---Folio 281-------
+ainsi obtenues correspondent aux surfaces réglées des congruences
+de droites; on peut les appeler \emph{surfaces canaux} de
+la congruence~\Eq{(1)}.
+
+Cherchons parmi ces surfaces canaux celles pour lesquelles
+chaque sphère est tangente à la sphère infiniment
+voisine. Ce sont en réalité des surfaces réglées à génératrices
+isotropes. Le cercle défini par les équations \Eq{(1)},~\Eq{(3)}
+doit se réduire à un couple de droites isotropes; le plan~\Eq{(3)}
+est tangent à la sphère~\Eq{(1)}. Ce qui donne la condition
+\[
+\Tag{(4)}
+\sum df^2 - dr^2 = 0,
+\]
+équation différentielle du \Ord{1}{re} ordre et du \Ord{2}{e} degré; il y a
+donc \Card{2} familles de sphères spéciales, le point de contact de
+l'une d'elles avec la sphère infiniment voisine étant défini
+par
+\[
+\Tag{(5)}
+\frac{x - f}{df} = \frac{y - g}{dg} = \frac{z - h}{dh}
+  = \frac{-r}{dr}\Add{,}
+\]
+$df, dg, dh$~sont les coefficients de direction du rayon du point
+de contact.
+
+\Illustration[2.75in]{289a}
+L'équation~\Eq{(4)} définit sur la surface~$(S)$ \Card{2} directions $\omega l, \omega l'$; soient $M, M'$~les
+points de contact de la
+sphère~$(\Sigma)$ correspondante
+avec la surface focale~$(F)$;
+la droite~$MM'$ est représentée
+par les \Card{2} équations~\Eq{(2)},
+ou encore, puisque les
+points~$M, M'$ sont sur
+tous les cercles caractéristiques, par les \Card{2} équations~\Eq{(3)}.
+%% -----File: 290.png---Folio 282-------
+qui correspondent aux enveloppes spéciales; or\Add{,} dans ce cas
+l'équation~\Eq{(3)} représente le plan tangent à la sphère en l'un
+des points~$I, I'$; donc les droites~$II', MM'$ sont polaires réciproques
+par rapport à la sphère~$(\Sigma)$. Si nous supposons cette
+sphère réelle, et si les points~$M, M'$ sont réels, $I,I'$~sont
+imaginaires; et inversement. Nous désignerons par~$D$ la droite~$MM'$,
+et par~$\Delta$ la droite~$ll'$; $\omega l, \omega l'$~sont dans le plan tangent
+en~$\omega$ à la surface~$(S)$; $MM'$~est perpendiculaire à ce plan tangent;
+les points~$M, M'$, et par suite les droites~$\omega M, \omega M'$ sont
+symétriques par rapport à ce plan tangent.
+
+Si nous remarquons maintenant que $\omega M$~est normale à la
+\Ord{1}{ère} nappe de la surface focale, et $\omega M'$~normale à la \Ord{2}{e}, nous
+voyons que l'on peut considérer~$\omega M$ comme rayon incident, $\omega M'$~comme
+rayon réfléchi sur la surface~$(S)$, et par suite, nous
+avons une congruence de normales qui se réfléchit sur la surface~$(S)$
+suivant une congruence de normales. La surface~$(S)$
+étant quelconque, donnons-nous une surface~$(F_1)$; nous pourrons
+toujours considérer les sphères ayant leurs centres sur~$(S)$
+et tangentes à~$(F_1)$; $(F_1)$~sera l'une des nappes focales de
+la congruence de sphères ainsi obtenues, et la congruence des
+normales à~$(F_1)$ se réfléchira sur~$(S)$ suivant la congruence
+des normales à~$(F_2)$ \Ord{2}{e} nappe focale. D'où le \emph{Théorème de
+Malus: Les rayons normaux à une surface quelconque se réfléchissent
+sur une surface quelconque suivant les normales à
+une nouvelle surface}.
+
+\Illustration[2.25in]{291a}
+Ce Théorème peut s'étendre aux rayons réfractés. Reprenons
+%% -----File: 291.png---Folio 283-------
+la construction d'Huygens. Considérons une sphère de
+centre~$\omega$; soit $\omega M$~le rayon incident,
+et la normale~$\omega N$ à la surface \DPtypo{réflechissante}{réfléchissante}.
+Construisons une \Ord{2}{e} sphère de centre~$\omega$ et dont le rayon
+soit dans le rapport~$n$ avec le rayon
+de la \Ord{1}{ère}. Considérons le plan tangent~$\omega T$
+à la surface réfléchissante.
+Au point~$M$ où le rayon incident rencontre la \Ord{1}{ère} sphère menons
+le plan tangent à cette sphère, qui rencontre le plan~$\omega l$
+suivant la droite~$T$, et par la droite~$T$ menons le plan~$TP$
+tangent à la \Ord{2}{e} sphère. En appelant~$i\Add{,} i'$ les angles de~$\omega M$
+et~$\omega P$ avec~$\omega N$, on a immédiatement
+\[
+\omega T = \frac{\omega M}{\sin i} = \frac{\omega P}{\sin i'}\Add{,}
+\]
+d'où
+\[
+\frac{\sin i'}{\sin i} = \frac{\omega P}{\omega M} = n\Add{,}
+\]
+$\omega P$~est le rayon réfracté. Partons alors d'une congruence de
+normales, soit $(F_1)$~la surface normale; pour construire les
+rayons réfractés, il faut considérer les sphères~$(\Sigma')$ concentriques
+à~$(\Sigma)$ et de rayon~$nr$; les points~$I, I'$ sont définis
+par les équations~\Eq{(5)}; or\Add{,} ces équations ne changent pas lorsqu'on
+remplace~$r$ par~$nr$, la droite~$II'$ est la même que précédemment,
+et le Théorème s'étend à la réfraction.
+
+\Section{Congruences spéciales.}
+{2.}{} A la congruence de sphères considérée nous avons
+associé \Card{4} congruences de droites: celle des droites~$\omega M$ normales
+à~$(F_1)$, celles des droites~$\omega M'$ normales à~$(F_2)$~celle des
+%% -----File: 292.png---Folio 284-------
+droites~$\Delta$, et celle des droites~$D$.
+
+Supposons~$MM'$~confondus; ils sont confondus aussi avec~$II'$;
+les \Card{2} nappes focales sont confondues; alors le lieu des
+points~$I\Add{,} I'$ confondus est une ligne de courbure de la surface~$(F)$
+et la sphère~$(\Sigma)$ est la sphère de courbure correspondante.
+\emph{La congruence de sphères est alors constituée par des sphères
+de courbure d'une même surface~$(F)$.}
+
+\emph{Réciproquement}. Soit une surface~$(F)$ et ses sphères de
+courbure d'une même famille, la surface~$(F)$ est surface focale
+double de la congruence de ces sphères de courbure.
+
+Toutes les congruences de droites considérées se réduisent
+ici à~$2$, celle des droites~$D$ tangentes à une famille de
+lignes de courbure, et celle des droites~$\Delta$~tangentes à l'autre
+famille. La surface~$(S)$ est alors l'une des nappes de
+la développée de la surface focale double. Aux lignes de
+courbure intégrales~de~\Err{\Eq{(H)}}{\Eq{(4)}} correspond sur la surface~$(S)$ une
+famille de géodésiques. On est ainsi conduit à la détermination
+des géodésiques de~$(S)$ en écrivant que l'équation~\Eq{(4)} a
+une racine double en~$du, dv$. Cette équation s'écrit
+\begin{gather*}
+E\, du^2 + 2F\, du\, dv + G\, dv^2
+  - \left(\frac{\dd r}{\dd u}\, du
+        + \frac{\dd r}{\dd v}\, dv \right)^2 = 0\Add{,} \\
+\left[ E - \left(\frac{\dd r}{\dd u}\right)^2 \right] du^2
+  + 2 \Biggl[ F - \frac{\dd r}{\dd u}\, \frac{\dd r}{\dd v} \Biggr] du\, dv
+  + \left[ G - \left(\frac{\dd r}{\dd v}\right)^2 \right] dv^2 = 0\Add{,}
+\end{gather*}
+pour qu'il y ait une racine double, il faut que
+\[
+\left[ E - \left(\frac{\dd r}{\dd u}\right)^2 \right]
+\left[ G - \left(\frac{\dd r}{\dd v}\right)^2 \right]
+  - \Biggl[ F - \frac{\dd r}{\dd u}\, \frac{\dd r}{\dd v} \Biggr]^2 = 0\Add{,}
+\]
+ou
+\[
+H^2 - \left[ E \left(\frac{\dd r}{\dd v}\right)^2
+  - 2F \frac{\dd r}{\dd u}\, \frac{\dd r}{\dd v}
+  + G \left(\frac{\dd r}{\dd u}\right)^2 \right] = 0\Add{,}
+\]
+équation aux dérivées partielles qui détermine~$r$. Ayant~$r$ on
+obtient la famille de géodésiques correspondante par l'intégration
+%% -----File: 293.png---Folio 285-------
+d'une équation différentielle ordinaire.
+
+\Section{Théorème de Dupin.}
+{3.}{} Supposons que la surface focale ait ses \Card{2} nappes
+distinctes, et cherchons ses lignes de courbure. Soient~$\lambda\Add{,} \mu\Add{,} \nu$
+les cosinus directeurs de~$\omega M$.
+\[
+\lambda = \frac{x - f}{r}, \qquad
+\mu = \frac{y - g}{r}, \qquad
+\nu = \frac{z - h}{r};
+\]
+d'où
+\[
+x = f + \lambda r, \qquad
+y = g + \mu r, \qquad
+z = h + \nu r.
+\]
+Portons ces valeurs de~$x,y,z$ dans les équations~\Eq{(2)}, elles
+deviennent
+\[
+\Tag{(6)}
+\sum \lambda \frac{\dd f}{\dd u} + \frac{\dd r}{\dd u} = 0, \qquad
+\sum \lambda \frac{\dd f}{\dd v} + \frac{\dd r}{\dd v} = 0.
+\]
+Soient~$i, i'$ les angles de $\omega M$~et~$\omega M'$ avec~$\omega N$, ces angles sont
+supplémentaires, $\cos i' = -\cos i$; si $l,m,n$~sont les cosinus
+directeurs de~$\omega N$\DPtypo{.}{,} \DPtypo{On}{on} a
+\[
+\Tag{(7)}
+\sum \lambda l - \cos i = 0.
+\]
+Calculons~$\cos i$. Dans le plan tangent à~$(S)$ soient $\omega U, \omega V$ les
+tangentes aux courbes $v = \cte$, $u = \cte[]$.
+
+\Illustration{293a}
+Les cosinus directeurs de~$\omega U$ sont
+\[
+\frac{1}{\sqrt{E}}\, \frac{\dd f}{\dd u}, \qquad
+\frac{1}{\sqrt{E}}\, \frac{\dd g}{\dd u}, \qquad
+\frac{1}{\sqrt{E}}\, \frac{\dd h}{\dd u};
+\]
+\DPtypo{Ceux}{ceux} de~$\omega V$ sont
+\[
+\frac{1}{\sqrt{G}}\, \frac{\dd f}{\dd v}, \qquad
+\frac{1}{\sqrt{G}}\, \frac{\dd g}{\dd v}, \qquad
+\frac{1}{\sqrt{G}}\, \frac{\dd h}{\dd v}.
+\]
+Soit $\omega \delta$~le segment directeur de~$\omega M$,
+ses projections orthogonales sur~$\omega U, \omega V$ sont:
+\[
+A = -\frac{1}{\sqrt{E}}\, \frac{\dd r}{\dd u}, \qquad
+B = -\frac{1}{\sqrt{G}}\, \frac{\dd r}{\dd v},
+\]
+sa projection sur~$\omega W$ est~$\cos i$. Soit $\theta$~l'angle de~$\omega U$~et~$\omega V$:
+\[
+\Cos \theta = \frac{F}{\sqrt{EG}}.
+\]
+%% -----File: 294.png---Folio 286-------
+\DPtypo{soient}{Soient} $U, V$~les coordonnées par rapport à~$\omega U, \omega V$ de la projection~$\delta'$
+de~$\delta$ sur le plan tangent; elles sont données par les
+équations
+\[
+U \cos\theta + V = B, \qquad
+U + V \cos\theta = A;
+\]
+d'où
+\[
+U \sin^2\theta = A - B \cos\theta, \qquad
+V \sin^2\theta = B - A \cos\theta;
+\]
+d'où encore
+\begin{multline*}
+\sin^{\Err{2}{4}}\theta \bigl[U^2 + V^2 + 2UV \cos\theta\bigr] \\
+  = (A - B \cos\theta)^2 + (B - A \cos\theta)^2
+  + 2 \cos\theta (A - B \cos\theta) (B - A \cos\theta)\DPtypo{.}{,}
+\end{multline*}
+ou
+\[
+\sin^2\theta \bigl[U^2 + V^2 + 2UV \cos\theta\bigr]
+  = A^2 - 2AB \cos\theta + B^2.
+\]
+Donc
+\[
+\omega \delta'{}^2 = U^2 + V^2 + 2UV \cos\theta
+  = \frac{1}{\sin^2\theta} \bigl[A^2 - 2AB \cos\theta + B^2\bigr],
+\]
+et
+\[
+1 = \overline{\omega\delta}^2
+  = \overline{\omega\delta'}^2 + \cos^2 i
+  = \cos^2 i + \frac{1}{\sin^2\theta} \bigl[A^2 - 2AB \cos\theta + B^2\bigr].
+\]
+Or,
+\[
+A^2 - 2AB \cos\theta + B^2
+  = \frac{1}{H^2}\, \Phi\left( \frac{\dd r}{\dd v}, -\frac{\dd r}{\dd u}\right),
+\]
+d'où la formule
+\[
+\sin^2 i = \frac{1}{H^2}\,
+  \Phi\left(\frac{\dd r}{\dd v}, -\frac{\dd r}{\dd u}\right).
+\]
+
+Ceci posé, les lignes de courbure de la surface focale
+sont définies par l'équation
+\[
+\begin{vmatrix}
+  dx & \lambda & d\lambda
+\end{vmatrix} = 0,
+\]
+ou
+\[
+\begin{vmatrix}
+  df + \lambda · dr + r · d\lambda & \lambda & d\lambda
+\end{vmatrix} = 0;
+\]
+qui se réduit à
+\[
+\begin{vmatrix}
+  df & \lambda & d\lambda
+\end{vmatrix} = 0.
+\]
+\DPtypo{multiplions}{Multiplions} par le déterminant
+$\begin{vmatrix}
+  \lambda & \dfrac{\dd f}{\dd u} & \dfrac{\dd f}{\dd v}
+\end{vmatrix}$
+qui n'est
+pas nul, la normale n'étant pas dans le plan tangent. L'équation
+devient
+\[
+\left\lvert
+\begin{array}{@{}lll@{}}
+\sum \lambda\,df & \sum \lambda^2 & \sum \lambda\, d\lambda \\
+\sum \mfrac{\dd f}{\dd u}\, df    & \sum \lambda \mfrac{\dd f}{\dd u} & \sum \mfrac{\dd f}{\dd u}\, d\lambda \\
+\sum \mfrac{\dd f}{\dd v}\, df    & \sum \lambda \mfrac{\dd f}{\dd v} & \sum \mfrac{\dd f}{\dd v}\, d\lambda
+\end{array}
+\right\rvert
+= 0,
+\]
+%% -----File: 295.png---Folio 287-------
+ou, en tenant compte de~\Eq{(6)}
+\[
+\begin{vmatrix}
+  -dr & 1 & 0  \\
+  \sum \mfrac{\dd f}{\dd u}\, df & -\mfrac{\dd r}{\dd u} & \sum \mfrac{\dd f}{\dd u}\, d\lambda \\
+  \sum \mfrac{\dd f}{\dd v}\, df & -\mfrac{\dd r}{\dd v} & \sum \mfrac{\dd f}{\dd v}\, d\lambda
+\end{vmatrix}
+= 0.
+\]
+Multiplions la \Ord{1}{ère} ligne par~$\dfrac{\dd r}{\dd u}$ et ajoutons à la \Ord{2}{e}, puis
+par~$\dfrac{\dd r}{\dd v}$ et ajoutons à la \Ord{3}{e}. Nous obtenons
+\[
+\begin{vmatrix}
+\sum \mfrac{\dd f}{\dd u}\, df - \mfrac{\dd r}{\dd u}\, dr & \sum \mfrac{\dd f}{\dd u}\, d\lambda \\
+\sum \mfrac{\dd f}{\dd v}\, df - \mfrac{\dd r}{\dd v}\, dr & \sum \mfrac{\dd f}{\dd v}\, d\lambda
+\end{vmatrix}
+=0.
+\]
+\DPtypo{les}{Les} éléments de la \Ord{1}{ère} colonne sont les demi-dérivées partielles
+par rapport à~$du\Add{,} dv$ de la forme quadratique
+\[
+\sum df^2 - dr^2 = \Phi_1 (du, dv),
+\]
+qui définit le couple des droites~$\omega I, \omega I'$. Voyons si les éléments
+de la \Ord{2}{e} colonne sont susceptibles d'une interprétation
+analogue. Si nous \DPtypo{differentions}{différentions}~\Eq{(6)} nous obtenons
+\[
+\sum \frac{\dd f}{\dd u}\, d\lambda
+  = -\sum \lambda\, d\left(\frac{\dd f}{\dd u}\right)
+                  - d\left(\frac{\dd r}{\dd u}\right);
+\]
+or,
+\[
+d\left(\frac{\dd r}{\dd u}\right)
+  = \Err{-}{}\frac{1}{2} · \frac{\dd(d^2r)}{\dd(du)},
+\]
+et
+\[
+\sum \lambda · d\left(\frac{\dd f}{\dd u}\right)
+  = \frac{1}{2}\, \frac{\dd(\sum \lambda · d^2f)}{\dd (du)}.
+\]
+Posons
+\[
+\scrA = \sum \lambda\, d^2 f, \qquad
+\scrB = \scrA + d^2r
+\]
+les différentielles secondes étant prises par rapport
+à~$u$~et~$v$;  et l'équation s'écrit,
+\[
+\begin{vmatrix}
+\mfrac{\dd \Phi_1}{\dd\, du} & \mfrac{\dd \scrB}{\dd\, du} \\
+\mfrac{\dd \Phi_1}{\dd\, dv} & \mfrac{\dd \scrB}{\dd\, dv}
+\end{vmatrix}
+= 0:
+\]
+les directions principales de~$(F)$ sont conjuguées harmoniques
+par rapport aux \Card{2} couples $\Phi_1 = 0$~et~$\scrB = 0$. Calculons~$\scrA$. Pour
+%% -----File: 296.png---Folio 288-------
+cela, éliminons $\lambda\Add{,} \mu\Add{,} \nu$ entre les équations \Eq{(6)}\Add{,}~\Eq{(7)} et
+\[
+\sum \lambda\, d^2 f - \scrA = 0;
+\]
+nous obtenons
+\[
+\begin{vmatrix}
+\mfrac{\dd f}{\dd u} & \mfrac{\dd f}{\dd v} & 1 & d^2 f \\
+\dots & \dots & \dots & \dots \\
+\dots & \dots & \dots & \dots \\
+\mfrac{\dd r}{\dd u} & \mfrac{\dd r}{\dd v} & -\cos i & -\scrA
+\end{vmatrix}
+= 0,
+\]
+ce qui donne en développant
+\[
+\scrA H + H \cos i \Psi(du, dv) + H \chi(du, dv) = 0;
+\]
+et:
+\[
+\scrB = d^2 r - \cos i \Psi(du, dv) - \chi(du, dv)
+  = -\Psi_{1}(du, dv) - \cos i \Psi(du, dv).
+\]
+
+Les lignes de courbure de la \Ord{2}{e} nappe sont tangentes aux
+directions conjuguées par rapport à $\Phi_{1} = 0$ et
+\[
+\scrB_{1} = -\Psi_{1}(du, dv) + \Psi(du, dv) = 0;
+\]
+pour que les lignes de courbure se correspondent sur les \Card{2} nappes,
+il faut et il suffit que le couple des directions
+principales soit conjugué par rapport aux \Card{3} couples
+\[
+\Phi_{1} = 0,\qquad
+\Psi_{1} - \cos i \Psi = 0,\qquad
+\Psi_{1} + \cos i \Psi = 0,
+\]
+ou par rapport aux couples
+\[
+\Phi_{1} = 0,\qquad
+\Psi_1 \Err{+ \cos i \Psi}{} = 0,\qquad
+\Psi = 0.
+\]
+\DPchg{Etant}{Étant} conjuguées par rapport à $\Psi = 0$, elles correspondent à
+des lignes conjuguées sur~$(S)$, d'où le \emph{Théorème de Dupin}: si
+les lignes de courbure se correspondent sur les \Card{2} nappes focales,
+les développables des normales correspondantes coupent
+la surface~$(S)$ suivant le même réseau conjugué; et \emph{réciproquement}.
+Donc \emph{la condition nécessaire et suffisante pour que
+les développables d'une congruence de normales se réfléchissent
+sur une surface suivant des développables est qu'elles
+%% -----File: 297.png---Folio 289-------
+déterminent sur la surface un réseau conjugué}.
+
+\Section{Congruence des droites \texorpdfstring{$D$}{D}.}
+{4.}{} Cherchons les développables de la congruence des
+droites~$D$; elles sont définies par l'équation
+\[
+\begin{vmatrix}
+dx & dy & dz \\
+ l &  m &  n \\
+dl & dm & dn
+\end{vmatrix}
+= 0;
+\]
+or
+\[
+x = f + r \lambda,\qquad
+y = g + r \mu,\qquad
+z = h + r \nu,
+\]
+et l'équation devient
+\[
+\begin{vmatrix}
+df + r\, d\lambda + \lambda\, dr & l & dl
+\end{vmatrix} = 0.
+\]
+Multiplions le \Ord{1}{er} membre par le déterminant non nul
+\[
+\begin{vmatrix}
+l & \mfrac{\dd f}{\dd u} & \mfrac{\dd f}{\dd v}
+\end{vmatrix}
+\]
+nous avons:
+\[
+\begin{vmatrix}
+r \sum l\, d \lambda + dr \sum \lambda l & 1 & 0 \\
+\sum \mfrac{\dd f}{\dd u}\, df + r \sum \mfrac{\dd f}{\dd u}\, d\lambda
+  + dr \sum \lambda \mfrac{\dd f}{\dd u} & 0 & \sum \mfrac{\dd f}{\dd u}\, dl \\
+\sum \mfrac{\dd f}{\dd v}\, df + r \sum \mfrac{\dd f}{\dd v}\, d\lambda
+  + dr \sum \lambda \mfrac{\dd f}{\dd v} & 0 & \sum \mfrac{\dd f}{\dd v}\, dl
+\end{vmatrix}
+= 0,
+\]
+ou
+\[
+\begin{vmatrix}
+\sum \mfrac{\dd f}{\dd u}\, df + r \sum \mfrac{\dd f}{\dd u}\, d\lambda
+  + dr \sum \lambda \mfrac{\dd f}{\dd u} & \sum \mfrac{\dd f}{\dd u}\, dl \\
+\sum \mfrac{\dd f}{\dd v}\, df + r \sum \mfrac{\dd f}{\dd v}\, d\lambda
+  + dr \sum \lambda \mfrac{\dd f}{\dd v} & \sum \mfrac{\dd f}{\dd v}\, dl
+\end{vmatrix}
+= 0.
+\]
+
+Les éléments de la dernière colonne sont les demi-dérivées
+partielles par rapport à~$du, dv$ de la forme $\Psi(du, dv) = \Err{}{-}\sum df · dl$.
+Quant aux éléments de la \Ord{1}{ère} ligne, remarquons que
+\[
+\sum \frac{\dd f}{\dd u}\, d\lambda = \frac{1}{2} · \frac{\dd \scrB}{\dd u},
+\qquad
+\sum \frac{\dd f}{\dd v}\, d\lambda = \frac{1}{2} · \frac{\dd \scrB}{\dd v},
+\]
+où
+\[
+\scrB = -\Psi_{1} - \cos i · \Psi.
+\]
+Enfin, si nous remarquons que les points $M, M'$ sont définis par
+les relations:
+\[
+\sum \lambda\, \frac{\dd f}{\dd u} + \frac{\dd r}{\dd u} = 0,\qquad
+\sum \lambda\, \frac{\dd f}{\dd v} + \frac{\dd r}{\dd v} = 0,
+\]
+%% -----File: 298.png---Folio 290-------
+nous avons:
+\[
+\sum \frac{\dd f}{\dd u}\, df
+  + dr \sum \lambda\, \frac{\dd f}{\dd u}
+  = \sum \frac{\dd f}{\dd u}\, df - \frac{\dd r}{\dd u}\, dr
+  = \frac{1}{2}\, \frac{d\Phi_{1}}{\dd\, du},\quad
+\text{et l'analogue};
+\]
+de sorte que les éléments de la \Ord{1}{ère} colonne sont les demi-dérivées
+partielles par rapport à~$du, dv$ de la forme $\Phi_{1} - r(\Psi_{1} + \Psi \cos i)$.
+
+\emph{Les développables de la congruence correspondent
+sur la surface~$(S)$ aux directions conjuguées par rapport
+à}
+\begin{alignat*}{2}
+\Psi &= 0,\qquad && \Phi_{1} - r [\Psi_{1} + \Psi \cos i] = 0, \\
+\intertext{ou par rapport à}
+\Psi &= 0, && \Phi_{1} - r \Psi_{1} = 0;
+\end{alignat*}
+le résultat ne change pas si on change $i$~en~$\pi - i$, et les
+développables de la congruence des droites~$D$ correspondent
+sur la surface~$(S)$ à un réseau conjugué.
+
+Considérons les plans focaux; un plan focal est parallèle
+à la direction $l, m, n$, et à la direction $dl, dm, dn$. Mais
+\[
+l^2 + m^2 + n^2 = 1,
+\]
+d'où
+\[
+l\, dl + m\, dm + n\, dn = 0;
+\]
+$dl, dm, dn$ correspondent à la direction du plan focal parallèle
+au plan tangent à la surface. Or\Add{,} les \Card{2} directions correspondant
+aux \Card{2} plans focaux, donc aux \Card{2} développables, étant
+conjuguées, on a
+\[
+\sum dl · \delta f = 0;
+\]
+le plan focal est perpendiculaire à la direction $\delta f, \delta g, \delta h$ qui
+correspond à l'autre plan focal. \emph{Chaque plan focal est perpendiculaire
+à la direction de la surface~$(S)$ correspondant à
+l'une des développables.}
+%% -----File: 299.png---Folio 291-------
+
+\Section{Congruence des droites \texorpdfstring{$\Delta$}{Delta}.}
+{5.}{} La droite~$\Delta$ est l'intersection des plans tangents à la sphère en~$M$ et à la surface~$(S)$ en~$\omega$
+\[
+\sum \lambda\,(X - f) - r = 0,  \qquad \sum l\,(X - f) = 0.
+\]
+Cherchons les développables. Exprimons que la droite précédente
+rencontre la droite infiniment voisine. Cela donne
+\[
+\sum d\lambda\, (X - f) - \sum \lambda\, df - dr = 0, \qquad
+\sum dl\, (X - f) - \sum l\, df = 0;
+\]
+conditions qui se simplifient en remarquant que $\sum l\, df = 0$, et
+$\sum \lambda\, df + dr = 0$. Il reste:
+\[
+\Tag{(1)}
+\sum d\lambda (X - f) = 0, \qquad
+\sum dl (X - f) = 0.
+\]
+Exprimons que les équations obtenues sont compatibles, nous
+avons l'équation qui définit les développables
+\[
+\Tag{(2)}
+\begin{vmatrix}
+  l & d\lambda & dl
+\end{vmatrix} = 0.
+\]
+Multiplions encore par le déterminant non nul
+\[
+\begin{vmatrix}
+  l & \dfrac{\dd f}{\dd u} & \dfrac{\dd f}{\dd v}
+\end{vmatrix},
+\]
+nous obtenons
+\[
+\begin{vmatrix}
+1 & \sum l\, d\lambda   &   0 \\
+0 & \sum d\lambda\, \mfrac{\dd f}{\dd u} & \sum dl\, \mfrac{\dd f}{\dd u} \\
+0 & \sum d\lambda\, \mfrac{\dd f}{\dd v} & \sum dl\, \mfrac{\dd f}{\dd v}
+\end{vmatrix}
+= 0,
+\]
+ou
+\[
+\begin{vmatrix}
+\sum \mfrac{\dd f}{\dd u}\, d\lambda & \sum \mfrac{\dd f}{\dd u}\, dl \\
+\sum \mfrac{\dd f}{\dd v}\, d\lambda & \sum \mfrac{\dd f}{\dd v}\, dl
+\end{vmatrix}
+= 0;
+\]
+les éléments de la \Ord{1}{ère} colonne sont les demi-dérivées partielles
+par rapport à~$du, dv$ de la forme~$\scrB = -\Psi_1 - \Psi \cos i$.
+Ceux de la \Ord{2}{e} colonne sont les demi-dérivées partielles de~$\Psi$.
+\emph{Les développables de la congruence des droites~$\Delta$ correspondent
+sur la surface~$(S)$ au réseau conjugué par rapport aux
+couples}
+\[
+\Psi = 0, \qquad \Psi_1 = 0.
+\]
+%% -----File: 300.png---Folio 292-------
+
+Quant aux points focaux, ils sont définis par les équations
+de~$\Delta$ et les équations~\Eq{(1)}, compatibles en vertu de la
+relation~\Eq{(2)}. On en déduit que les directions joignant~$\omega$ aux
+points focaux sont définies par les relations
+\[
+\sum  l · \delta f = 0,\qquad
+\sum dl · \delta f = 0,\qquad
+\sum d\lambda · \delta f = 0;
+\]
+la \Ord{1}{ère} exprime que ces droites sont dans le plan tangent à~$(S)$,
+la \Ord{2}{e} que ce sont les tangentes conjuguées de~$(S)$ qui
+correspondent aux développables.
+
+\Illustration{300a}
+Supposons que les \Card{2} congruences précédentes se correspondent
+par développables. Les \Card{2} réseaux conjugués déterminés
+sur la surface~$(S)$ sont confondus; il faut alors que les \Card{3} couples
+$\Psi = 0$, $\Psi_{1} = 0$, $\Phi_{1} - r \Psi_{1} = 0$, ou $\Psi = 0$, $\Psi_{1} = 0$, $\Phi_{1} = 0$ appartiennent
+à une même involution, et alors les lignes de
+courbure se correspondent sur les \Card{2} nappes de la surface~$(F)$.
+Nous avons alors sur la surface~$(S)$ un réseau conjugué qui
+correspond aux développables des \Card{4} congruences
+$\omega M, \omega M', D, \Delta$. Les points
+focaux~$F\Add{,} F'$ de~$\Delta$ sont sur les tangentes
+aux \Card{2} courbes conjuguées qui passent
+par~$\omega$, les droites $MF, MF'$ sont les
+tangentes aux lignes de courbure en~$M$
+de la surface~$(F)$, la droite~$D$ est perpendiculaire au plan~$F \omega F'$,
+et ses plans focaux sont perpendiculaires à $\omega F$~et~$\omega F'$.
+Les développables de la congruence des droites~$D$ coupent les
+\Card{2} nappes de l'enveloppe~$(F)$ suivant leurs lignes de courbure.
+%% -----File: 301.png---Folio 293-------
+
+\Section{Le système triple de \DPtypo{Ribeaucour}{Ribaucour}.}
+{6.}{} Plaçons-nous dans ce dernier cas; soit $(\gamma)$ une des
+courbes conjuguées de la surface~$(S)$; quand $\omega$~décrit~$(\gamma)$, le
+point~$M$ décrit une ligne de courbure~$(K)$ de la surface~$(F)$
+qui sera tangente à~$MF$, et la droite~$\Delta$ enveloppe une courbe~$(c)$
+lieu de~$F'$. Considérons la sphère~$(\sigma)$ de centre~$F'$ et passant
+par~$M$; cette sphère a pour enveloppe une surface canal~$(E)$;
+la sphère~$(\sigma)$ ayant son rayon~$MF'$ perpendiculaire à~$MF$
+est constamment tangente à la courbe~$(K)$, donc le point~$M$ est
+un point du cercle caractéristique~$(H)$; le plan de ce cercle
+est perpendiculaire à la droite~$\Delta$ tangente à~$(c)$, son centre~$H$
+sera le pied de la perpendiculaire abaissée de~$M$ sur~$\Delta$; ce
+cercle sera donc orthogonal à la sphère~$(\Sigma)$ au point~$M$ et au
+point~$M'$ symétrique par rapport au plan~$F \omega F'$; et la surface~$(E)$
+est engendrée par le cercle orthogonal à la sphère~$(\Sigma)$
+aux points $M, M'$; ce cercle tangent en~$M$ à~$\omega M$ reste orthogonal
+à la ligne de courbure~$(K)$, or\Add{,} il est ligne de courbure sur la surface~$(E)$,
+donc $(K)$~est aussi ligne de courbure sur la surface~$(E)$.
+Si nous faisons varier~$(K)$ nous obtenons une famille
+de surfaces~$(E)$ qui seront toutes orthogonales à $(F_{1}),(F_{2})$, et
+qui les couperont suivant leurs lignes de courbure. Si maintenant
+nous prenons sur~$(F)$ le \Ord{2}{e} système de lignes de courbure,
+nous devrons considérer les sphères de centres~$F$ et passant
+par~$M$; le cercle caractéristique sera encore le cercle~$(H)$;
+de plus, $FM$~et~$F'M$ étant perpendiculaires, les sphères $(\sigma)\Add{,} (\sigma')$ %[** TN: Erratum corrected]
+sont orthogonales, donc aussi leurs enveloppes $(E)\Add{,} (E')$. Nous
+avons donc \Card{2} familles de surfaces canaux qui se coupent
+orthogonalement suivant des lignes de courbure, les cercles~$(H)$;
+%% -----File: 302.png---Folio 294-------
+donc elles appartiennent à un système triple orthogonal;
+autrement dit les cercles~$(H)$ sont orthogonaux à une famille
+de surfaces, à laquelle appartiennent $(F), (F')$; et ils établissent
+une correspondance entre les points~$M, M'$, donc entre
+les lignes de courbure de ces surfaces. Donc \emph{lorsque les cercles~$(H)$
+d'une congruence sont orthogonaux aux \Card{2} nappes focales~$(F_1)\Add{,} (F_2)$
+d'une congruence de sphères, et s'ils établissent
+une correspondance entre les lignes de courbure de ces \Card{2} nappes,
+ils sont orthogonaux à une infinité de surfaces sur lesquelles
+les lignes de courbure se correspondent; ces surfaces
+appartiennent à un système triple orthogonal dont les deux
+autres familles sont constituées par les surfaces canaux engendrées
+par ceux des cercles~$(H)$ qui s'appuient sur une des
+lignes de courbure de~$(F_1)$~ou~$(F_2)$}. De telles congruences de
+cercles s'appellent \emph{systèmes cycliques}.
+
+\Section{Congruences de cercles et systèmes cycliques.}
+{7.}{} Considérons une famille de $\infty^{2}$~cercles, et cherchons
+s'il existe des surfaces normales à tous ces cercles.
+Soit~$K$ l'un d'eux, $C(x_0, y_0, z_0)$~son centre, $\rho$~son rayon,
+$x_0, y_0, z_0, \rho$~étant fonctions de \Card{2} paramètres~$u, v$.
+Pour définir le plan
+de ce cercle nous définirons \Card{2} directions
+rectangulaires $CA(a, b, c)$~et~$CB(a', b', c')$ passant par le centre du
+cercle, et nous fixerons la position
+%% -----File: 303.png---Folio 295-------
+d'un point~$M$ sur le cercle par l'angle~$(CA, CM) = t$. Les coordonnées
+de~$M$ par rapport au système~$CAB$ sont $\rho \cos t$,~$\rho \sin t$, et
+ses coordonnées~$x, y, z$ sont
+\begin{alignat*}{4}%[** TN: Rebroken]
+x &= x_0 &&+ \rho (a \cos t &&+ a' \sin t) &&= x_0 + \rho \alpha', \\
+y &= y_0 &&+ \rho (b \cos t &&+ b' \sin t) &&= y_0 + \rho \beta', \\
+z &= z_0 &&+ \rho (c \cos t &&+ c' \sin t) &&= z_0 + \rho \gamma'.
+\end{alignat*}
+
+\Illustration[1.5in]{302a}
+\noindent Cherchons à déterminer~$t$ de façon que la surface lieu du
+point correspondant admette pour normale la tangente au cercle
+au point~$M$, dont nous désignerons par $\alpha, \beta, \gamma$ les cosinus directeurs.
+Nous obtenons la condition
+\[
+\sum \alpha\, dx = 0,
+\]
+équation aux différentielles totales des surfaces cherchées.
+Développons cette équation; $\alpha, \beta, \gamma$ sont les projections du segment
+directeur de la direction~$CM'$ correspondant à~$t + \frac{\pi}{2}$
+\[
+\alpha = -a \sin t + a' \cos t, \qquad
+\beta  = -b \sin t + b' \cos t, \qquad
+\gamma = -c \sin t + c' \cos t\DPtypo{;}{.}
+\]
+Maintenant
+\begin{alignat*}{5}%[** TN: Rebroken, filled in last two equations]
+dx &= dx_0 &&+ \alpha' · d\rho &&+ \rho \alpha · dt &&+ \rho (\cos t · da &&+ \sin t · da'), \\
+dy &= dy_0 &&+ \beta'  · d\rho &&+ \rho \beta  · dt &&+ \rho (\cos t · db &&+ \sin t · db'), \\
+dz &= dz_0 &&+ \gamma' · d\rho &&+ \rho \gamma · dt &&+ \rho (\cos t · dc &&+ \sin t · dc');
+\end{alignat*}
+d'où
+\begin{gather*}
+\sum \alpha\, dx
+  = \sum \alpha · dx_0 + \rho · dt
+  + \rho \left[\cos t\sum \alpha\, da + \sin t\sum \alpha\, da'\right] = 0, \\
+%
+\sin t \sum a\, dx + \cos t\sum a'\, dx_0 + \rho\,dt
+  + \rho \left[\cos^2 t \sum a'\, da - \sin^2 t\sum a\, da'\right] = 0\Add{.}
+\end{gather*}
+Mais
+\[
+\sum aa'= 0,
+\]
+d'où en différentiant
+\[
+\sum a\, da' + \sum a'\, da = 0;
+\]
+et l'équation devient
+\begin{gather*}
+-\sin t\sum a\, dx_0 + \cos t\sum a'\, dx_0
+  + \rho\, dt - \rho\sum a\, da' = 0, \\
+\Tag{(3)}
+dt = \sum a\, da' + \frac{1}{\rho}\sum a\, dx_0 \sin t
+  - \frac{1}{\rho} \sum a'\, dx_0 \cos t.
+\end{gather*}
+%% -----File: 304.png---Folio 296-------
+Posons
+\begin{gather*}
+\Tag{(4)}
+\tg \frac{t}{2} = w, \\
+t = 2 \arctg w,
+\end{gather*}
+Nous obtenons
+\[
+\frac{2\, dw}{1 + w^2}
+  = \sum a\, da' + \frac{1}{\rho} \sum a\, dx_0 · \dfrac{2w}{1 + w^2}
+  - \frac{1}{\rho} \sum a'\, dx_0 \frac{1 - w^2}{1 + w^2},
+\]
+ou
+\[
+\Tag{(5)}
+2\, dw = (1 + w^2) \sum a\, da' + \frac{2w}{\rho} \sum a\, dx_0
+  + \frac{w^2 - 1}{\rho} \sum a'\, dx_0,
+\]
+équation qui jouit de propriétés analogues à celles de l'équation
+de Riccati. Elle peut se mettre sous la forme
+\[
+dw = A\, du + A'\, dv + w (B\, du + B'\, dv) + w^2 (C\, du + C'\, dv),
+\]
+qui se décompose en \Card{2} équations aux dérivées partielles:
+\[
+\Tag{(6)}
+\frac{\dd w}{\dd u} = A + B w + C w^2, \qquad
+\frac{\dd w}{\dd v} = A' + B' w + C' w^2;
+\]
+d'où la condition nécessaire et suffisante d'intégrabilité
+\begin{multline*}%[** TN: Added break]
+\frac{\dd A}{\dd v} + w \frac{\dd B}{\dd v} + w^2 \frac{\dd C}{\dd v}
+  + (B + 2Cw) (A' + B' w + C' w^2) \\
+  - \left[ \frac{\dd A'}{\dd u}
+    + w\frac{\dd B'}{\dd u} + w^2\frac{\dd \Err{c}{C}'}{\dd u}\right]
+  - (B' + 2C' w) (A + Bw + Cw^2) = 0.
+\end{multline*}
+
+Toute intégrale du système~\Eq{(6)} satisfait à cette condition,
+qui est de la forme
+\[
+L + Mw + Nw^2 = 0.
+\]
+Si cette condition n'est pas identiquement satisfaite, il ne
+peut y avoir d'autres solutions que celles de l'équation
+précédente, qui en admet \Card{2}. Si l'on veut qu'il y en ait une
+infinité, cette condition doit être identiquement satisfaite,
+et comme elle est du \Ord{2}{e} degré, il suffit qu'elle soit satisfaite
+par \Card{3} fonctions. Les conditions pour qu'il en soit
+ainsi sont
+\begin{alignat*}{5}%[** TN: Rebroken]
+L &= \frac{\dd A}{\dd v} &&- \frac{\dd A'}{\dd u} &&+ &&BA' - AB' &&= 0, \\
+M &= \frac{\dd B}{\dd v} &&- \frac{\dd B'}{\dd u} &&+ 2(&&CA' - AC') &&= 0\DPtypo{.}{,} \\
+N &= \frac{\dd C}{\dd v} &&- \frac{\dd C'}{\dd u} &&+ &&CB'- BC' &&= 0\Add{.}
+\end{alignat*}
+%% -----File: 305.png---Folio 297-------
+\emph{Si les cercles sont normaux à \Card{3} surfaces, ils sont normaux à
+une infinité de surfaces.}
+
+Il est facile de construire des cercles normaux à \Card{2} surfaces,
+car il existe des sphères tangentes aux \Card{2} surfaces, et
+les cercles orthogonaux à ces sphères aux points de contact
+sont normaux aux \Card{2} surfaces. Si les lignes de courbure se
+correspondent sur les \Card{2} surfaces, on a un système cyclique,
+système de cercles normaux à $\infty^{1}$~surfaces. Réciproquement,
+supposons \Card{2} surfaces normales aux cercles, les conditions
+d'intégrabilité se réduiront à une seule; d'autre part, si on
+a une enveloppe de sphères, pour exprimer que les lignes de
+courbure se correspondent sur les \Card{2} nappes, on obtient aussi
+une seule condition. Cherchons à montrer que ces conditions
+sont identiques.
+
+Supposons donc qu'il existe une surface normale à tous
+ces cercles, supposons qu'elle corresponde à $t = 0$, ou $w = 0$,
+l'équation~\Eq{(5)} admet la solution $w = 0$, d'où la condition
+\[
+\sum a\, da' - \frac{1}{\rho} \sum a'\, dx_0 = 0;
+\]
+et l'équation devient
+\[
+\Tag{(7)}
+dw = w^2 \sum a\, da' + \frac{w}{\rho} \sum a\, dx_0.
+\]
+Soit $M_0(x,y,z)$ le point correspondant à $t = 0$
+\begin{alignat*}{3}%[** TN: Filled in last two columns]
+x    &= x_0 + \rho a, & y   &= y_0 + \rho b, & z    &= z_0 + \rho c, \\
+x_0  &= x   - \rho a, & y_0 &= y   - \rho b, & z_0  &= z   - \rho c, \\
+dx_0 &= dx - \rho\, da - a\, d\rho,\quad &
+dy_0 &= dy - \rho\, db - b\, d\rho,\quad &
+dz_0 &= dz - \rho\, dc - c\, d\rho;
+\end{alignat*}
+d'où
+\[
+\sum a\, dx_0 = \sum a\, dx - d\rho.
+\]
+Si maintenant nous considérons la normale~$(l,m,n)$ en~$M_0$ à~$(\Sigma)$,
+c'est la tangente au cercle, et \Eq{(7)}~devient
+%% -----File: 306.png---Folio 298-------
+\[
+dw = w^2 \sum a\, dl + \frac{w}{\rho} \left(\sum a\, dx - d\rho\right),
+\]
+ou
+\[
+\frac{dw}{w} + \frac{d\rho}{\rho} = w \sum a\, dl + \frac{1}{\rho} \sum a\, dx.
+\]
+
+\Illustration[2.25in]{306a}
+Nous introduisons ainsi la quantité
+\[
+\Tag{(8)}
+\rho w = r,
+\]
+et nous obtenons
+\begin{align*}
+\frac{dr}{r} &= \frac{r}{\rho} \sum a\, dl + \frac{1}{\rho} \sum a\, dx, \\
+dr &= \frac{r^2}{\rho} \sum a\, dl + \frac{r}{\rho} \sum a\, dx.
+\end{align*}
+Or
+\[
+r = \rho \tg \frac{t}{2},
+\]
+ce qui montre que $r$~est le rayon de la sphère~$(\Sigma)$ tangente
+aux surfaces lieux de $M$~et~$M_0$; son centre est le point~$\omega$ intersection
+des tangentes au cercle en $M$~et~$M_0$.
+
+Supposons maintenant qu'il existe une \Ord{2}{e} surface normale
+aux cercles. Posons
+\begin{gather*}
+\Tag{(9)}
+\frac{1}{r} = S, \\
+dr = -r^2 · dS;
+\end{gather*}
+et l'équation devient
+\begin{gather*}
+-r^2 · dS = \frac{r^2}{\rho} \sum a\, dl + \frac{r}{\rho} \sum a\, dx, \\
+\Tag{(10)}
+dS + \frac{S}{\rho} \sum a\, dx + \frac{1}{\rho} \sum a\, dl = 0.
+\end{gather*}
+Soit $S_1$\DPtypo{,}{} la solution connue
+\[
+dS_1 + \frac{S_1}{\rho} \sum a\, dx + \frac{1}{\rho} \sum a\, dl = 0,
+\]
+d'où en retranchant
+\begin{gather*}
+d(S - S_1) + \frac{S - S_1}{\rho} \sum a\, dx = 0, \\
+\Tag{(11)}
+d\log (S - S_1) = -\frac{1}{\rho} \sum a\, dx.
+\end{gather*}
+Pour que l'équation ait d'autres intégrales, il faut que $\dfrac{1}{\rho} \sum a\, dx$
+%% -----File: 307.png---Folio 299-------
+soit différentielle exacte. Or\Add{,} nous avons
+\[
+\frac{\dd S_1}{\dd u} + \frac{S_1}{\rho} \sum a \frac{\dd x}{\dd u}
+  + \frac{1}{\rho} \sum a \frac{\dd l}{\dd u} = 0, \quad
+\frac{\dd S_1}{\dd v} + \frac{S_1}{\rho} \sum a \frac{\dd x}{\dd v}
+  + \frac{1}{\rho} \sum a \frac{\dd l}{\dd v} = 0.
+\]
+Supposons que les lignes coordonnées soient lignes de courbure.
+Les formules d'Olinde Rodrigues donnent
+\begin{alignat*}{3}%[** TN: Filled in missing columns in two systems below]
+\frac{\dd l}{\dd u} &= -\frac{1}{R}\, \frac{\dd x}{\dd u}, \qquad &
+\frac{\dd m}{\dd u} &= -\frac{1}{R}\, \frac{\dd y}{\dd u}, \qquad &
+\frac{\dd n}{\dd u} &= -\frac{1}{R}\, \frac{\dd z}{\dd u}, \\
+%
+\frac{\dd l}{\dd v} &= -\frac{1}{R'}\, \frac{\dd x}{\dd v}, \qquad &
+\frac{\dd m}{\dd v} &= -\frac{1}{R'}\, \frac{\dd y}{\dd v}, \qquad &
+\frac{\dd n}{\dd v} &= -\frac{1}{R'}\, \frac{\dd z}{\dd v}.
+\end{alignat*}
+Posons
+\[
+- \frac{1}{R} = T,  \qquad  - \frac{1}{R'} = T',
+\]
+\begin{alignat*}{3}
+\frac{\dd l}{\dd u} &= T \frac{\dd x}{\dd u}, \qquad &
+\frac{\dd m}{\dd u} &= T \frac{\dd y}{\dd u}, \qquad &
+\frac{\dd n}{\dd u} &= T \frac{\dd z}{\dd u}, \\
+%
+\frac{\dd l}{\dd v} &= T' \frac{\dd x}{\dd v}, \qquad &
+\frac{\dd m}{\dd v} &= T' \frac{\dd y}{\dd v}, \qquad &
+\frac{\dd n}{\dd v} &= T' \frac{\dd z}{\dd v};
+\end{alignat*}
+Alors
+\[
+\sum a \frac{\dd l}{\dd u} = T \sum a \frac{\dd x}{\dd u}, \qquad
+\sum a \frac{\dd l}{\dd v} = T'\sum a \frac{\dd x}{\dd v},
+\]
+et les conditions pour que $S_1$\DPtypo{,}{} soit intégrale deviennent
+\[
+\frac{\dd S_1}{\dd u} + (S_1 + T)  \frac{\displaystyle\sum a \dfrac{\dd x}{\dd u}}{\rho} = 0,
+\qquad
+\frac{\dd S_1}{\dd v} + (S_1 + T') \frac{\displaystyle\sum a \dfrac{\dd x}{\dd v}}{\rho} = 0;
+\]
+d'où
+\[
+-\frac{1}{\rho} \sum a\, dx
+  = \frac{1}{S_1 + T}\,  \frac{\dd S_1}{\dd u}\, du
+  + \frac{1}{S_1 + T'}\, \frac{\dd S_1}{\dd v}\, dv\Add{.}
+\]
+Exprimons que le \Ord{2}{e} membre est une différentielle exacte,
+nous aurons l'équation aux dérivées partielles des systèmes
+cycliques
+\[
+\Tag{(12)}
+\Omega
+  = \frac{\dd}{\dd v} \left(\frac{1}{S_1 + T}\,  \frac{\dd S_1}{\dd u}\right)
+  - \frac{\dd}{\dd u} \left(\frac{1}{S_1 + T'}\, \frac{\dd S_1}{\dd v}\right)
+  = 0.
+\]
+Montrons que \emph{cette condition exprime que les lignes de courbure
+se correspondent sur les surfaces $M_0, M_1$}. D'après le
+Théorème de Dupin, pour qu'il en soit ainsi il faut et il
+suffit que ces lignes de courbure correspondent à un réseau
+conjugué sur la surface lieu de~$\omega$. Soient $X,Y,Z$ les coordonnées
+de~$\omega$:
+\[
+X = x + \frac{1}{S} l, \qquad
+Y = y + \frac{1}{S} m, \qquad
+Z = z + \frac{1}{S} n;
+\]
+pour que sur cette surface les courbes $u = \cte$, $v = \cte$ forment
+%% -----File: 308.png---Folio 300-------
+un réseau conjugué, il faut que
+\[
+\Tag{(13)}
+\begin{vmatrix}
+  \mfrac{\dd^2 X}{\dd u\, \dd v} &
+  \mfrac{\dd X}{\dd u} &
+  \mfrac{\dd X}{\dd v}
+\end{vmatrix} = 0.
+\]
+Mais
+\begin{align*}
+\frac{\dd X}{\dd u}
+  &= \frac{\dd x}{\dd u} + \frac{T}{S}\, \frac{\dd x}{\dd u}
+  + l\frac{\dd \left(\dfrac{1}{s}\right)}{\dd u}
+  = \left(1 + \frac{T}{S}\right) \frac{\dd x}{\dd u}
+  + l \frac{\dd \left(\dfrac{1}{s}\right)}{\dd u},
+  &&\text{\dots, \qquad \dots,} \\
+\frac{\dd X}{\dd v}
+  &= \left(1 + \frac{T'}{S}\right) \frac{\dd x}{\dd v}
+  + l\frac{\dd \left(\dfrac{1}{s}\right)}{\dd v},
+  &&\text{\dots, \qquad \dots;}
+\end{align*}
+relations qu'on peut encore écrire
+\begin{alignat*}{3}
+\frac{\dd X}{\dd u} &= \frac{S + T}{S^2}
+  &&\left[S \frac{\dd x}{\dd u} - \frac{1}{S + T}\, \frac{\dd s}{\dd u} l\right],
+  &&\text{\dots, \qquad \dots,} \\
+\frac{\dd X}{\dd v} &= \frac{S + T'}{S^2}
+  &&\left[S \frac{\dd x}{\dd v} - \frac{1}{S + T'}\, \frac{\dd s}{\dd v} l\right],
+  \qquad&&\text{\dots, \qquad \dots,}
+\end{alignat*}
+dans le déterminant~\Eq{(13)} nous pouvons remplacer $\dfrac{\dd^2 X}{\dd u\, \dd v}$ par
+\[%[** TN: Not displayed in original]
+\frac{\dd}{\dd v} \left(M \frac{\dd X}{\dd u}\right)
+  - \frac{\dd}{\dd u} \left(N \frac{\dd X}{\dd v}\right)
+\]
+à condition que $M - N \neq 0$; nous prendrons
+$M = \dfrac{S^2}{S + T}$ et $N = \dfrac{S^2}{S + T'}$; nous avons alors à vérifier la relation
+\[
+\begin{vmatrix}
+\mfrac{\dd S}{\dd v} · \mfrac{\dd x}{\dd u}
+  - \mfrac{\dd S}{\dd u}\, \mfrac{\dd x}{\dd v}
+  - \mfrac{T'}{S + T}\, \mfrac{\dd S}{\dd u} · \mfrac{\dd x}{\dd v}
+  + \mfrac{T}{S + T'}\, \mfrac{\dd S}{\dd v} · \mfrac{\dd X}{\dd u}
+  - \Omega l \\
+S \mfrac{\dd x}{\dd u} - \mfrac{1}{S + T}\,  \mfrac{\dd S}{\dd u}\, l \\
+S \mfrac{\dd X}{\dd v} - \mfrac{1}{S + T'}\, \mfrac{\dd S}{\dd v}\, l
+\end{vmatrix} = 0\DPtypo{;}{.}
+\]
+Multiplions la \Ord{2}{e} ligne par $-\dfrac{S + T + T'}{S(S + T')}\, \dfrac{\dd S}{\dd v}$, la \Ord{3}{e} par
+$\dfrac{S + T + T'}{S(S + T)}\, \dfrac{\dd S}{\dd u}$ et ajoutons à la \DPtypo{1ère}{\Ord{1}{ère}}, nous obtenons
+\[
+\Omega S^2 \begin{vmatrix}
+  1 & \mfrac{\dd x}{\dd u} & \mfrac{\dd x}{\dd v}
+\end{vmatrix} = 0,
+\]
+or\Add{,} le déterminant n'est pas nul, $S$~non plus, donc $\Omega = 0$ et
+réciproquement. Les conditions sont identiques.
+
+\Section{Transformation de contact de \DPtypo{Ribeaucour}{Ribaucour}.}
+{}{Remarque.} Considérons une sphère fixe de centre~$\omega$, et
+les cercles~$(K)$ orthogonaux à cette sphère; considérons une
+surface~$(S)$, un de ses points~$M$, et l'élément de contact en ce
+point; il y a un cercle~$(K)$ et un seul passant par~$M$ et normal
+%% -----File: 309.png---Folio 301-------
+en ce point à la surface~$(S)$. Donc à la surface~$(S)$ correspond
+une congruence de cercles qui lui sont orthogonaux; de
+plus ces cercles étant orthogonaux à la sphère~$(\omega)$ en \Card{2} points
+sont orthogonaux à \Card{3} surfaces\Add{;} ils constituent donc un système
+cyclique. Soient $P\Add{,} P'$ les points où le cercle~$(K)$ rencontre la
+sphère; déterminons sur ce cercle le point~$M'$ tel que $\Ratio{M}{M'}{P}{P'} = \cte[]$.
+Le lieu du point~$M'$ est une surface normale à~$(K)$, puisque
+l'équation~\Eq{(5)} a mêmes propriétés que l'équation de Riccati\Add{.}
+A l'élément de contact de la surface~$(S)$ au point~$M$ correspond
+ainsi un élément de contact d'une autre surface; les lignes
+de courbure se correspondent sur les \Card{2} surfaces, et nous avons
+ainsi un groupe de transformations de contact conservant les
+lignes de courbure.
+
+Ces résultats subsistent si on prend les cercles~$(K)$
+normaux à un plan fixe.
+
+\Section{Surfaces de Weingarten.}
+{8.}{} Nous avons considéré des congruences de sphères telles
+que les lignes de courbure se correspondent sur les \Card{2} nappes
+focales. Aux sphères, la transformation de S.~Lie fait
+correspondre des droites, et aux lignes de courbure correspondent
+les lignes asymptotiques. Nous aurons donc à considérer
+des congruences de droites telles que les asymptotiques
+se correspondent sur les \Card{2} nappes focales. Nous nous
+bornerons au cas où la congruence est une congruence de normales,
+et le problème revient ainsi à chercher les surfaces
+telles que les asymptotiques se correspondent sur les \Card{2} nappes
+de la développée.
+
+Soit donc une surface~$(\Sigma)$ sur laquelle nous prendrons
+les lignes de courbure pour lignes coordonnées; soient $l,m,n$
+%% -----File: 310.png---Folio 302-------
+les cosinus directeurs de la normale, $R, R'$ les rayons de
+courbure principaux. Les \Card{2} nappes de la développée ont pour
+équations
+\begin{alignat*}{3}
+\Tag{(S)}
+X &= x + Rl, & Y &= y + Rm, & Z &= z + Rn; \\
+\Tag{(S')}
+X' &= x + R'l, \qquad & Y' &= y + R'm, \qquad & Z' &= z + R'n.
+\end{alignat*}
+Cherchons les asymptotiques de~$(S), (S')$; et exprimons que les
+équations différentielles en $u, v$ sont les mêmes. Ici les lignes
+coordonnées formant un réseau orthogonal et conjugué, on
+a
+\begin{gather*}
+ds^{2} = E\, du^{2} + G\, dv^{2}, \\
+\sum l\, d^{2}x = L\, du^{2} + N\, dv^{2};
+\end{gather*}
+et
+\[
+\frac{1}{R} = \frac{L}{E}, \qquad
+\frac{1}{R'} = \frac{N}{G},
+\]
+d'où
+\[
+\sum l\, d^{2} x = \frac{E}{R}\, du^{2} + \frac{G}{R'}\, dv^{2}.
+\]
+Les formules d'O.~Rodrigues donnent
+\begin{alignat*}{3}%[** TN: Filled in last two columns]
+\frac{\dd x}{\dd u} &= -R\, \frac{\dd l}{\dd u}, &
+\frac{\dd y}{\dd u} &= -R\, \frac{\dd m}{\dd u}, &
+\frac{\dd z}{\dd u} &= -R\, \frac{\dd n}{\dd u}, \\
+\intertext{d'où}
+\frac{\dd l}{\dd u} &= -\frac{1}{R}\, \frac{\dd x}{\dd u}, &
+\frac{\dd m}{\dd u} &= -\frac{1}{R}\, \frac{\dd y}{\dd u}, &
+\frac{\dd n}{\dd u} &= -\frac{1}{R}\, \frac{\dd z}{\dd u}, \\
+\intertext{et}
+\frac{\dd l}{\dd v} &= -\frac{1}{R'}\, \frac{\dd x}{\dd v}, \qquad &
+\frac{\dd m}{\dd v} &= -\frac{1}{R'}\, \frac{\dd y}{\dd v}, \qquad &
+\frac{\dd n}{\dd v} &= -\frac{1}{R'}\, \frac{\dd z}{\dd v};
+\end{alignat*}
+et par conséquent
+\begin{align*}%[** TN: Rebroken]
+dX &= dx + R\, dl + l\, dR \\
+   &= \frac{\dd x}{\dd u}\, du + \frac{\dd x}{\dd v}\, dv
+  - R \left(\frac{1}{R} · \frac{\dd x}{\dd u} du
+    + \frac{1}{\DPtypo{R}{R'}}\, \frac{\dd x}{\dd v} dv\right) + l\, dR \\
+   &= \left(1 - \frac{R}{R'}\right) \frac{\dd x}{\dd v} dv + l\, dR,
+\end{align*}
+formules qui montrent que la normale à~$(S)$ a pour coefficients
+de direction $\dfrac{\dd x}{\dd u}, \dfrac{\dd y}{\dd u}, \dfrac{\dd z}{\dd u}$.
+
+On a donc sur cette surface~$(S)$
+\[
+ds^{2} = \left(1 - \frac{R}{R'}\right)^{2} G\, dv^{2} + dR^{2},
+\]
+ce qui met en évidence sur la surface~$(S)$ une famille de géodésiques
+$v = \cte$, et leurs trajectoires orthogonales $R = \cte[]$.
+%% -----File: 311.png---Folio 303-------
+L'équation \DPtypo{differentielle}{différentielle} des asymptotiques est
+\begin{align*}
+&\sum dl · dX = 0, \\
+\intertext{ou}
+&\sum d\left(\frac{\dd x}{\dd u}\right) · dX = 0.
+\end{align*}
+Développons cette équation. Le coefficient de $\left(1 - \dfrac{R}{R'}\right) · dv$ est
+\[
+\sum \frac{\dd x }{ \dd v}\, d\left(\frac{\dd x}{\dd u}\right)
+  = du \sum \frac{\dd x}{\dd v} · \frac{\dd^{2} x}{\dd u^{2}}
+  + dv \sum \frac{\dd x}{\dd v} · \frac{\dd^{2} x}{\dd u\, \dd v};
+\]
+\DPtypo{Or}{or} on a
+\[
+\sum \frac{\dd x}{\dd u} · \frac{\dd x}{\dd v} = 0;
+\]
+d'où
+\[
+\sum \frac{\dd x}{\dd v} · \frac{\dd^{2} x}{\dd u^{2}}
+  = -\sum \frac{\dd x}{\dd u} · \frac{\dd^{2} x}{\dd u\, \dd v}
+  = -\frac{1}{2} · \frac{\dd E}{\dd v},
+\]
+et
+\[
+\sum \frac{\dd x}{\dd v} · \frac{\dd^{2} x}{\dd u\, \dd v}
+  = \frac{1}{2} · \frac{\dd G}{\dd u}.
+\]
+Le coefficient de~$dR$ est
+\[
+\sum l\, d\left(\frac{\dd x}{\dd u}\right)
+  = \sum l\, \frac{\dd^{2} x}{\dd u^{2}} · du
+  + \sum l\, \frac{\dd^{2} x}{\dd u\, \dd v}\, dv
+  = \frac{E}{R}\, du,
+\]
+d'où l'équation aux asymptotiques
+\[
+\frac{1}{2}\left(1 - \frac{R}{R'}\right)
+  \left[- \frac{\dd E}{\dd v}\, du\, dv
+        + \frac{\dd G}{\dd u}\, dv^{2}\right]
+      + \frac{E}{R}\, dR\, du = 0.
+\]
+Les courbes $u = \cte$, $v = \cte$ correspondent à des courbes conjuguées
+sur la surface~$(S)$, donc le coefficient de~$du\, dv$ dans
+l'équation précédente est nul:
+\[
+\Tag{(1)}
+-\frac{1}{2} \left(1 - \frac{R}{R'}\right) \frac{\dd E}{\dd v}
+  + \frac{E}{R}\, \frac{\dd R}{\dd v} = 0;
+\]
+et l'équation devient
+\[
+\frac{1}{2} \left(1 - \frac{R}{R'}\right) \frac{\dd G}{\dd u}\, dv^{2}
+  + \frac{E}{R}\, \frac{\dd R}{\dd u}\, du^{2} = 0.
+\]
+De même sur la surface~$(S')$ on obtiendra la condition
+\[
+\Tag{(2)}
+-\frac{1}{2} \left(1 - \frac{R'}{R}\right) · \frac{\dd G}{\dd u}
+  + \frac{G}{R'}\, \frac{\dd R'}{\dd u} = 0,
+\]
+de sorte que l'équation aux asymptotiques de~$(S)$ peut s'écrire
+\[
+-\frac{G}{R'{}^2}\, \frac{\dd R'}{\dd u}\, dv^{2}
+  + \frac{E}{R^{2}}\, \frac{\dd R}{\dd u}\, du^{2} = 0,
+\]
+ou
+\[
+G\, \frac{\dd\left(\dfrac{1}{R'}\right)}{\dd u}\, dv^{2}
+  - E\, \frac{\dd\left(\dfrac{1}{R}\right)}{\dd u}\, du^{2} = 0;
+\]
+%% -----File: 312.png---Folio 304-------
+et de même pour~$(S')$
+\[
+E\, \frac{\dd \left(\dfrac{1}{R}\right)}{\dd v}\, du^{2}
+  - G\, \frac{\dd \left(\dfrac{1}{R'}\right)}{\dd v}\, dv^{2} = 0.
+\]
+Pour que ces équations soient identiques, il faut et il suffit
+que l'on ait
+\[
+\begin{vmatrix}
+\mfrac{\dd \left(\dfrac{1}{R}\right)}{\dd u} &
+\mfrac{\dd \left(\dfrac{1}{R'}\right)}{\dd u} \\
+\mfrac{\dd \left(\dfrac{1}{R}\right)}{\dd v} &
+\mfrac{\dd \left(\dfrac{1}{R'}\right)}{\dd v}
+\end{vmatrix}
+= 0,
+\]
+c'est-à-dire que $\dfrac{1}{R}$~soit fonction de~$\dfrac{1}{R'}$. \emph{Les rayons de courbure
+sont fonctions l'un de l'autre \(\DPtypo{Ribeaucour}{Ribaucour}\).} Ces surfaces
+s'appellent \emph{surfaces de Weingarten, ou surfaces~$W$}. Les surfaces
+minima en sont un cas particulier $(R + R' = 0)$.
+
+Supposons que nous partions d'une surface~$(W)$: $R'$~est
+fonction de~$R$, et la condition~\Eq{(2)} montre que
+\[
+\frac{\dd \log G}{\dd u} = \Psi(R)\, \frac{\dd R}{\dd u},
+\]
+d'où
+\begin{gather*}
+\log G = \chi(R) + \theta(v), \\
+G = e^{\chi(R)}\, e^{\theta(v)} = F(R)\, K(v);
+\end{gather*}
+et sur la développée
+\[
+ds^{2} = \Theta^{2}(R)\, K(v)\, dv^{2} + dR^{2}.
+\]
+Posons
+\[
+\sqrt{K(v)}\, dv = dV,
+\]
+nous avons
+\[
+ds^{2} = dR^{2} + \Theta^{2}(R)\, dV^{2},
+\]
+forme caractéristique de l'élément d'arc des surfaces de révolution
+rapportées aux méridiens et aux parallèles. Si nous
+rapportons la méridienne à l'arc, ses équations sont
+\begin{alignat*}{3}
+x &= \Theta (s), & y &= 0,   & z &= \Theta_{1}(s);\\
+\intertext{et celles de la surface de révolution sont}
+x &= \Theta(s)\cos V, \qquad & y &= \Theta(s)\sin V, \qquad &z &= \Theta_{1}(s).
+\end{alignat*}
+%% -----File: 313.png---Folio 305-------
+\emph{On voit ainsi que les développées de toute surface~$W$ sont applicables
+sur des surfaces de révolution, les méridiens correspondant
+à une famille de géodésiques et les parallèles à
+leurs trajectoires orthogonales.}
+
+\Paragraph{Application.} Supposons la surface~$W$ à courbure totale
+constante. En changeant d'unité on peut toujours écrire
+\begin{align*}
+&RR' = - 1, \\
+&R' = - \frac{1}{R};
+\end{align*}
+la condition~\Eq{(2)} s'écrit
+\begin{gather*}
+\left(1 + \frac{1}{R^{2}}\right) \frac{\dd G}{\dd u}
+  = -\frac{2G}{R}\, \frac{\dd R}{\dd u}, \\
+\frac{\dd \log G}{\dd u}
+  = -\frac{2R}{R^{2} + 1}\, \frac{\dd R}{\dd u}
+  = -\frac{\dd \log (R^{2} + 1)}{\dd u}, \\
+G = \frac{1}{R^{2} + 1}\, K(v), \\
+dS^{2} = (R^{2} + 1) · dV^{2} + dR^{2}.
+\end{gather*}
+Posons
+\[
+\Theta(R) = \sqrt{R^{2} + 1}\Add{,}
+\]
+la méridienne de la surface de révolution est donc telle que l'on ait
+\[
+x = \sqrt{s^{2} + 1},
+\]
+d'où
+\[
+s = \sqrt{x^{2} - 1}.
+\]
+Cherchons~$z$.
+\begin{align*}
+dx^{2} + dz^{2} &= ds^{2} = dx^{2} · \frac{x^{2}}{x^{2} - 1}, \\
+dz^{2} &= \frac{dx^{2}}{x^{2} - 1}, \\
+dz &= \frac{dx}{\sqrt{x^{2} - 1}},
+\end{align*}
+\vspace*{-\belowdisplayskip}%
+\vspace*{-\abovedisplayskip}%
+\begin{gather*}
+z = \DPtypo{L}{\log}(x + \sqrt{x^{2} - 1}),\vphantom{\bigg|} \\
+x + \sqrt{x^{2} - 1} = e^{z},
+\end{gather*}
+d'où
+\[
+x - \sqrt{x^{2} - 1} = e^{-z};
+\]
+d'où
+\[
+x = \frac{1}{2}\, (e^{z} + e^{-z})\Add{,}
+\]
+%% -----File: 314.png---Folio 306-------
+\emph{les \Card{2} nappes de la développée d'une surface à courbure totale
+constante sont applicables sur l'alysséide}. c\Add{.}à\Add{.}d.\ sur la surface
+engendrée par une chaînette qui tourne autour de sa base.
+
+
+\ExSection{XIII}
+
+\begin{Exercises}
+\item[54.] Soit $S$ une surface quelconque et $\Pi$~un plan quelconque. On
+considère toutes les sphères~$U$ ayant leurs centres sur~$S$ et
+coupant le plan~$\Pi$ sous un angle constant~$\phi$ tel que l'on ait
+$\cos\phi = \dfrac{1}{\DPtypo{K}{k}}$. Soit $S'$ la surface déduite de~$S$ en réduisant les
+%[** TN: Fraction has imaginary value, but also appears in typeset edition]
+ordonnées de~$S$ perpendiculaires à~$\Pi$ dans le rapport $\dfrac{\sqrt{1-k^2}}{1}$.
+
+Les sphères~$U$ enveloppent une surface à deux nappes. Montrer
+que leurs lignes de courbure correspondent point par point à
+celles de~$S'$. Examiner le cas où $S$~est du second degré.
+
+\item[55.] De chaque point~$M$ d'une surface~$S$ comme centre, on décrit un
+cercle~$K$ situé dans le plan tangent à~$S$, et dont le rayon soit
+égal à une constante donnée.
+
+%[** Regularized formatting of parts]
+\Primo. Déterminer les familles de
+$\infty^{1}$~cercles~$K$ qui engendrent une surface sur laquelle ces cercles
+soient lignes de courbure. Lieux des centres des sphères
+dont une telle surface est l'enveloppe.
+
+\Secundo. Trouver la condition
+nécessaire et suffisante pour que les cercles~$K$ forment
+%% -----File: 328.png---Folio 320-------
+un système cyclique. Cette condition étant supposée
+remplie, soit~$S_{1}$, l'une des surfaces normales aux cercles~$K$;
+montrer que les lignes de courbure de~$S_{1}$, correspondent à celles
+de~$S$, quand on fait correspondre à chaque point~$M$ de~$S$
+le point~$M_{1}$ du cercle~$K$ correspondant où $S_{1}$~est normal à~$K$.
+
+\Tertio. Montrer que $S_{1}$~a une courbure totale constante, et que
+la congruence de droites qui a~$S,S_{1}$ pour surfaces focales est
+une congruence de normales.
+
+\Quarto. Soit $C$~l'un des centres de
+courbure principaux de~$S$ en~$M$, et $C_{1}$~le centre de courbure
+principal de~$S_{1}$ en~$M_{1}$, qui correspond à~$C$. \DPchg{Etudier}{Étudier} la congruence
+des droites~$CC_{1}$.
+
+\item[56.] \DPchg{Etant}{Étant} donnée une surface~$S$, on désigne par~$C$ l'une quelconque
+des lignes de courbure de l'une des familles, par~$C'$ l'une
+quelconque des lignes de courbure de l'autre famille, de
+sorte qu'en un point~$M$ de~$S$ se croisent une courbe~$C$ et une
+courbe~$C'$. Soient $\omega, \omega'$ les centres de courbure principaux
+correspondant à ces deux courbes; et soient $G,G'$ les centres
+de courbure géodésique de ces deux courbes.
+
+% [** TN: Regularized formatting of parts]
+\Primo. Que peut-on
+dire des congruences définies respectivement par les quatre
+droites $MG, MG', G\omega, G'\omega'$?
+
+\Secundo. Soit $(\gamma)$~le cercle osculateur
+à~$C$ en~$M$. Démontrer que $(\gamma)$~engendre une surface canal
+quand $M$~décrit une courbe~$C'$. Trouver les sphères dont cette
+surface canal est l'enveloppe.
+
+\Tertio. Montrer que si $S$~fait
+partie de l'une des familles d'un système triple orthogonal,
+les cercles osculateurs aux trajectoires orthogonales des
+surfaces de cette famille, construits aux divers points de~$S$
+forment un système cyclique.
+
+\item[57.] Soit $O$ un point fixe, et $S$~une surface quelconque; en un point
+quelconque~$M$ de~$S$ on mène le plan tangent~$P$ et de~$O$ on abaisse
+la perpendiculaire sur~$P$; soit $H$ non pied.
+
+%[** TN: Regularized formatting of parts]
+\Primo. Trouver les
+courbes de~$S$ qui, en chacun de leurs points~$M$, admettent $MH$~pour
+normale.
+
+\Secundo. Soit $HI$ la médiane du triangle~$OHM$; la
+congruence des droites~$HI$ est une congruence de normales.
+Trouver les surfaces normales à toutes ces droites. Montrer
+que leurs lignes de courbure correspondent à un réseau de
+courbes conjuguées décrites par~$M$ sur~$S$.
+
+\Tertio. Soit $K$ le
+point où le plan perpendiculaire à~$MO$ rencontre~$MH$; et soit~$(\gamma)$
+le cercle de centre~$K$, passant en~$O$, et \DPtypo{situe}{situé} dans le
+plan~$MOK$. Les cercles~$(\gamma)$ forment un système cyclique.
+
+\item[58.] De chaque point~$M$ du \DPchg{paraboloide}{paraboloïde}
+\[
+\Tag{(P)}
+xy - az = 0,
+\]
+comme centre, on décrit une sphère~$\Sigma$ tangente au plan~$\DPtypo{xoy}{xOy}$.
+Soit $A$ le point de contact de~$\Sigma$ avec ce plan et $B$~le second
+point de contact de~$\Sigma$ avec son enveloppe.
+
+%[** TN: Regularized formatting of parts]
+\Primo. Quelles courbes
+doit décrire~$M$ sur~$(P)$ pour que $AS$~engendre une développable?
+Ces courbes forment sur~$P$ un réseau conjugué, et leurs
+tangentes en chaque point~$M$ sont perpendiculaires aux plans
+focaux de la congruence engendrée par~$AS$.
+
+\Secundo. Déterminer
+les lignes de courbure de l'enveloppe de~$\Sigma$; les normales
+menées à cette enveloppe le long de chaque ligne de courbure
+découpent sur~$(P)$ un réseau \DPtypo{conjugue}{conjugué}.
+
+\Tertio. On considère le
+cercle~$C$ normal à~$\Sigma$ en $A$~et~$B$. Montrer qu'il y a une infinité
+de surfaces normales à tous les cercles~$C$, et les déterminer.
+
+\Quarto. Montrer que ces surfaces forment l'une des familles d'un
+système triple orthogonal, et achever de déterminer ce
+système.
+\end{Exercises}
+
+%[** TN: Exercises moved to the end of the respective chapters]
+%% -----File: 315.png---Folio 307-------
+% Chapitre II: #8 -- #11
+%% -----File: 316.png---Folio 308-------
+% Chapitre III: #12 -- #14
+%% -----File: 317.png---Folio 309-------
+% #14 -- #17
+%% -----File: 318.png---Folio 310-------
+% Chapitre IV: #18 -- #20; Chapitre V: #21 -- #26
+%% -----File: 319.png---Folio 311-------
+% #26 -- #28
+%% -----File: 320.png---Folio 312-------
+% #28; Chapitre VI: #29 -- #31
+%% -----File: 321.png---Folio 313-------
+% #31; Chapitre VII: #32 -- #36
+%% -----File: 322.png---Folio 314-------
+% #36 -- #39
+%% -----File: 323.png---Folio 315-------
+% #39 -- Chapitre VIII: #40 -- #41
+%% -----File: 324.png---Folio 316-------
+% #41 -- #42
+%% -----File: 325.png---Folio 317-------
+% Chapitre IX: #43 -- #44; Chapitre X: #45 -- #47
+%% -----File: 326.png---Folio 318-------
+% #47 -- #49; Chapitre XI: #50 -- #52; Chapitre XII: #53
+%% -----File: 327.png---Folio 319-------
+% #53; Chapitre XIII: #54 -- #55
+%% -----File: 329.png---Folio 321-------
+% #55 -- #56
+%% -----File: 330.png---Folio 322-------
+% #57 -- #58
+%% -----File: 331.png---Folio 323-------
+% #58
+
+\iffalse%%[** TN: Raw DP formatter code for errata]
+\Section{Errata\Add{.}}
+
+\DPnote{[** d^o means dito, used
+\newcommand{\dito}{\qquad d$^\text{o}$\qquad}]}
+
+\begin{tabular}{ccll}
+Page & Ligne && Lire \\
+[x]7  & 6      & Au lieu de $- \dfrac{1}{6R^2} ds$ & $- \dfrac{1}{6R^2} ds^3$ \\
+[x]7  & Figure & [Illustration] & [Illustration] \\
+[x]7  & 15     & Au lieu de $T\ 0$ &  $T < 0$ \\
+[x]11 & avant dernière & \dito  le lieu des sphères & le lieu des centres des sphères. \\
+[x]16 & 7      & \dito  $- \dfrac{1}{6R} ds^3$ & $- \dfrac{1}{6R^2} ds^3$ \\
+[x]17 & 8      & \dito  secteur  & vecteur \\
+[x]19 & 15     & \dito  $\sum \dfrac{\dd x}{\dd u} \dfrac{\dd x}{\dd v^2}$ & $\sum \dfrac{\dd x}{\dd u} \dfrac{\dd x}{\dd v}$ \\
+[x]20 & 8 en remontant & \dito  $(S') (S')$ & $(S) (S')$ \\
+[x]21 & 3      & \dito            $x = F$   & $x = F_1$ \\
+[x]   &        & \phantom{\dito } $y = G_2$ & $y = G_1$ \\
+[x]   &        & \phantom{\dito } $z = H$   & $z = H_1$ \\
+[x]21 & 7 en remontant & \dito  $\dfrac{E_1}{E} = \dfrac{F_1}{F} = \dfrac{G_1}{G}$ & $\dfrac{E}{E_1} = \dfrac{F}{F_1} = \dfrac{G}{G_1}$ \\
+[x]22 & 13, 14 & \dito  $dd(uv),d\beta(uv)$ & $d\alpha(u,v),d\beta(u,v)$ \\
+[x]29 & 9 en remontant & \dito  $\dfrac{a\sin\theta}{R}$ & $- \dfrac{a\sin\theta}{R}$ \\
+[x]30 & 9      & \dito  $\sum a, \dfrac{d^2x}{ds}$ & $\sum a, \dfrac{d^{2}x}{ds^2}$ \\
+[x]31 & 3 en remontant & \dito  $\Omega^2 (du\,d^{2}v - \dots$ & $- H^2 (du\,d^{2}v$ \\
+[x]33 & Figure & la lettre M &  Manque au sommet de l'angle $\theta$. \\
+[x]35 & dernière & Au lieu de $1$  & $l$ \\
+[x]37 & 15     & \dito  $r\,d^{2}x$ & $r\,dx^2$ \\
+[x]37 & 10 en remontant & \dito   elipse  & ellipse \\
+\end{tabular}
+%% -----File: 332.png---Folio 324-------
+\begin{tabular}{ccll}
+\emph{Page} & \emph{Ligne} &&  \emph{Lire} \\
+[x]41  & dernière        & Au lieu de $= - c$  &  $= - c^2$ \\
+[x]49  & 3 en remontant  & \dito $h()$  & $h (u)$ \\
+[x]53  & 4 en remontant  & \dito pour coordonnées  &  pour lignes coordonnées \\
+[x]57  & 3 en remontant  & \dito $E\, du + F'\, dv$  &  E'\,du + F'\,dv \\
+[x]74  & 3               & \dito $A' < 0$  &   $A A'< 0$ \\
+[x]75  & 5 en remontant  & \dito $\dots dv\,dv - \dots dv\,du \dots$  &  $d\gamma \dots d\gamma$ \\
+[x]87  & 3 en remontant  & \dito $MG$ fait avec $MG$  &  $MG_1$ fait \dots \\
+[x]94  & 1               & \dito par $u,v,w,\dots a,s,$  &  par $u',v',w',\dots$ a~$s$, \\
+[x]114 & 8 en remontant  & \dito $z = (v)\phi$  &   $z = \phi(v)$ \\
+[x]142 & 4 en remontant  & \dito $\dfrac{1}{du} \dfrac{2du}{1+w^2}$  &  $\dfrac{1}{du} \dfrac{2dw}{1+w^2}$ \\
+[x]152 & 10              & \multicolumn{2}{l}{Mettre le chiffre (3) à l'équation précédente} \\
+[x]156 & 12 en remontant & Au lieu de sur $(\phi)$    &   sur $(\phi')$ \\
+[x]163 & 10 en remontant & \dito   $\dfrac{-r\,dr}{dr}$  &  $\dfrac{-r\,dr}{dr^2}$ \\
+[x]165 & Figure          & $(T)$  &  $(\Gamma)$ \\
+[x]169 & 2 en remontant  & \dito   $+\zeta\left(-\dfrac{a}{R} - \dfrac{a''}{T} -\right)ds$   &  $+\eta \left(-\dfrac{a}{R} - \dfrac{a''}{T}\right)ds$ \\
+[x]175 & 10 en remontant & \dito   résolues en $x,y$  &  résolues en $x_1,y_1$ \\
+[x]176 & 2               & \dito   avec $0,x$  &  $0x_1$  \\
+[x]188 & 2               & \dito   $\rho = \cte$  &  $\rho = 0$ \\
+[x]189 &                 & \dito   $M M$  &   $M M_1$  \\
+[x??]189 & 8 en remontant  & \dito  $x = x + \rho x \text{etc}\dots$  &  $= x + \rho x,$ \\
+[x]191 & 12              & \dito   $x$  &  $x_1$
+\end{tabular}
+%% -----File: 333.png---Folio 325-------
+\begin{tabular}{ccll}
+\emph{Page} & \emph{Ligne} &&  \emph{Lire} \\
+[x]191 &  10 en remontant &  Au lieu de $\dfrac{\dd Q}{\dd \lambda}$  &  $\dfrac{\dd Q}{\dd \lambda} \dfrac{\dd x}{\dd \mu$ \\
+[x]191 &  7  en remontant &  \dito  $\dfrac{\dd x}{\dd \mu$  &  $\dfrac{\dd x'}{\dd \mu}$ \\
+[x]191 &  7  en remontant &  \dito   car $\dfrac{\dd^2 x}{\dd \lambda\, \dd \mu}$ etc\dots  &  car $\dfrac{\dd\DPtypo{'}{^2} x}{\dd \lambda\, \dd \mu} = \dfrac{\dd x_1}{\dd \mu}$ et $\dfrac{\dd x}{\dd \lambda} = x_1$ \\
+[x]193 &  4               &  \dito   $(S)$  &  (5) \\
+[x]19[** TN: sic, presumed 193] &  12     &  \dito   $x_1$  &  $x$  \\
+[x]199 &  6 en remontant  &  \dito   $(S_1)$  &  $(S'_1)$  \\
+[x]199 &  2 en remontant  &  \dito   l'homothétique de~$M$  &  l'homothétique de~$M_1$ \\
+[x]200 &  3               &  \dito   au rayon~$OM$  &   au rayon~$OM_1$  \\
+[x]200 &  7               &  \dito   $OM$  &   $OM_1$  \\
+[x]200 &  8               &  \dito   $(S)$ (au commencement de la ligne)  &  $(S_1)$  \\
+[x]204 &  14              &  \dito   $= 0 \lambda$ &  $= 0$ \\
+[x]207 &  6 en remontant  &  \dito   $f(x,y,z,U}{W,V}{W) = 0$  &  $f(x,y,z,-U}{W,-V}{W) = 0$ \\
+[x?]212 &  9               &  \dito   les développables de l'une des familles  &  ces développables \\
+[x]212 &  11              &  \dito   indépendant de la développable     &      indépendant de la congruence \\
+[x]222 &  13              &  \dito   $p = cY-bz$ \dots  &  ajouter: ou encore $p = yz-zY$ et l'équation écrite à la ligne
+ 18 sera $\chi (X-x, Y-y, Z-z, yZ-zY, zX-xZ, xY-yX) = 0$ \\
+[x]223 &  11 en remontant &  \dito   $p = yz' - zy$  &  $p = yz' - zy'$ \\
+[x]    &                  &          $q = zx' - xz$  &  $q = zx'- xz'$ \\
+[x]224 &  9               &  \dito   $p_{ik} = \begin{vmatrix}x & x_k \\ y & y_k \end{matrix}$ & $p_{ik} = \begin{vmatrix}x_i & x_k \\ y_i & y_k \end{matrix}$
+\end{tabular}
+%% -----File: 334.png---Folio 326-------
+\begin{tabular}{ccll}
+\emph{Page} & \emph{Ligne} &&  \emph{Lire} \\
+[x]232 &  8               & Au lieu de $+ c R$  &  $+ C R$ \\
+[x]243 &  4               & \dito    focal &  polaire \\
+[x]270 &  1               & \dito   $\dfrac{\dd \Psi}{dy} + \dfrac{\dd \Psi}{dz} = 0$  &  $\dfrac{\dd \Psi}{\dd y} + q \dfrac{\dd \Psi}{\dd z} = 0$ \\
+[x]271 &  5 en remontant  & \dito   intégralité &  intégrabilité \\
+[x]277 &  6               & \dito   $\sum (x-x)^2$  & $\sum (x-x_0)^2$ \\
+[x]277 &  4 en remontant  & \dito   $dC = \dfrac{dz_0}{2R_0}$  &  $dC = \dfrac{dz_0}{2R}$ \\
+[x]278 &  6 en remontant  & \dito   du groupe des rayons &  du groupe des transformations par rayons\dots \\
+[x]278 &  1 en remontant  & \dito   $-\dfrac{(x^2+y^2+z^2+R^2)}{4R^2 (e-\lambda)}$  &  $-\dfrac{(x^2+y^2+z^2+R^2)^2}{4R^2 (e-\lambda)}$ \\
+[x]284 &  11 en remontant &  \dito   $(H)$  &  (4) \\
+[x]285 &  Figure          & [Illustration] &  [Illustration] \\
+[x]286 &  7               & \dito   $\sin^2 \theta [$  &  $\sin^4 \theta [$ \\
+[x]287 &  8 en remontant  & \dito   $= -\dfrac{1}{2}$.  &  $= \dfrac{1}{2}$. \\
+[x]288 &  9 en remontant  & \dito   $\Psi_1 + \cos i \Psi = 0$  &  $\Psi_1 = 0$ \\
+[x]289 &  6 en remontant  & \dito   $\sum df · dl$.  &  $-\sum df · dl$. \\
+[x]293 &  4 en remontant  & \dito   $('\sigma) (\sigma)$  & $(\sigma) (\sigma')$ \\
+[x]296 &  13              & \dito   $w^2 \dfrac{\dd c'}{\dd u}$  &  $w^2 \dfrac{\dd C' }{\dd u}]$
+\end{tabular}
+\fi
+
+
+%%%% LICENSE %%%%
+\backmatter
+\pagenumbering{Alph}
+\phantomsection
+\pdfbookmark[0]{License.}{License}
+\fancyhead[C]{LICENSE}
+\SetPageNumbers
+
+\begin{PGtext}
+End of Project Gutenberg's Leçons de Géométrie Supérieure, by Ernest Vessiot
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+% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %
+%                                                                         %
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+% *** END OF THIS PROJECT GUTENBERG EBOOK LEÇONS DE GÉOMÉTRIE SUPÉRIEURE ***
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diff --git a/35052-t/images/src/015a.eepic b/35052-t/images/src/015a.eepic
new file mode 100644
index 0000000..4e5a03e
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diff --git a/35052-t/images/src/060a.eepic b/35052-t/images/src/060a.eepic
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diff --git a/35052-t/images/src/062a.eepic b/35052-t/images/src/062a.eepic
new file mode 100644
index 0000000..9c57f70
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diff --git a/35052-t/images/src/088a.eepic b/35052-t/images/src/088a.eepic
new file mode 100644
index 0000000..a4c9be3
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new file mode 100644
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new file mode 100644
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diff --git a/35052-t/images/src/141a.eepic b/35052-t/images/src/141a.eepic
new file mode 100644
index 0000000..ea83af6
--- /dev/null
+++ b/35052-t/images/src/141a.eepic
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diff --git a/35052-t/images/src/146a.eepic b/35052-t/images/src/146a.eepic
new file mode 100644
index 0000000..13a06f4
--- /dev/null
+++ b/35052-t/images/src/146a.eepic
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diff --git a/35052-t/images/src/151a.eepic b/35052-t/images/src/151a.eepic
new file mode 100644
index 0000000..ae93b41
--- /dev/null
+++ b/35052-t/images/src/151a.eepic
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diff --git a/35052-t/images/src/151b.eepic b/35052-t/images/src/151b.eepic
new file mode 100644
index 0000000..7a439a5
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diff --git a/35052-t/images/src/153a.eepic b/35052-t/images/src/153a.eepic
new file mode 100644
index 0000000..9df433f
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index 0000000..da7d25c
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new file mode 100644
index 0000000..78bc1d5
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new file mode 100644
index 0000000..8a96957
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diff --git a/35052-t/images/src/169a.eepic b/35052-t/images/src/169a.eepic
new file mode 100644
index 0000000..9cc819e
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diff --git a/35052-t/images/src/176a.eepic b/35052-t/images/src/176a.eepic
new file mode 100644
index 0000000..75c580f
--- /dev/null
+++ b/35052-t/images/src/176a.eepic
@@ -0,0 +1,281 @@
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diff --git a/35052-t/images/src/183a.eepic b/35052-t/images/src/183a.eepic
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new file mode 100644
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diff --git a/35052-t/images/src/210a.eepic b/35052-t/images/src/210a.eepic
new file mode 100644
index 0000000..458b718
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diff --git a/35052-t/images/src/289a.eepic b/35052-t/images/src/289a.eepic
new file mode 100644
index 0000000..8089e78
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diff --git a/35052-t/images/src/291a.eepic b/35052-t/images/src/291a.eepic
new file mode 100644
index 0000000..d8b7901
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index 0000000..7b0ec5d
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new file mode 100644
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new file mode 100644
index 0000000..c9b785c
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diff --git a/35052-t/images/src/preamble.tex b/35052-t/images/src/preamble.tex
new file mode 100644
index 0000000..05a5886
--- /dev/null
+++ b/35052-t/images/src/preamble.tex
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+%% Upright capital letters in math mode
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+\DeclareMathSymbol{H}{\mathalpha}{operators}{`H}
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+\DeclareMathSymbol{L}{\mathalpha}{operators}{`L}
+\DeclareMathSymbol{M}{\mathalpha}{operators}{`M}
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diff --git a/35052-t/old/35052-t.tex b/35052-t/old/35052-t.tex
new file mode 100644
index 0000000..9688732
--- /dev/null
+++ b/35052-t/old/35052-t.tex
@@ -0,0 +1,17138 @@
+% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %
+%                                                                         %
+% Project Gutenberg's Leçons de Géométrie Supérieure, by Ernest Vessiot   %
+%                                                                         %
+% This eBook is for the use of anyone anywhere at no cost and with        %
+% almost no restrictions whatsoever.  You may copy it, give it away or    %
+% re-use it under the terms of the Project Gutenberg License included     %
+% with this eBook or online at www.gutenberg.org                          %
+%                                                                         %
+%                                                                         %
+% Title: Leçons de Géométrie Supérieure                                   %
+%        Professées en 1905-1906                                          %
+%                                                                         %
+% Author: Ernest Vessiot                                                  %
+%                                                                         %
+% Editor: Anzemberger                                                     %
+%                                                                         %
+% Release Date: January 24, 2011 [EBook #35052]                           %
+%                                                                         %
+% Language: French                                                        %
+%                                                                         %
+% Character set encoding: ISO-8859-1                                      %
+%                                                                         %
+% *** START OF THIS PROJECT GUTENBERG EBOOK LEÇONS DE GÉOMÉTRIE SUPÉRIEURE ***
+%                                                                         %
+% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %
+
+\def\ebook{35052}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%                                                                  %%
+%% Packages and substitutions:                                      %%
+%%                                                                  %%
+%% book:     Required.                                              %%
+%% inputenc: Standard DP encoding. Required.                        %%
+%% babel:    French language features. Required.                    %%
+%%                                                                  %%
+%% calc:     Infix arithmetic. Required.                            %%
+%%                                                                  %%
+%% ifthen:   Logical conditionals. Required.                        %%
+%%                                                                  %%
+%% amsmath:  AMS mathematics enhancements. Required.                %%
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+%%                                                                  %%
+%%                                                                  %%
+%% Producer's Comments:                                             %%
+%%                                                                  %%
+%%   Minor spelling and punctuation corrections are marked with     %%
+%%   \DPtypo{original}{corrected}. Errata listed in the original    %%
+%%   typed manuscript are applied with \Err{}{}. Punctuation added  %%
+%%   for uniformity is marked with \Add{}. Spelling modernizations  %%
+%%   are marked with \DPchg{}{}. Other changes are [** TN: noted]   %%
+%%   in this file.                                                  %%
+%%                                                                  %%
+%%   The original typed manuscript contained an unusually large     %%
+%%   number of abbreviations, errors, and inconsistencies. To the   %%
+%%   extent feasible, these have been regularized. Particularly,    %%
+%%                                                                  %%
+%%   1. In Chapter 3, there were two sections "3". Section numbers  %%
+%%      3--7 were incremented to 4--8.                              %%
+%%                                                                  %%
+%%   2. The original used numerals for both cardinals and ordinals. %%
+%%      The \Card{} and \Ord{}{} macros convert these to words.     %%
+%%                                                                  %%
+%%   3. Exercises were moved to the end of the respective chapters. %%
+%%                                                                  %%
+%% PDF pages: 244                                                   %%
+%% PDF page size: A4 (210 × 297 mm)                                 %%
+%% PDF document info: filled in                                     %%
+%% 50 PDF diagrams.                                                 %%
+%%                                                                  %%
+%% Summary of log file:                                             %%
+%% * Six harmless overfull hboxes.                                  %%
+%% * One underfull vbox, sixteen underfull hboxes.                  %%
+%%                                                                  %%
+%%                                                                  %%
+%% Compile History:                                                 %%
+%%                                                                  %%
+%% January, 2011: adhere (Andrew D. Hwang)                          %%
+%%              texlive2007, GNU/Linux                              %%
+%%                                                                  %%
+%% Command block:                                                   %%
+%%                                                                  %%
+%%     pdflatex x3                                                  %%
+%%                                                                  %%
+%%                                                                  %%
+%% January 2011: pglatex.                                           %%
+%%   Compile this project with:                                     %%
+%%   pdflatex 35052-t.tex ..... THREE times                         %%
+%%                                                                  %%
+%%   pdfTeXk, Version 3.141592-1.40.3 (Web2C 7.5.6)                 %%
+%%                                                                  %%
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+\usepackage[french]{babel}    % the language
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+\usepackage{calc}[2005/08/06]
+
+\usepackage{ifthen}[2001/05/26]  %% Logical conditionals
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+\usepackage{amsmath}[2000/07/18] %% Displayed equations
+\usepackage{amssymb}[2002/01/22] %% and additional symbols
+\usepackage{mathrsfs}[1996/01/01]%% AMS script fonts
+
+\usepackage{alltt}[1997/06/16]   %% boilerplate, credits, license
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+\usepackage{array}[2005/08/23]   %% extended array/tabular features
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+\usepackage{indentfirst}[1995/11/23]
+
+\usepackage{graphicx}[1999/02/16]%% For a diagram,
+\usepackage{wrapfig}[2003/01/31] %% wrapping text around it,
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+% for running heads; no package date available
+\usepackage{fancyhdr}
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+\usepackage[body={5.6in,9.5in},hmarginratio=2:3]{geometry}[2002/07/08]
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+\providecommand{\ebook}{00000}    % Overridden during white-washing
+\usepackage[pdftex,
+  hyperfootnotes=false,
+  pdfkeywords={Andrew D. Hwang, Laura Wisewell,
+               Project Gutenberg Online Distributed Proofreading Team,
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+  pdfpagemode=UseNone, % default value
+  bookmarks=true,      % default value
+  linktocpage=false,   % default value
+  pdfpagelayout=TwoPageRight,
+  pdfdisplaydoctitle,
+  pdfpagelabels=true,
+  bookmarksopen=true,
+  bookmarksopenlevel=1,
+  colorlinks=true,
+  linkcolor=black]{hyperref}[2007/02/07]
+
+% Set title, author here to avoid numerous hyperref warnings from accents
+\hypersetup{pdftitle={The Project Gutenberg eBook \#\ebook:%
+    L'\texorpdfstring{Leçons de Géométrie Supérieure}{Lecons de Geometrie Superieure}},
+  pdfauthor={Ernest Vessiot}}
+
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%% COMMANDS %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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+\setlength{\headheight}{15pt}
+
+\newcommand{\SetPageNumbers}{\fancyhead[RO,LE]{\thepage}}
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+\newcommand{\SetHead}[1]{%
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+\AtBeginDocument{%
+  \renewcommand{\contentsname}{%
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+%[** TN: Using centered headings instead of marginal notes.]
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+
+\newcommand{\ExSection}[1]{%
+  \SetHead{EXERCICES}
+  \section*{\large\Heading\MakeUppercase{Exercices.}}
+  \pdfbookmark[1]{Exercices.}{Exercices.#1}
+}
+
+\newcommand{\Chapitre}[2]{%
+  \cleardoublepage
+  \thispagestyle{empty}
+  \SetPageNumbers
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+  \section*{\LARGE\Heading\MakeUppercase{Chapitre #1}.}
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+    \addcontentsline{toc}{chapter}{Chapitre~I. #2}%
+    \SetHead{CHAPITRE~I.}%
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+    \SetHead{CHAPITRE~#1.}%
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+
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+    \protect\subsection*{\protect\centering\protect\normalsize\protect#1}%
+  }
+}
+
+%\Section[ToC entry]{Centered Heading.}{Number.}{Run-in heading}
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+    \Input[#1]{#2}
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+  \ignorespaces
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+\newcommand{\Figure}[2][2in]{%
+  \begin{figure}[hbt]
+    \centering\Input[#1]{#2}
+  \end{figure}%
+  \ignorespaces
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+
+\newcommand{\Figures}[3][2in]{%
+  \begin{figure}[hbt]
+    \centering\Input[#1]{#2}\hfil\Input[#1]{#3}
+  \end{figure}%
+  \ignorespaces
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+
+\newenvironment{Exercises}{%
+  \begin{list}{}{%
+      \setlength{\leftmargin}{\parindent}%
+      \setlength{\labelwidth}{\parindent}%
+      \setlength{\listparindent}{\parindent}%
+      \small%
+    }%
+  }{%
+  \end{list}%
+  \tb
+  \normalsize%
+}
+
+
+% Change from the book's list of errata
+\newcommand{\Err}[2]{#2}
+
+% Changes and notes made for stylistic or notational consistency
+\newcommand{\DPtypo}[2]{#2} % presumed error
+\newcommand{\DPchg}[2]{#2}  % modernization of spelling
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+\begin{document}
+
+\pagestyle{empty}
+\pagenumbering{alph}
+
+%%%% PG BOILERPLATE %%%%
+\phantomsection
+\pdfbookmark[0]{PG Boilerplate.}{Boilerplate}
+
+\begin{center}
+\begin{minipage}{\textwidth}
+\small
+\begin{PGtext}
+Project Gutenberg's Leçons de Géométrie Supérieure, by Ernest Vessiot
+
+This eBook is for the use of anyone anywhere at no cost and with
+almost no restrictions whatsoever.  You may copy it, give it away or
+re-use it under the terms of the Project Gutenberg License included
+with this eBook or online at www.gutenberg.org
+
+
+Title: Leçons de Géométrie Supérieure
+       Professées en 1905-1906
+
+Author: Ernest Vessiot
+
+Editor: Anzemberger
+
+Release Date: January 24, 2011 [EBook #35052]
+
+Language: French
+
+Character set encoding: ISO-8859-1
+
+*** START OF THIS PROJECT GUTENBERG EBOOK LEÇONS DE GÉOMÉTRIE SUPÉRIEURE ***
+\end{PGtext}
+\end{minipage}
+\end{center}
+
+\clearpage
+
+
+%%%% Credits and transcriber's note %%%%
+\begin{center}
+\begin{minipage}{\textwidth}
+\begin{PGtext}
+Produced by Andrew D. Hwang, Laura Wisewell, Pierre Lacaze
+and the Online Distributed Proofreading Team at
+http://www.pgdp.net (The original copy of this book was
+generously made available for scanning by the Department
+of Mathematics at the University of Glasgow.)
+\end{PGtext}
+\end{minipage}
+\end{center}
+\vfill
+
+\begin{minipage}{0.85\textwidth}
+\small
+\pdfbookmark[0]{Note sur la Transcription.}{Note sur la Transcription}
+\subsection*{\centering\normalfont\scshape%
+\normalsize\MakeLowercase{Notes sur la transcription}}%
+
+\raggedright
+  Ce livre a été réalisé à l'aide d'un manuscrit dactylographié, dont
+  les images ont été fournies par le Département des Mathématiques de
+  l'Université de Glasgow.
+  \bigskip
+
+  Des modifications mineures ont été apportées à la présentation,
+  l'orthographe, la ponctuation et aux notations mathématiques. Le
+  fichier \LaTeX\ source contient les notes de ces corrections.
+\end{minipage}
+
+%% -----File: 001.png---Folio xx-------
+\clearpage
+\frontmatter
+\setlength{\TmpLen}{18pt}%
+\begin{center}
+\small PUBLICATIONS DU LABORATOIRE DE MATHÉMATIQUES \\[\TmpLen]
+\textsf{\bfseries De l'Université de Lyon}
+
+\vfill
+% [** Decoration]
+
+\textbf{\LARGE LEÇONS} \\[2\TmpLen]
+\footnotesize DE \\[3\TmpLen]
+\textbf{\Huge GÉOMÉTRIE SUPÉRIEURE} \\[2\TmpLen]
+\textsf{\bfseries Professées en 1905--1906} \\[2\TmpLen]
+\Large PAR M. E. VESSIOT \\[2\TmpLen]
+\textsc{\small Rédigées par M. ANZEMBERGER}
+\vfill
+% [** Decoration]
+\vfill
+\setlength{\TmpLen}{9pt}
+\footnotesize IMPRIMERIES RÉUNIES \\[\TmpLen]
+\scriptsize ANCIENNES MAISONS \\[\TmpLen]
+\normalsize DELAROCHE ET SCHNEIDER \\[\TmpLen]
+\scriptsize\textsf{\bfseries 8, rue Rachais} \\[\TmpLen]
+\footnotesize BUREAUX $\bigl\{$%
+\settowidth{\TmpLen}{\scriptsize\textit{85, rue de la République}}%
+\parbox[l]{\TmpLen}{\scriptsize\itshape%
+    85, rue de la République \\
+    9, quai de l'Hôpital} \\[9pt]
+\textsf{\bfseries\footnotesize LYON}
+\end{center}
+%% -----File: 002.png---Folio xx-------
+\clearpage
+\pagestyle{fancy}
+\fancyhf{}
+\thispagestyle{empty}
+\SetPageNumbers
+\SetHead{TABLE DES MATIÉRES}
+\tableofcontents
+
+\iffalse
+TABLE  DES  MATIERES.
+
+Pages
+
+CHAPITRE I.--REVISION DES POINTS ESSENTIELS DE LA THEORIE DES
+COURBES GAUCHES ET DES SURFACES DEVELOPPABLES:
+
+I.--Courbes gauches:
+
+1. Trièdre de Serret-Frenet............................  1
+
+2. Formules de Serret-Frenet...........................  2
+
+3. Courbure et Torsion.................................  4
+
+4. Discussion. Centre de Courbure .....................  5
+
+5. Signe de la torsion. Forme de la courbe.............  6
+
+6. Mouvement du trièdre de Serret-Frenet...............  8
+
+7. Calcul de la courbure...............................  9
+
+8. Calcul de la Torsion................................ 10
+
+9. Sphère osculatrice.................................. 11
+
+II.--Surfaces développables:
+
+10. Propriétés générales............................... 12
+
+11. Réciproques........................................ 15
+
+12. Surface rectifiante. Surface polaire............... 16
+
+CHAPITRE II.--SURFACES.
+
+1. Courbes tracées sur une surface. Longueurs d'arc et
+angles................................................. 19
+
+2. Déformation et représentation conforme.............. 20
+
+3. Les directions conjuguées et la forme \Sigma ld^2x.. 24
+
+4. Formules fondamentales pour une courbe de la surface 27
+%% -----File: 003.png---Folio xx-------
+
+CHAPITRE III.--ETUDE DES ELEMENTS FONDAMENTAUX DES COURBES
+D'UNE SURFACE.
+
+1. Courbure normale................................... 33
+
+2. Variations de la courbure normale.................. 35
+
+3. Lignes minima...................................... 40
+
+3. Lignes asymptotiques............................... 43
+
+4. Surfaces minima.................................... 48
+
+5. Lignes de courbure................................. 50
+
+6. Courbure géodésique. Propriétés des géodésiques.... 52
+
+7. Torsion géodésique. Théorèmes de Joachimsthal...... 57
+
+CHAPITRE IV.--LES SIX INVARIANTS.--LA COURBURE TOTALE.
+
+1. Les six invariants................................. 61
+
+2. Les conditions d'intégrabilité..................... 66
+
+3. Courbure totale.................................... 69
+
+4. Coordonnées orthogonales et isothermes............. 71
+
+5. Relations entre la courbure totale et la courbure
+géodésique............................................ 74
+
+CHAPITRE V.--SURFACES REGLEES.
+
+1. Surfaces développables............................. 80
+
+2. Développées des courbes gauches.................... 84
+
+3. Lignas de courbure................................. 87
+
+4. Développement d'une surface développable sur un plan
+Réciproque............................................ 89
+
+5. Lignes géodésiques d'une surface développable...... 93
+
+6. Surfaces réglées gauches--trajectoires orthogonales
+des génératrices...................................... 97
+
+7. Cône directeur. Point central. Ligne de striction.. 98
+%% -----File: 004.png---Folio xx-------
+
+8. Variations du plan tangent le long d'une génératrice.. 101
+
+9. Elément linéaire...................................... 106
+
+10. La forme \Sigma ld^2x et les lignes asymptotiques.... 110
+
+11. Lignes de courbure................................... 118
+
+12. Centre de courbure géodésique........................ 118
+
+CHAPITRE VI.--CONGRUENCES DE DROITES[**.]
+
+1. Points et plans focaux................................ 121
+
+2. Développables de la congruence. Examen des divers
+cas possibles. Cas singuliers ........................... 129
+
+3. Sur le point de vue corrélatif. Congruences de
+Koenigs. Surfaces de Joachimsthal........................ 136
+
+4. Détermination des développables d'une congruence...... 145
+
+CHAPITRE VII.--CONGRUENCES DE NORMALES.
+
+1. Propriété caractéristique des congruences de normales. 150
+
+2. Relations entre une surface et sa développée. Surface
+canal. Cyclide de Dupin. Cas singulier................... 153
+
+3. Etude des surfaces enveloppes de sphères. Correspondance
+entre les droites et les sphères. Equation de
+la cyclide de Dupin. Surface canal isotrope.............. 158
+
+4. Lignes de courbure et lignes asymptotiques. Bandes
+asymptotiques et bandes de courbure...................... 164
+
+5. Lignes de courbure des enveloppes de sphères.......... 168
+
+6. Cas où l'une des nappes de la développée est une
+développable............................................. 171
+
+CHAPITRE VIII.--LES CONGRUENCES DE DROITES ET LES CORRESPONDANCES
+ENTRE DEUX SURFACES.
+
+1. Nouvelle représentation des congruences............... 181
+
+2. Emploi des coordonnées homogènes...................... 183
+%% -----File: 005.png---Folio xx-------
+
+3. Correspondance entre les points M, M_1 de deux
+surfaces, telle que les développables de la
+congruence des droites MM_1 coupent les deux
+surfaces suivant deux réseaux conjugués homologues.... 189
+
+4. Correspondance par plans tangents parallèles....... 197
+
+CHAPITRE IX.--COMPLEXES DE DROITES.
+
+1. Eléments fondamentaux d'un complexe de droites..... 201
+
+2. Surfaces du complexe............................... 205
+
+3. Complexes spéciaux. Surface des singularités. Surfaces
+et courbes des complexes spéciaux..........  211
+
+4. Surfaces normales aux droites du complexe.......... 218
+
+CHAPITRE X.--COMPLEXES LINEAIRES.
+
+1. Généralités sur les complexes algébriques............ 220
+
+2. Coordonnées homogènes................................ 221
+
+3. Complexe linéaire.................................... 226
+
+4. Faisceau de complexes linéaires...................... 226
+
+5. Complexes linéaires en involution.................... 228
+
+6. Droites conjuguées................................... 230
+
+7. Réseau de complexes linéaires........................ 235
+
+8. Courbes d'un complexe linéaire. Leurs propriétés..... 236
+
+9. Surfaces normales aux droites d'un complexe linéaire. 240
+
+10. Surfaces réglées d'un complexe linéaire............. 243
+
+CHAPITRE XI.--TRANSFORMATIONS DUALISTIQUES. TRANSFORMATION DE
+SOPHUS LIE.
+
+1. Eléments de contact et multiplicités................. 245
+
+2. Transformations de contact. Transformations dualistiques.... 249
+%% -----File: 006.png---Folio xx-------
+
+3. Transformation de Sophus Lie........................ 255
+
+4. Transformation des droites en sphères............... 260
+
+5. Transformation des lignes asymptotiques............. 263
+
+6. Transformation des lignes de courbure............... 265
+
+CHAPITRE XII.--SYSTEMES TRIPLES ORTHOGONAUX.
+
+1. Théorème de Dupin................................... 268
+
+2. Equation aux dérivées partielles de Darboux......... 269
+
+3. Systèmes triples orthogonaux contenant une surface.. 274
+
+4. Systèmes triples orthogonaux contenant une famille de
+plans.................................................. 275
+
+5. Systèmes triples orthogonaux contenant une famille de
+sphères ............................................... 275
+
+CHAPITRE [** VIII missing].--CONGRUENCES DE SPHERES ET SYSTEMES CYCLIQUES.
+
+1. Généralités......................................... 280
+
+2. Congruences spéciales............................... 283
+
+3. Théorème de Dupin................................... 285
+
+4. Congruence des droites D............................ 289
+
+5. Congruence des droites \Delta....................... 291
+
+6. Le système triple de Ribaucour...................... 293
+
+7. Congruences de cercles et systèmes cycliques.
+Transformation de contact de Ribaucour................. 294
+
+8. Surfaces de Weingarten.............................. 301
+
+EXERCICES.                                             307
+\fi
+%% -----File: 007.png---Folio xx-------
+
+
+\Preface
+
+Ces leçons ont été professées en 1905--1906, pour répondre
+au programme spécial d'Analyse Mathématique de l'Agrégation.
+Elles ont été autographiées à la demande de mes étudiants, et
+rédigées par l'un d'eux.
+
+Peut-être pourront-elles être utiles aux étudiants désireux
+de s'initier à la géométrie supérieure, et leur être
+une bonne préparation à l'étude des livres de M.~Darboux et
+des mémoires originaux.
+
+J'ai supposé connus seulement les principes les plus
+simples de la théorie du contact; j'ai repris les points essentiels
+de la théorie des courbes gauches et de la théorie
+des surfaces, en mettant en évidence le rôle essentiel des
+formules de Frenet et des deux formes quadratiques différentielles
+de Gauss.
+
+L'objet principal de mes leçons était l'étude des systèmes
+de droites, et leur application à la théorie des surfaces.
+Il était naturel d'y joindre l'étude des systèmes de
+sphères, que j'ai poussée jusqu'aux propriétés élémentaires,
+si attrayantes, des systèmes cycliques de Ribaucour. J'ai
+insisté sur la correspondance des droites et des sphères, je
+l'ai éclairée par l'emploi des notions d'éléments de contact
+et de multiplicités, qui est également utile dans la théorie
+des congruences de droites; j'ai montré comment elle se traduisait
+par la transformation de contact de Lie.
+%% -----File: 008.png---Folio xx-------
+
+J'ai cherché à développer les diverses questions par
+la voie la plus naturelle et la plus analytique; voulant
+montrer à mes élèves comment la recherche méthodique, la discussion
+approfondie des questions même les plus simples,
+l'étude attentive et l'interprétation des résultats conduisent
+aux \DPtypo{consequences}{conséquences} les plus intéressantes.
+
+\null\hfil\hfil
+\parbox[c]{2in}{\centering
+Le 1\textsuperscript{er} Juin 1906. \\
+\textsc{E.~Vessiot}.}\hfil
+%% -----File: 009.png---Folio 1-------
+
+\mainmatter
+
+\Chapitre{Premier}{Révision des points essentiels de la théorie des Courbes Gauches et des Surfaces \DPtypo{Developpables}{Développables}.}
+
+\SubChap{I. Courbes Gauches.}
+
+\Section{Trièdre de Serret-Frenet\Add{.}}
+{1.}{} Les coordonnées d'un point d'une courbe gauche peuvent
+s'exprimer en fonction d'un paramètre~$t$
+\[
+x = f(t)\Add{,}\qquad y = g(t)\Add{,}\qquad z = h(t)\Add{.}
+\]
+Nous considérerons dans une telle courbe la \emph{tangente}, qui a
+pour paramètres directeurs $\dfrac{dx}{dt}$, $\dfrac{dy}{dt}$, $\dfrac{dz}{dt}$ et le \emph{plan osculateur}
+qui contient la tangente $\left(\dfrac{dx}{dt}, \dfrac{dy}{dt}, \dfrac{dz}{dt}\right)$ et l'accélération
+$\left(\dfrac{d^2x}{dt^2}, \dfrac{d^2y}{dt^2}, \dfrac{d^2z}{dt^2}\right)$ et dont par suite les coefficients sont les
+déterminants du \Ord{2}{e} ordre déduits du tableau
+\[
+\begin{Vmatrix}
+\mfrac{dx}{dt}     & \mfrac{dy}{dt}     & \mfrac{dz}{dt} \\
+\mfrac{d^2x}{dt^2} & \mfrac{d^2y}{dt^2} & \mfrac{d^2z}{dt^2}
+\end{Vmatrix}
+\]
+
+\Paragraph{Remarque.} Si on change de paramètre, en posant $t = \phi(u)$\Add{,}
+l'accélération nouvelle $\left(\dfrac{d^2x}{du^2}, \dfrac{d^2y}{du^2}, \dfrac{d^2z}{du^2}\right)$ est toujours dans le
+plan osculateur.
+
+Considérons en un point~$M$ d'une courbe la tangente~$MT$,
+la normale située dans le plan osculateur, ou \emph{normale principale}~$MN$,
+et la normale~$MB$ perpendiculaire au plan osculateur,
+ou \emph{binormale}. Ces \Card{3} droites forment un trièdre trirectangle
+que nous \DPtypo{appelerons}{appellerons} \emph{trièdre de Serret ou de Frenet}. L'une de
+ses faces, celle déterminée par la tangente et la normale principale,
+est le plan osculateur; celle déterminée par la normale
+principale et la binormale est le plan normal; enfin celle
+déterminée par la tangente et la binormale s'appelle le \emph{plan
+%% -----File: 010.png---Folio 2-------
+rectifiant}.
+
+Prenons sur la courbe une origine des arcs quelconques,
+et un sens des arcs croissants également quelconque. La différentielle
+de l'arc~$s$ est donnée par la formule
+\[
+ds^2 = dx^2 + dy^2 + dz^2
+\]
+d'où
+\[
+\frac{ds}{dt}
+  = ± \sqrt{\left(\frac{dx}{dt}\right)^2
+          + \left(\frac{dy}{dt}\right)^2
+          + \left(\frac{dz}{dt}\right)^2}
+\]
+et
+\[
+\left(\frac{dx}{ds}\right)^2
+  + \left(\frac{dy}{ds}\right)^2
+  + \left(\frac{dz}{ds}\right)^2 = 1\Add{,}
+\]
+$\dfrac{dx}{ds}$, $\dfrac{dy}{ds}$, $\dfrac{dz}{ds}$ sont ainsi les cosinus directeurs d'une des directions
+de la tangente, celle qui correspond au sens des arcs
+croissants; soient $a\Add{,} b\Add{,} c$ ces cosinus directeurs, nous avons
+\[
+\Tag{(1)}
+a = \frac{dx}{ds}\Add{,}\qquad
+b = \frac{dy}{ds}\Add{,}\qquad
+c = \frac{dz}{ds}\Add{.}
+\]
+
+Nous prendrons sur la normale principale une direction
+positive arbitraire de cosinus directeurs $a'\Add{,} b'\Add{,} c'$ et sur
+la binormale une direction positive de cosinus directeurs $a''\Add{,} b''\Add{,} c''$
+telle que le trièdre constitué par ces \Card{3} directions
+ait même disposition que le trièdre de coordonnées. On a alors
+\[
+\begin{vmatrix}
+a   & b   & c \\
+a'  & b'  & c' \\
+a'' & b'' & c''
+\end{vmatrix}
+= 1
+\]
+et chaque élément de ce déterminant est égal à son coefficient.
+
+\Section{Formules de Serret-Frenet\Add{.}}
+{2.}{} Il existe entre ces cosinus directeurs et leurs différentielles
+des relations importantes. Nous avons en effet
+\[
+a^2 + b^2 + c^2 = 1
+\]
+d'où en dérivant par rapport à~$s$
+\[
+\sum a\, \frac{da}{ds} = 0\Add{.}
+\]
+Mais d'après les relations~\Eq{(1)} on a
+\[
+\frac{da}{ds} = \frac{d^2x}{ds^2}\Add{,}\qquad
+\frac{db}{ds} = \frac{d^2y}{ds^2}\Add{,}\qquad
+\frac{dc}{ds} = \frac{d^2z}{ds^2}\Add{,}
+\]
+et la relation précédente s'écrit:
+%% -----File: 011.png---Folio 3-------
+\[
+\sum a\, \frac{d^2x}{\DPtypo{ds}{ds^2}} = 0\Add{.}
+\]
+
+La direction de coefficients directeurs
+\[
+\frac{d^2x}{ds^2},\quad
+\frac{d^2y}{ds^2},\quad
+\frac{d^2z}{ds^2}\quad \text{ou}\quad
+\frac{da}{ds},\quad
+\frac{db}{ds},\quad
+\frac{dc}{ds}
+\]
+est donc perpendiculaire à la tangente; d'autre part elle est
+dans le plan osculateur, c'est donc la normale principale, et
+on a des relations de la forme:
+\[
+\Tag{(2)}
+\frac{da}{ds} = \frac{1}{R}\, a'\Add{,}\qquad
+\frac{db}{ds} = \frac{1}{R}\, b'\Add{,}\qquad
+\frac{dc}{ds} = \frac{1}{R}\, c'\Add{.}
+\]
+On en déduit, pour le facteur $\dfrac{1}{R}$,
+\[
+\Tag{(3)}
+\frac{1}{R} = \sum a'\, \frac{da}{ds}\Add{.}
+\]
+De ces relations\DPnote{(2)} on tire, en multipliant par $a''\Add{,} b''\Add{,} c''$ et
+ajoutant
+\[
+\sum a''\, \frac{da}{ds} = 0\Add{.}
+\]
+D'autre part on a
+\[
+\sum aa'' = 0
+\]
+d'où en dérivant
+\[
+\sum a\, \frac{da''}{ds} + \sum a''\, \frac{da}{ds} = 0
+\]
+et par suite
+\[
+\sum a\, \frac{da''}{ds} = 0\Add{.}
+\]
+On a d'ailleurs
+\[
+\sum a''{}^2 = 1
+\]
+d'où
+\[
+\sum a''\, \frac{da''}{ds} = 0
+\]
+et les deux relations précédentes montrent que la direction
+$\dfrac{da''}{ds}, \dfrac{db''}{ds}, \dfrac{dc''}{ds}$ est perpendiculaire à la tangente et à la binormale.
+C'est donc encore la normale principale, et on a des
+relations de la forme\DPtypo{;}{:}
+\[
+\Tag{(4)}
+\frac{da''}{ds} = \frac{1}{T}\, a'\Add{,}\qquad
+\frac{db''}{ds} = \frac{1}{T}\, b'\Add{,}\qquad
+\frac{dc''}{ds} = \frac{1}{T}\, c'\Add{.}
+\]
+On en déduit, pour le facteur $\dfrac{1}{T}$,
+\[
+\Tag{(5)}
+\frac{1}{T} = \sum a'\, \frac{da''}{ds}\Add{.}
+\]
+Enfin de la relation
+\[
+\sum a'a'' = 0
+\]
+on tire
+\[
+\sum a'\, \frac{da''}{ds} + \sum a''\, \frac{da'}{ds} = 0\Add{,}
+\]
+ou
+\[
+\sum a''\, \frac{da'}{ds} = -\sum a'\, \frac{da''}{ds} = -\frac{1}{T}\Add{.}
+\]
+%% -----File: 012.png---Folio 4-------
+De la relation
+\[
+\sum a'a = 0
+\]
+on tire de même
+\[
+\sum a\, \frac{da''}{ds} = -\sum a'\, \frac{da}{ds} = -\frac{1}{R}\Add{,}
+\]
+et enfin de
+\[
+\sum a'{}^2 = 0
+\]
+on tire
+\[
+\sum a'\, \frac{da'}{ds} = 0\Add{.}
+\]
+On a ainsi \Card{3} équations en $\dfrac{da'}{ds}, \dfrac{db'}{ds}, \dfrac{dc'}{ds}$,
+\begin{align*}
+\sum a\, \frac{da'}{ds} &= -\frac{1}{R}\Add{,} \\
+\sum a'\, \frac{da'}{ds} &= 0\Add{,} \\
+\sum a''\, \frac{da'}{ds} &= -\frac{1}{T}\Add{,}
+\end{align*}
+et l'on en tire
+\[
+\Tag{(6)}
+\frac{da'}{ds} = -\frac{a}{R} - \frac{a''}{T},\qquad
+\frac{db'}{ds} = -\frac{b}{R} - \frac{b''}{T},\qquad
+\frac{dc'}{ds} = -\frac{c}{R} - \frac{c''}{T}.
+\]
+
+Les \Card{3} groupes de relations \Eq{(2)}\Add{,}~\Eq{(4)}\Add{,}~\Eq{(6)} constituent \emph{les
+formules de Serret ou de Frenet}.
+
+\Section{Courbure et \DPtypo{Torsion}{torsion}\Add{.}}
+{3.}{Interprétation de~$R$.} Considérons le point~$t$ de coordonnées
+$a\Add{,} b\Add{,} c$. Les formules~\Eq{(2)} expriment une propriété de la
+courbe lieu de ces points; cette courbe est tracée sur une
+sphère de rayon~$1$, on l'appelle \emph{indicatrice
+sphérique} de la courbe~$(C)$,
+et les formules~\Eq{(2)} montrent que \emph{la
+tangente en~$t$ à l'indicatrice sphérique
+est parallèle à la normale
+principale en~$M$ à la courbe~$C$}. Soit $u$
+l'arc de cette indicatrice compté à
+partir d'une origine arbitraire dans
+un sens également arbitraire, on aura
+\[
+\frac{da}{du} = ea',\qquad
+\frac{db}{du} = eb',\qquad
+\frac{dc}{du} = ec',\qquad (e = ±1)
+\]
+
+\Illustration[1.75in]{012a}
+\noindent d'où, en tenant compte des formules~\Eq{(2)}
+\[
+\frac{1}{R} = e\, \frac{du}{ds}\Add{.}
+\]
+%% -----File: 013.png---Folio 5-------
+
+Considérons alors les points $t$,~$t'$ correspondant aux
+points $M$,~$M'$; $\dfrac{du}{ds}$~est la limite du rapport $\dfrac{\arc tt'}{\arc MM'}$ quand $M'$~se
+rapproche indéfiniment de~$M$. L'arc~$tt'$ étant infiniment petit
+peut être remplacé par l'arc de grand cercle correspondant,
+qui n'est autre que la mesure de l'angle~$tOt'$ des \Card{2} tangentes
+infiniment voisines; c'est \emph{l'angle de contingence}; cette limite
+s'appelle la \emph{courbure} de la courbe au point~$C$; $R$~est le \emph{rayon
+de courbure}.
+
+\Paragraph{Interprétation de~$T$.} Pour interpréter~$T$, on \DPtypo{considèrera}{considérera}
+de même le lieu du point~$b$ de coordonnées $a\Add{,} b\Add{,} c$, ou \emph{deuxième
+indicatrice sphérique}. On pourra remarquer que d'après les
+formules \Eq{(2)}\Add{,}~\Eq{(4)}, \emph{les tangentes en $t$,~$b$ aux deux indicatrices
+sont parallèles à la normale principale en~$M$}. Si $v$~est l'arc
+de cette deuxième indicatrice sphérique, on trouvera comme
+précédemment que
+\[
+\frac{1}{T} = e'\, \frac{dv}{ds}\qquad (e' = ±1)
+\]
+et que $\dfrac{1}{T}$ est la limite du rapport de l'angle des plans
+osculateurs en $M$,~$M'$ à l'arc~$MM'$; c'est la \emph{torsion} en~$M$, et $T$~est
+le \emph{rayon de torsion}.
+
+\emph{Les deux indicatrices sont polaires réciproques sur la
+sphère.}
+
+
+\Section{Discussion. Centre de courbure.}
+{4.}{} Les cosinus directeurs que nous avons introduits dépendent
+de \Card{3} hypothèses arbitraires sur la disposition du
+trièdre de coordonnées, le sens des arcs croissants, et le sens
+positif choisi sur la normale principale. Si nous changeons
+ces hypothèses, et si nous désignons par $e_1, e_2, e_3$ des nombres
+égaux à~$±1$, $s$~sera remplacé par~$e_1s$, $a\Add{,} b\Add{,} c$ par $e_1a, e_1b, e_1c$;
+$a'\Add{,} b'\Add{,} c'$ par $e_2a, e_2b, e_2c$; et enfin, d'après les relations
+%% -----File: 014.png---Folio 6-------
+\[
+a'' = e_3 (bc' - cb'),\qquad
+b'' = e_3 (ca' - ac'),\qquad
+c'' = e_3 (ab' - ba'),
+\]
+$a''\Add{,} b''\Add{,} c''$ seront remplacés par $e_1e_2e_3a'', e_1e_2e_3b'', e_1e_2e_3c''$. Les
+formules~\Eq{(2)} donnent alors
+\[
+\frac{e_1\, da}{e_1\, ds} = \frac{1}{R}\, e_2a',\qquad \text{etc}\ldots,
+\]
+c'est à dire $R$~se change en~$e_2R$; et son signe ne dépend que
+de la direction positive choisie sur la normale principale.
+
+Donc le point~$C$ de la normale principale, tel que l'on
+%[** TN: MC has an overline accent; only instance, omitting]
+ait $MC = R$ ($R$~étant défini algébriquement comme précédemment),
+est un élément géométrique attaché à la courbe donnée. Ce
+point~$C$ s'appelle \emph{centre de courbure en~$M$}.
+
+Voyons maintenant~$T$. Les formules~\Eq{(4)} donnent
+\begin{align*}
+\frac{e_1e_2e_3\, da''}{\DPtypo{e}{e_1}\, ds}
+  &= \frac{1}{T}\, e_2a'\Add{,}\qquad \text{etc.} \\
+\intertext{ou}
+\frac{e_3\, da''}{ds} &= \frac{1}{T}\, a'\Add{,}\qquad \text{etc.}
+\end{align*}
+
+Donc $T$~se change en~$e_3T$; et le signe de~$T$ dépend uniquement
+de la disposition du trièdre de coordonnées. Il n'y a
+donc pas lieu de définir un centre de torsion.
+
+\Section{Signe de la torsion. Forme de la courbe.}
+{5.}{} Pour interpréter le signe de~$T$, nous allons étudier
+la rotation d'un plan passant par la tangente~$MT$ et par un
+point~$M'$ de la courbe infiniment voisin. Rapportons la courbe
+au trièdre de Serret, la tangente étant~$OX$, la normale principale~$OY$,
+la binormale~$OZ$. Alors $a = 1$, $a' = 0$, $a'' = 0$, $b = 0$,
+$b' = 1$, $b'' = 0$, $c = 0$, $c' = 0$, $c'' = 1$. Nous allons chercher
+les développements des coordonnées d'un point de la courbe
+infiniment voisin de~$M$ suivant les puissances croissantes de~$ds$,
+($ds$~étant l'arc de la courbe compté à partir du point~$O$).
+
+Nous avons
+\begin{align*}%[** TN: Added elided equations]
+X &= \frac{ds}{1}\, \frac{dx}{ds}
+   + \frac{ds^2}{2}\, \frac{d^2x}{ds^2}
+   + \frac{ds^3}{6}\, \frac{d^3x}{ds^3} + \dots\Add{,} \\
+Y &= \frac{ds}{1}\, \frac{dy}{ds}
+   + \frac{ds^2}{2}\, \frac{d^2y}{ds^2}
+   + \frac{ds^3}{6}\, \frac{d^3y}{ds^3} + \dots\Add{,} \\
+Z &= \frac{ds}{1}\, \frac{dz}{ds}
+   + \frac{ds^2}{2}\, \frac{d^2z}{ds^2}
+   + \frac{ds^3}{6}\, \frac{d^3z}{ds^3} + \dots\Add{.}
+\end{align*}
+%% -----File: 015.png---Folio 7-------
+
+Or:
+\begin{align*}
+\frac{dx}{ds} &= a = 1\Add{,} \\
+\frac{d^2x}{ds^2} &= \frac{da}{ds} = \frac{a'}{R} = 0\Add{,} \\
+\frac{d^3x}{ds^3} &= \frac{d^2a}{ds^2}
+  = \frac{1}{R}\, \frac{da'}{ds} + \frac{d\left(\dfrac{1}{R}\right)}{ds}\, a'
+  = \frac{1}{R} \left(-\frac{a}{R} - \frac{a''}{T}\right) - \frac{1}{R^2}\, a'\, \frac{dR}{ds}
+  = -\frac{1}{R^2}\Add{,}
+\end{align*}
+et de même pour les autres coordonnées. On trouve ainsi
+\[
+\Tag{(7)}%[** TN: Added brace]
+\left\{
+\begin{aligned}
+X &= ds &                    &-\frac{1}{6R^2}\, \Err{ds}{ds^3} + \dots\Add{,} \\
+Y &=    &\frac{1}{2R}\, ds^2 &-\frac{1}{6R^2}\, \frac{dR}{ds}\, ds^3 + \dots\Add{,} \\
+Z &=    &                    &-\frac{1}{6RT}\, ds^3 + \dots\Add{.}
+\end{aligned}
+\right.
+\]
+Tels sont les \DPtypo{developpements}{développements} des coordonnées, du point~$M'$ voisin
+de~$M$.
+
+Le plan que nous considérons passe par la tangente; le
+sens de sa rotation est donné par le signe de~$\dfrac{Z}{Y}$, coefficient
+angulaire de sa trace sur le plan des~$YZ$. Or,
+\[
+\frac{Z}{Y} = -\frac{ds}{3T}\, \bigl[1 + ds\, (\dots\dots)\bigr]\Add{.}
+\]
+
+\Illustration[3in]{015a}
+Ce coefficient angulaire est positif si $T\Err{\ }{<}0$, pour $s$~croissant,
+c'est à dire si le point se déplace dans la direction
+de la tangente; le plan va alors tourner dans le sens positif.
+Le point~$M'$ étant au-dessus du
+plan des~$\DPtypo{xy}{XY}$, l'arc~$MM'$ de la courbe
+est en avant du plan~$XZ$, si $T<0$; il
+est au contraire en arrière si $T>0$.
+
+Les formules~\Eq{(7)} permettent de
+représenter les projections de la
+courbe sur les \Card{3} faces du trièdre de
+Serret dans le voisinage du point~$M$.
+%% -----File: 016.png---Folio 8-------
+Nous supposerons pour faire ces projections $R > 0$ et $T < 0$.
+
+La considération des formules~\Eq{(7)} prises deux à deux
+montre que sur le plan rectifiant~$(XZ)$ la projection a au
+point~$m_1$ un point d'inflexion, la tangente inflexionnelle étant~$OX$.
+Sur le plan osculateur, la projection a au point~$m$ un point
+ordinaire, la tangente étant~$OX$; enfin sur le plan normal~$(Y\Add{,}Z)$
+la projection a en~$m_2$ un point de rebroussement, la tangente de
+rebroussement étant~$OY$.
+
+
+\Section{Mouvement du trièdre de Serret-Frenet.}
+{6.}{Remarque.} Considérons un point~$P$ invariablement lié
+au trièdre de Serret, et soient $X\Add{,} Y\Add{,} Z$ ses coordonnées constantes
+par rapport à ce trièdre; soient $\xi, \eta, \zeta$ les coordonnées de
+ce point par rapport à un système d'axes fixes. Lorsque le sommet
+du trièdre de Serret décrit la courbe donnée, les projections
+de la vitesse du point~$P$ sur les axes fixes sont, en remarquant
+que l'on a
+\begin{gather*}
+\begin{alignedat}{4}
+\xi   &= x &&+ aX &&+ a'Y &&+ a''Z, \\
+\eta  &= y &&+ bX &&+ b'Y &&+ b''Z, \\
+\zeta &= z &&+ cX &&+ c'Y &&+ c''Z\Add{,}
+\end{alignedat} \\[6pt]
+\begin{alignedat}{4}
+\frac{d\xi}{dt}
+  &= \frac{dx}{dt} &&+ X\frac{da}{dt} &&+ Y\frac{da'}{dt} &&+ Z\frac{da''}{dt},\\
+\frac{d\eta}{dt}
+  &= \frac{dy}{dt} &&+ X\frac{db}{dt} &&+ Y\frac{db'}{dt} &&+ Z\frac{db''}{dt},\\
+\frac{d\zeta}{dt}
+  &= \frac{dz}{dt} &&+ X\frac{dc}{dt} &&+ Y\frac{dc'}{dt} &&+ Z\frac{dc''}{dt}\Add{,}
+\end{alignedat}
+\end{gather*}
+ou encore
+\begin{align*}%[** TN: Added elided equations]
+\frac{d\xi}{dt}
+  &= \frac{ds}{dt}\, a + X\, \frac{a'}{R} + Y \left(-\frac{a}{R} - \frac{a''}{T}\right) + Z\, \frac{a'}{T}, \\
+\frac{d\eta}{dt}
+  &= \frac{ds}{dt}\, b + X\, \frac{b'}{R} + Y \left(-\frac{b}{R} - \frac{b''}{T}\right) + Z\, \frac{b'}{T}, \\
+\frac{d\zeta}{dt}
+  &= \frac{ds}{dt}\, c + X\, \frac{c'}{R} + Y \left(-\frac{c}{R} - \frac{c''}{T}\right) + Z\, \frac{c'}{T}\Add{.}
+\end{align*}
+
+Les projections de la vitesse sur les axes mobiles sont
+alors
+\begin{alignat*}{4}
+V_x &= a\, \frac{d\xi}{dt} &&+ b\, \frac{d\eta}{dt} &&+ c\, \frac{d\zeta}{dt}
+     &&= \frac{ds}{dt} \left(1 - \frac{Y}{R}\right)\Add{,} \\
+V_y &=  a'\, \frac{d\xi}{dt} &&+ b'\, \frac{d\eta}{dt} &&+ c'\, \frac{d\zeta}{dt}
+     &&= \frac{ds}{dt} \left(\frac{X}{R} + \frac{Z}{T}\right)\Add{,} \\
+V_z &=  a''\, \frac{d\xi}{dt} &&+ b''\, \frac{d\eta}{dt} &&+ c''\, \frac{d\zeta}{dt}
+     &&= -\frac{ds}{dt}\, \frac{Y}{T}\Add{,}
+\end{alignat*}
+$\dfrac{ds}{dt}$ est la vitesse du sommet du trièdre. Si nous ne considérons
+que la vitesse de rotation, nous savons que, si $p\Add{,} q\Add{,} r$ sont les
+composantes de la rotation instantanée sur les axes mobiles,
+%% -----File: 017.png---Folio 9-------
+on a
+\[
+V_x = qZ - rY\Add{,}\qquad
+V_y = rX - pZ\Add{,}\qquad
+V_z = pY - qX\Add{,}
+\]
+et nous trouvons ainsi, en identifiant avec les expressions
+précédentes (dans l'hypothèse $t = s$)
+\[
+p = -\frac{1}{T}\Add{,}\qquad
+q = 0\Add{,}\qquad
+r = \frac{1}{R}\Add{,}
+\]
+ce qui montre qu'\emph{à chaque instant, la rotation instantanée
+est dans le plan rectifiant et a pour composantes suivant la
+tangente et la binormale la torsion et la courbure}.
+
+Si l'on suppose le trièdre de Serret transporté à l'origine,
+il tourne autour de son sommet, l'axe instantané de rotation
+est dans le plan rectifiant, et le mouvement du trièdre
+est obtenu par le roulement de ce plan sur un certain cône.
+
+
+\Section[Calcul de la courbure.]{Calcul de~$R$.}
+{7.}{} Reprenons la formule~\Eq{(3)}
+\[
+\frac{1}{R} = \sum a'\, \frac{da}{ds}\Add{.}
+\]
+Nous avons
+\[
+a = \frac{dx}{ds},
+\]
+d'où
+\[
+\frac{da}{ds} = \frac{ds\, d^2x - dx\, d^2s}{\DPtypo{ds^2}{ds^3}}.
+\]
+Soit maintenant
+\[
+A = dy\, d^2z - dz\, d^2y\DPtypo{.}{,}\qquad
+B = dz\, d^2x - dx\, d^2z,\qquad
+C = dx\, d^2y - dy\, d^2x,
+\]
+et posons
+\[
+\sqrt{A^2 + B^2 + C^2} = D.
+\]
+$A\Add{,} B\Add{,} C$ sont les coefficients du plan osculateur, et par suite
+les cosinus directeurs de la binormale sont
+\[
+a'' = \frac{A}{D}\Add{,}\qquad
+b'' = \frac{B}{D}\Add{,}\qquad
+c'' = \frac{C}{D}\Add{,}
+\]
+et les cosinus directeurs de la normale principale, perpendiculaire
+aux deux droites précédentes, sont
+\begin{align*}% [** TN: Elided equations not added]
+a' &= \frac{B\, dz - C\, dy}{D\, ds}
+    = \frac{\DPtypo{dx^2}{dx^2}\, (dz^2 + dy^2) - dx\, (dz\, \DPtypo{dz^2}{d^2z} + dy\, \DPtypo{dy^2}{d^2y})}{D\, ds} \\
+   &= \frac{\DPtypo{dx^2}{d^2x}·ds^2 - dx·ds\, \DPtypo{ds^2}{d^2s}}{D\, ds}
+    = \frac{ds\, d^2x - dx\, d^2s}{D}\Add{,} \\[6pt]
+b' &= \dots \qquad  c' = \dots\Add{,}
+\end{align*}
+\iffalse%%%%[** TN: Code for elided equations]
+b' &= \frac{C\, dx - A\, dz}{D\, ds}
+    = \frac{dy^2\, (dx^2 + dz^2) - dy\, (dx\, d^2x + dz\, d^2z)}{D\, ds} \\
+   &= \frac{d^2y·ds^2 - dy·ds\, d^2s}{D\, ds}
+    = \frac{ds\, d^2y - dy\, d^2s}{D}\Add{,} \\[6pt]
+c' &= \frac{A\, dy - B\, dx}{D\, ds}
+    = \frac{dz^2\, (dy^2 + dx^2) - dz\, (dy\, d^2y + dx\, d^2x)}{D\, ds} \\
+   &= \frac{d^2z·ds^2 - dz·ds\, d^2s}{D\, ds}
+    = \frac{ds\, d^2z - dz\, d^2s}{D}\Add{,}
+\fi %%%% End of code for elided equations
+%% -----File: 018.png---Folio 10-------
+et alors
+\[
+\frac{1}{R} = \sum a'\, \frac{da}{ds}
+  = \sum \frac{B\, dz - C\, dy}{D\DPtypo{}{\,ds}}\,
+    \frac{ds\, d^2x - dx\, d^2s}{\DPtypo{ds^2}{ds^3}}
+\]
+ce qui peut s'écrire
+%[** TN: Original exponents unclear, but math checked by hand.]
+\begin{align*}
+\frac{1}{R}
+  &= \frac{1}{D\, ds^3}  \sum d^2x\, (B\, dz - C\, dy)
+   - \frac{d^2s}{D\, ds^4} \sum dx\, (B\, dz - C\, dy) \\
+\intertext{et se réduit à:}
+\frac{1}{R}
+  &= \frac{1}{D\, ds^3} \sum d^2x\, (B\, dz - C\, dy)
+   = \frac{1}{D\, ds^3}
+  \begin{vmatrix}
+  dx   &  dy  &  dz  \\
+  d^2x & d^2y & d^2z \\
+    A  &  B   &  C
+  \end{vmatrix} = \frac{D}{ds^3}\Add{,}
+\end{align*}
+d'où enfin:
+\[
+\frac{1}{R} = \frac{\sqrt{\sum (dy\, d^2z\DPtypo{_}{-} dz\, d^2y)^2}}
+                   {(dx^2 + dy^2 + dz^2)^{\tfrac{3}{2}}}.
+\]
+
+
+\Section[Calcul de la torsion]{Calcul de~$T$.}
+{8.}{} On aura de même
+\begin{align*}
+\frac{1}{T}
+  &= \sum a'\, \frac{da''}{ds}
+   = \sum \frac{B\, dz \DPtypo{_}{-} C\, dy}{D·ds}\,
+          \frac{D·dA - A\, dD}{D^2\, ds} \\
+\intertext{ce qui peut s'écrire}
+\frac{1}{T}
+  &= \frac{1}{D^2\, \DPtypo{ds}{ds^2}} \sum dA\, (B\, dz - C\, dy)
+   - \frac{dD}{\DPtypo{D^2}{D^3}\, ds^2} \sum A\, (B\, dz - C\, dy)
+\end{align*}
+et se réduit à
+\begin{align*}
+\frac{1}{T}
+  &= \frac{1}{D^2\, ds^2} \sum dA\, (B\, dz - C\, dy)
+   = \frac{1}{D^2\, ds} \sum (dy\, d^3z - dz\, d^3y) (ds\, d^2x - dx\, d^2s) \\
+\intertext{ou}
+\frac{1}{T}
+  &= \frac{1}{D^2} \sum d^2x (dy\, d^3z - dz\, d^3y)
+   - \frac{d^2s}{D^2\, ds} \sum dx\, (dy\, d^3z - dz\, d^3y); \\
+\intertext{la \Ord{2}{e} somme est nulle, et il reste}
+\frac{1}{T}
+  &= \frac{1}{\DPtypo{D}{D^2}} \sum d^2x\, (dy\, d^3z - dz\, d^3y)
+   = -\frac{1}{D^2}
+     \begin{vmatrix}
+     dx &  dy  &  dz \\
+     d^2x & d^2y & d^2z \\
+     d^3x & d^3y & d^3z
+  \end{vmatrix}
+\end{align*}
+avec
+\[
+D^2 = \sum (dy\, d^2z - dz\, d^2y)^2\Add{.}
+\]
+
+\Paragraph{Remarque.} Pour que la torsion d'une courbe soit constamment
+nulle, il faut et il suffit que l'on ait constamment
+\[
+\begin{vmatrix}
+  dx &  dy  &  dz \\
+  d^2x & d^2y & d^2z \\
+  d^3x & d^3y & d^3z
+\end{vmatrix} = 0,
+\]
+ce qui exige que $x, y, z$ soient liés par une relation linéaire,
+à coefficients constants, c'est-à-dire que la courbe soit plane.
+Ainsi \emph{les courbes à torsion constamment nulle sont des
+%% -----File: 019.png---Folio 11-------
+courbes planes.}
+
+\Section{Sphère osculatrice.}
+{9.}{} Cherchons les sphères qui ont en~$M$, avec la courbe
+considérée, un contact du second ordre. Le centre $(x_0\Add{,} y_0\Add{,} z_0)$ et
+le rayon~$R_0$ d'une telle sphère sont, d'après la théorie du
+contact, déterminés par les équations suivantes, que nous développons
+au moyen des formules de Serret-Frenet:
+\begin{align*}
+&\sum (x - x_0)^2 - R_0^2 = 0, \\
+&\frac{d}{ds} \left\{ \sum (x - x_0)^2 - R^2 \right\} = 0,\quad
+  \text{\DPchg{c.à.d.}{cést-à-dire}}\quad \sum a(x - x_0) = 0, \\
+&\frac{d^2}{ds^2} \left\{ \sum (x - x_0)^2 - R^2 \right\} = 0,\quad
+  \text{\DPchg{c.à.d.}{cést-à-dire}}\quad 1 + \frac{1}{R} \sum a' (x - x_0) = 0.
+\end{align*}
+
+Si on prend le trièdre de Serret-Frenet pour trièdre
+de coordonnées, comme on l'a fait plus haut, elles se réduisent
+à
+\[
+\sum X_0 - R_0^2 = 0,\qquad X_0 = 0,\qquad Y_0 = -R;
+\]
+et l'équation générale des sphères cherchées est, $Z_0$~restant
+arbitraire,
+\[
+X^2 + Y^2 + Z^2 - 2RY - 2Z_0Z = 0\Add{.}
+\]
+
+C'est un faisceau de sphères, dont fait partie le plan
+osculateur $Z = 0$. On vérifie ainsi la propriété de contact
+du plan osculateur.
+
+Le cercle commun à toutes ces sphères est, de plus,
+d'après la théorie du contact des courbes, celui qui a un
+contact du second ordre avec la courbe, \DPchg{c.à.d.}{cést-à-dire} le \emph{cercle
+osculateur}. Les équations sont
+\[
+Z = 0,\qquad X^2 + Y^2 - 2RY = 0,
+\]
+\DPchg{c.à.d.}{cést-à-dire} qu'il est dans le plan osculateur, a pour centre
+le centre de courbure~$C$ ($X = 0$, $Y = R$), et passe en~$M$. Le
+lieu des \Err{}{centres des }sphères considérées est l'axe du cercle osculateur.
+
+Parmi toutes ces sphères, il y en a une qui a un contact
+%% -----File: 020.png---Folio 12-------
+du troisième ordre avec la courbe. On l'obtient en introduisant
+la condition nouvelle:
+\begin{gather*}
+\frac{d^3}{ds^3} \left\{ \sum (x - x_0)^2 - R^2 \right\} = 0\Add{,} \\
+\intertext{\DPchg{c.à.d.}{cést-à-dire}}
+-\frac{1}{R^2}\, \frac{dR}{ds} \sum a'(x - x_0)
+  -\frac{1}{R} \left\{ \frac{1}{R} \sum a(x - x_0) + \frac{1}{T} \sum a''(x - x_0) \right\} = 0,
+\end{gather*}
+qui se réduit, avec les axes particuliers employés, à
+\[
+Z_0 = -T\, \frac{dR}{ds}.
+\]
+Le centre de cette \emph{sphère osculatrice} est donc défini par les
+formules:
+\[
+X_0 = 0,\qquad Y_0 = -R,\qquad Z_0 = -T\, \frac{dR}{ds}.
+\]
+Et son rayon est donné par
+\[
+R_0^2 = R^2 + T^2\, \frac{dR^2}{ds^2}.
+\]
+
+
+\SubChap{II. Surfaces développables.}
+
+\Section{Propriétés générales.}
+{10.}{} Une courbe gauche est le lieu de $\infty^{1}$~points; corrélativement
+nous \DPtypo{considèrerons}{considérerons} une surface développable, enveloppe
+de $\infty^{1}$~plans; la caractéristique de l'un de ces plans
+correspond corrélativement à la tangente en un point de la
+courbe, puisqu'elle est l'intersection de deux plans infiniment
+voisins.
+
+Soit
+\[
+\Tag{(1)}
+uX + vY + wZ + h = 0,
+\]
+l'équation générale des plans considérés, de sorte que $u, v, w, h$
+désignent des fonctions données d'un paramètre~$t$.
+
+Les caractéristiques ont, d'après la théorie des enveloppes,
+pour équations générales,
+\[
+\Tag{(2)}
+\left\{
+\begin{aligned}
+&uX + uY + wZ + h = 0\Add{,} \\
+&du·X + dv·Y + dw·Z + dh = 0.
+\end{aligned}
+\right.
+\]
+
+La surface développable, enveloppe des plans~\Eq{(1)}, est,
+%% -----File: 021.png---Folio 13-------
+d'après la théorie des enveloppes, le lieu des droites~\Eq{(2)},
+qui en sont, par conséquent, les génératrices rectilignes;
+et, toujours d'après la théorie des enveloppes, chacun des
+plans~\Eq{(1)} est tangent à la surface tout le long de la génératrice~\Eq{(2)}
+correspondant à la même valeur de~$t$.
+
+Considérons alors la courbe~$(C)$, lieu des points $(x,y,z)$
+définis par les équations:
+\[
+\Tag{(3)}
+\left\{
+\begin{aligned}
+&ux + vy + wz + h = 0\Add{,} \\
+&\DPtypo{x\, du}{u\, dx} + v\, dy + w\, dz + dh = 0\Add{,} \\
+&\DPtypo{x\, d^2u}{u\, d^2x} + v\, d^2y + w\, d^2z + d^2h = 0\Add{.}
+\end{aligned}
+\right.
+\]
+
+L'un quelconque de ses points~$M$ est sur la droite~\Eq{(2)},
+correspondant à la même valeur de~$t$, et, par conséquent, dans
+le plan~\Eq{(1)} correspondant. Cherchons la tangente à~$(C)$ en~$M$.
+Il faut différentier les équations~\Eq{(3)}; différentiant chacune
+des deux premières, en tenant compte de la suivante, nous
+trouvons
+\[
+\Tag{(4)}
+\left\{
+\begin{aligned}
+&u·dx +  v·dy +  w·dz = 0\Add{,} \\
+&du·dx + dv·dy + dw·dz = 0,
+\end{aligned}
+\right.
+\]
+ce qui exprime que la direction de la tangente est la même
+que celle de la droite~\Eq{(2)}. Donc les tangentes à~$(C)$ sont
+les génératrices de la développable.
+
+Cherchons encore le plan osculateur à~$(C)$ en~$M$. Il doit
+passer par la tangente, et être parallèle à la direction $(d^2x, d^2y, d^2z)$.
+Or\Add{,} si on différentie la première des équations~\Eq{(4)},
+en tenant compte de la seconde, on trouve
+\[
+u·d^2x + v·d^2y + w·d^2z = 0,
+\]
+ce qui montre que le plan~\Eq{(1)} satisfait à ces conditions.
+Donc le plan osculateur de~$(C)$ est le plan qui enveloppe la
+développable.
+
+$(C)$~s'appelle l'\emph{arête de rebroussement} de la développable
+%% -----File: 022.png---Folio 14-------
+
+Donc \emph{toute développable est l'enveloppe des plans osculateurs
+de son arête de rebroussement, et est engendrée par
+les tangentes à son arête de rebroussement}.%--
+
+\Paragraph{Remarques.} Nous avons fait implicitement diverses hypothèses.
+D'abord que le déterminant des équations~\Eq{(3)} n'est pas nul.
+S'il l'est, on a
+\[
+\begin{vmatrix}
+u  &  v  &  w  \\
+du &  dv &  dw \\
+d^2u & d^2v & d^2w
+\end{vmatrix} = 0\Add{,}
+\]
+ce qui exprime que $u, v, w$ sont liés par une relation linéaire
+homogène à coefficients constants; \DPchg{c.à.d.}{cést-à-dire} que les plans~\Eq{(1)}
+sont parallèles à une droite fixe. Dans ce cas, les droites~\Eq{(2)}
+sont parallèles à cette même direction, et la surface est
+un \emph{cylindre}. Dans ce cas figure, comme \emph{cas singulier}, celui où
+tous les plans~\Eq{(1)} passent par une droite fixe, qui est alors
+l'enveloppe.
+
+\DPchg{Ecartant}{Écartant} ce cas, nous avons admis qu'il y avait un lieu
+des points~$M$. Ceci suppose que $M$~n'est pas fixe. S'il en était
+ainsi les équations~\Eq{(3)} étant vérifiées par les coordonnées
+de ce point fixe, les plans~\Eq{(1)} passeraient par ce point fixe,
+ainsi que les droites~\Eq{(2)}. L'enveloppe serait un \emph{cône}.
+
+\DPchg{Ecartons}{Écartons} encore ce cas. Nous avons admis encore que les
+droites~\Eq{(2)} engendraient une surface. Mais cela n'est en défaut
+que si elles sont toutes confondues, ce qui est le cas
+singulier \DPtypo{déja}{déjà} examiné.
+
+Remarquons enfin que la courbe~$(C)$ est \DPtypo{forcèment}{forcément} gauche,
+car si elle était plane, son plan étant son plan osculateur
+unique, et nos raisonnements ne cessant pas de s'appliquer,
+tous les plans~\Eq{(1)} seraient confondus. Il n'y aurait donc
+pas $\infty^{1}$~plans~\Eq{(1)}.
+%% -----File: 023.png---Folio 15-------
+
+
+\Section{Réciproques.}
+{11.}{Réciproquement les plans osculateurs en tous les
+points d'une courbe gauche enveloppent une développable.} En
+effet, si nous reprenons les notations du §1, le plan osculateur
+en un point~$x\Add{,}y\Add{,}z$ d'une courbe a pour équation
+\[
+\sum a'' (X - x) = 0.
+\]
+Sa caractéristique est représentée par l'équation précédente
+et
+\[
+\sum  \frac{da''}{ds}\, (X - x) - \sum a''\, \frac{dx}{ds} = 0;
+\]
+mais on a
+\[
+\sum a''\, \frac{dx}{ds} = 0,\qquad \frac{da''}{ds} = \frac{1}{T}\, a';
+\]
+les équations de la caractéristique sont donc
+\[
+\sum a'(X - x) = 0,\qquad \sum a''(X - x) = 0.
+\]
+Et, si on prend comme trièdre de coordonnées le trièdre de
+Serret-Frenet, elles se réduisent à
+\[
+Y = 0,\qquad Z = 0.
+\]
+\emph{Donc la \DPtypo{caracteristique}{caractéristique} du plan osculateur en un point d'une
+courbe gauche est la tangente à cette courbe}, et l'enveloppe
+de ce plan est bien une surface développable. L'arête de rebroussement
+a pour équations
+\[
+\sum a''(X - x) = 0,\quad
+\sum a' (X - x) = 0,\quad
+\sum \frac{da'}{ds}\, (X - x) - \sum a'\, \frac{dx}{ds} = 0.
+\]
+
+Considérons la \DPtypo{3éme}{\Ord{3}{ème}}~équation; remarquons que l'on a
+\[
+\sum a'\, \frac{dx}{ds} = 0,\qquad
+\frac{da'}{ds} = -\left(\frac{a}{R} + \frac{a''}{T}\right);
+\]
+elle s'écrit alors
+\[
+\sum \left(\frac{a}{R} + \frac{a''}{T}\right) (X - x) = 0,
+\]
+ou encore, en tenant compte de la \Ord{1}{ère} équation
+\[
+\sum a(X - x) = 0.
+\]
+Nous avons ainsi \Card{3} équations linéaires et homogènes en $X - x$,
+$Y - y$, $Z - z$, dont le déterminant est~$1$; donc
+\[
+X - x = 0,\qquad
+Y - y = 0,\qquad
+Z - \DPtypo{Z}{z} = 0;
+\]
+\DPtypo{l'arète}{l'arête} de rebroussement est la courbe elle-même.
+%% -----File: 024.png---Folio 16-------
+
+\Paragraph{Remarque.} Le nom \DPtypo{d'arète}{d'arête} de rebroussement provient de ce fait
+que la \emph{section de la développable par le plan normal en~$M$ à
+l'arête de rebroussement présente au point~$M$ un point de rebroussement}.
+En effet, rapportons la courbe au trièdre de Serret
+relatif au point~$M$: les coordonnées d'un point de la courbe
+voisin du point~$M$ sont, d'après les formules établies au~\no5
+\begin{alignat*}{3}
+x &= ds &{}- \frac{1}{\Err{6R}{6R^2}}\, ds^3 &+ \dots, \\
+y &=    &  \frac{1}{2R}\, ds^2 &- \frac{1}{6R^2}\, \frac{dR}{ds}\, ds^3 + \dots, \\
+z &=    &                      &- \frac{1}{6RT}\, ds^3 + \dots\Add{.}
+\end{alignat*}
+Les coordonnées d'un point de la tangente au point~$x\Add{,}y\Add{,}z$ sont
+\begin{align*}
+X &= x + \lambda\, \frac{dx}{ds}
+   = \!\left(ds - \frac{1}{6R^2}\, ds^3 + \dots\right)
+       + \lambda\! \left(1 - \frac{1}{2R^2}\, ds^2 + \dots\right)\Add{\!,} \\
+Y &= y + \lambda\, \frac{dy}{ds}
+   = \!\left(\!\frac{1}{2R}\, ds^2 - \frac{1}{6R^2}\, \frac{dR}{ds}\, ds^3 + \dots\!\right)
+       + \lambda\! \left(\!\frac{1}{R}\, ds - \frac{1}{2R^2}\, \frac{dR}{ds}\, ds^2 + \dots\!\right)\Add{\!,} \\
+Z &= z + \lambda\, \frac{dz}{ds}
+   = \!\left(-\frac{1}{6RT}\, ds^3 + \dots\right)
+       + \lambda\! \left(-\frac{1}{2RT}\, ds^2 + \dots\right)\Add{\!.}
+\end{align*}
+Prenons l'intersection de cette tangente avec le plan normal
+$X = 0$, nous avons
+\[
+\lambda = -\frac{ds + \dots}{1 + \dots} = -ds + \dots
+\]
+et la courbe d'intersection \DPtypo{à}{a} pour équations
+%
+\begin{align*}
+&\smash{\raisebox{-0.25in}{\Input[1.5in]{024a}}}&&
+\begin{aligned}[b]
+Y &= -\frac{1}{2R}\, ds^2 + \dots\Add{,} \\
+Z &= \frac{1}{3RT}\, ds^3 + \dots\Add{.}
+\end{aligned}
+\end{align*}
+On voit qu'elle a au point~$M$ un point de rebroussement, la
+tangente de rebroussement étant la normale principale.%--
+
+\Section{Surface rectifiante. Surface polaire\Add{.}}
+{12.}{Remarques.} Cherchons les surfaces développables enveloppes
+des faces du trièdre de Serret dans une courbe \DPtypo{gauch}{gauche}~$(C)$.
+Nous venons de voir que \emph{le plan osculateur enveloppe la
+surface développable qui admet pour arête de rebroussement~$(C)$}\Add{.}
+%% -----File: 025.png---Folio 17-------
+
+Considérons maintenant le plan rectifiant
+\[
+\sum a'(X - x) = 0
+\]
+la caractéristique est représentée par l'équation précédente
+et par
+\[
+\frac{1}{R} \sum a(X - x) + \frac{1}{T} \sum a''(X - x) = 0\Add{.}
+\]
+Si on prend les axes de Serret ces équations deviennent
+\[
+Y=0,\qquad \frac{1}{R}\, X + \frac{1}{T}\, Z = 0,
+\]
+la caractéristique contient le point $Y = 0$, $X = -\dfrac{1}{T}$, $Z = \dfrac{1}{R}$,
+extrémité du \Err{secteur}{vecteur} qui représente la rotation instantanée
+du trièdre; \emph{c'est l'axe instantané de rotation du trièdre de
+Serret}. Son lieu s'appelle la \emph{surface rectifiante}. Elle contient
+la courbe~$(\DPtypo{c}{C})$.
+
+Considérons enfin le plan normal
+\[
+\sum a(X - x) = 0;
+\]
+la \Ord{2}{e} équation de la caractéristique est
+\[
+\sum \frac{da}{ds}\, (X - x) - \sum a\, \frac{dx}{ds} = 0,
+\]
+ou
+\[
+\frac{1}{R} \sum a'(X - x) - 1 = 0.
+\]
+Cette caractéristique s'appelle la \emph{droite polaire}, et son
+lieu s'appelle la \emph{surface polaire}.
+
+Prenant de nouveau les axes de Serret, les équations de la
+droite polaire deviennent
+\[
+X = 0,\qquad Y = R;
+\]
+Elle se confond donc avec \emph{l'axe du cercle osculateur}.
+
+Si nous cherchons le point d'intersection de la droite
+polaire avec l'arête de rebroussement de la surface polaire,
+nous avons les \Card{3} équations
+%% -----File: 026.png---Folio 18-------
+\[
+\sum a(X - x) = 0,\quad
+\sum a'(X - x) - R = 0,\quad
+\frac{1}{T} \sum a''(X - x) + \frac{dR}{ds} = 0\Add{,}
+\]
+qui deviennent, en prenant les axes de Serret,
+\[
+X = 0,\qquad Y = R,\qquad Z = -\frac{1}{T}\, \frac{dR}{ds}.
+\]
+
+Or\Add{,} ce sont les coordonnées du centre de la sphère osculatrice.
+(Voir \No9).
+
+Donc \emph{le point ou la droite polaire touche son enveloppe
+est le centre de la sphère osculatrice à la courbe~$(\DPtypo{c}{C})$.
+On peut dire encore que la courbe~$(\DPtypo{c}{C})$ est\DPtypo{}{ la} trajectoire orthogonale
+des plans osculateurs au lieu des centres de ses sphères
+osculatrices}.
+
+
+\ExSection{I}
+
+\begin{Exercises}
+\item[1.] Trouver l'axe instantané de rotation et de glissement
+du trièdre de Serret.
+
+\item[2.] Trouver les hélices circulaires osculatrices à une
+courbe gauche. Déterminer celle de ces hélices qui a même torsion
+que la courbe \DPtypo{donnee}{donnée}.
+
+\item[3.] Approfondir les relations entre une courbe et le lieu
+des centres de ses sphères osculatrices (courbure, torsion,
+\DPtypo{élement}{élément} d'arc).
+
+\item[4.] Chercher la condition nécessaire et suffisante pour
+qu'une courbe soit une courbe \DPtypo{spherique}{sphérique}.
+
+\item[5.] Déterminer toutes les courbes satisfaisant aux relations:
+\[
+\frac{dR}{ds} = F(R),\qquad T = G(R),
+\]
+où $F$~et~$G$ sont des fonctions données.
+
+\item[6.] Déterminer toutes les courbes à courbure constante.
+
+\item[7.] Déterminer toutes les courbes à torsion constante.
+\end{Exercises}
+
+%Voir les énoncés, page 18.
+%% -----File: 027.png---Folio 19-------
+
+
+\Chapitre{II}{Surfaces.}
+
+\Section[Courbes tracées sur une surface. Longeurs d'arc et angles.]
+{Le $ds^2$ de la surface, et les angles.}
+{1.}{Courbes tracées sur une surface. Longueurs d'arc et
+angles.} Les coordonnées d'un point d'une surface peuvent
+s'exprimer en fonction de deux paramètres arbitraires
+\[
+\Tag{(S)}
+x = f(u\Add{,} v),\qquad
+y = g(u\Add{,} v),\qquad
+z = h(u\Add{,} v);
+\]
+$u\Add{,} v$ sont les \emph{coordonnées curvilignes} d'un point de la surface~$(S)$.
+On définira une courbe~$(c)$ de la surface en établissant
+une relation entre $u, v$; ou, ce qui revient au même, en exprimant
+$u, v$ en fonction d'un même paramètre~$t$
+\[
+\Tag{(c)}
+u = \phi(t),\qquad
+v = \psi(t).
+\]
+La tangente à cette courbe a pour paramètres directeurs
+\[
+\Tag{(1)}
+dx = \frac{\dd x}{\dd u}\, du + \frac{\dd x}{\dd v}\, dv,\qquad
+dy = \frac{\dd y}{\dd u}\, du + \frac{\dd y}{\dd v}\, dv,\qquad
+dz = \frac{\dd z}{\dd u}\, du + \frac{\dd z}{\dd v}\, dv;
+\]
+la tangente est déterminée par les différentielles $du, dv$.
+
+L'élément d'arc a pour expression:
+\[
+\Tag{(2)}
+ds^2 = dx^2 + dy^2 + dz^2
+  = E\, du^2 + 2F\, du\, dv + G\, dv^2 = \Phi(du, dv)
+\]
+en posant
+\[
+E = \sum \left(\frac{\dd x}{\dd u}\right)^2,\qquad
+F = \sum \frac{\dd x}{\dd u}\, \frac{\dd x}{\Err{\dd v^2}{\dd v}},\qquad
+G = \sum \left(\frac{\dd x}{\dd v}\right)^2.
+\]
+
+Imaginons \Card{2} courbes passant par un même point $(u, v)$ de la
+surface; soient $du, dv$ les différentielles correspondant à l'une
+d'elles; $\delta u, \delta v$ celles correspondant à l'autre; $ds, \delta s$ les différentielles
+des arcs correspondants. Si $V$~est l'angle des deux
+courbes, nous avons
+\[
+\Cos V = \sum \frac{dx·\delta x}{ds·\delta s};
+\]
+or,
+\begin{align*}
+\sum dx·\delta x
+  &= \sum \left(\frac{\dd x}{\dd u}\, du       + \frac{\dd x}{\dd v}\,  dv\right)
+          \left(\frac{\dd x}{\dd u}\, \delta u + \frac{\dd x}{\dd \DPtypo{u}{v}}\, \delta v\right) \\
+  &=  E\, du\, \delta u + F (du\, \delta v + dv\, \delta u) + G\, dv\, \delta v;
+\end{align*}
+%% -----File: 028.png---Folio 20-------
+c'est une forme polaire de la forme quadratique $\Phi(du, dv)$
+et on a
+\[
+\Tag{(3)}
+\Cos V = \frac{1}{2}\, \frac{\delta u\, \dfrac{\dd\Phi(du, dv)}{\dd·du}
+                           + \delta v\, \dfrac{\dd\Phi(du, dv)}{\dd·dv}}
+   {\sqrt{\Phi(du, dv) · \Phi(\delta u, \delta v)}}
+\]
+
+Pour que les deux courbes soient orthogonales, il faut et
+il suffit que $\Cos V = 0$, ou
+\[
+\Tag{(4)}
+E\, du·\delta u + F(du·\delta v + dv·\delta u) + G·dv·\delta v = 0.
+\]
+En particulier, cherchons à quelles conditions les courbes coordonnées
+$u = \Cte$ et $v = \Cte$ forment un réseau orthogonal;
+alors $dv = 0$, $\DPtypo{du}{\delta u} =0$, la condition précédente se réduit à
+\[
+F\, du\, \delta v = 0,
+\]
+et comme $du\Add{,} \DPtypo{dv}{\delta v}$ ne sont pas constamment nuls, on a $F = 0$. Dans
+ce cas, le carré de l'élément d'arc prend la forme
+\[
+ds^2 = E\, du^2 + G\, dv^2.
+\]
+
+\Paragraph{Remarque.} Si on définit la surface par une équation de
+la forme
+\[
+Z = f(x\Add{,} y)
+\]
+en désignant comme d'habitude par $p, q$ les dérivées partielles
+de~$Z$ par rapport à $x, y$, on a
+\[
+ds^2 = dx^2 + dy^2 + (p\, dx + q\, dy)^2
+  = (1 + p^2)\, ds^2 + 2pq\, dx\, dy + (1 + q^2)\, dy^2\Add{,}
+\]
+\DPtypo{C.à.d.}{\DPchg{c.à.d.}{cést-à-dire}}
+\[
+E = 1 + p^2,\qquad F = pq,\qquad G = 1 + q^2.
+\]
+
+
+\Section{Déformation et représentation conforme.}
+{2.}{Surfaces applicables. Représentations conformes.}
+Considérons deux surfaces $(\Err{S'}{S})\Add{,} (S')$
+\begin{alignat*}{3}
+\Tag{(S)}
+x &= f(u, v),   &y &= g(u, v),   &z &= h(u\Add{,} v) \\
+\Tag{(S')}
+x &= F(u', v'),\qquad &y &= G(u', v'),\qquad &z &= H(u', v')
+\end{alignat*}
+on peut établir une correspondance point par point entre ces
+deux surfaces, et cela d'une infinité de manières. Il suffit
+de poser
+\[
+u' = \phi(u, v),\qquad
+v' = \psi(u\Add{,} v),
+\]
+les fonctions $\psi, \phi$ étant quelconques; à condition toutefois que
+%% -----File: 029.png---Folio 21-------
+les équations précédentes soient résolubles en $u, v$. Les équations
+de la surface~$(S')$ peuvent alors se mettre sous la forme
+\[
+\Tag{(S')}%[** TN: Duplicate label, re-expresses earlier equation (S')]
+x = \Err{F}{F_{1}}(u\Add{,} v),\qquad
+y = \Err{G}{G_{1}}(u\Add{,} v),\qquad
+z = \Err{H}{H_{1}}(u\Add{,} v),
+\]
+ce qui revient à dire que les points correspondants correspondent
+aux mêmes systèmes de valeurs des paramètres.
+
+Soient les éléments d'arcs sur ces \Card{2} surfaces
+\begin{alignat*}{3}
+ds^2   &= E\,   du^2 &&+ 2F\,   du·dv &&+ G\,   dv^2 \\
+ds_1^2 &= E_1\, du^2 &&+ 2F_1\, du·dv &&+ G_1\, dv^2
+\end{alignat*}
+Supposons ces éléments d'arc identiques, $E \equiv E_1$, $F \equiv F_1$, $G \equiv G_1$.
+Si alors $u, v$ sont exprimés en fonction du paramètre~$t$, les arcs
+des deux courbes correspondantes sur les deux surfaces compris
+entre \Card{2} points correspondants ont tous deux pour expression
+\[%[** TN: Not displayed in original]
+\int_{t_0}^{t_1} \sqrt{E\, du^2 + 2F\, du\, dv + G\, dv^2},
+\]
+$t_0, t_1$ étant les valeurs de~$t$ correspondant
+aux extrémités. Réciproquement, si les arcs homologues
+de deux courbes homologues sur les deux surfaces ont même
+longueur, les éléments d'arc sont identiques sur les deux surfaces.
+On dit que les deux surfaces sont \emph{applicables} l'une
+sur l'autre, ou résultent l'une de l'autre par \emph{déformation}.
+
+Dans cette correspondance, la fonction~$\Phi$ étant la même
+pour les \Card{2} surfaces, la formule~\Eq{(3)} montre que les angles se
+conservent. Mais la réciproque n'est pas vraie. L'expression
+de~$\Cos V$ est homogène et du \Ord{1}{er} degré en $E\Add{,} F\Add{,} G$; pour que les angles
+de deux courbes homologues soient égaux, il faut et il suffit
+que
+\[
+\Err{\frac{E_1}{E} = \frac{F_1}{F} = \frac{G_1}{G}}
+    {\frac{E}{E_1} = \frac{F}{F_1} = \frac{G}{G_1}} = \chi(u,v),
+\]
+ce rapport étant indépendant de $du, dv$. On dit dans ce cas qu'il
+y a \emph{représentation conforme} des deux surfaces l'une sur l'autre.
+
+
+\Section{Problème de la représentation conforme.}
+{}{\DPchg{Etant}{Étant} données deux surfaces, il est toujours possible d'établir
+entre elles une représentation conforme.} Ceci revient
+à dire que l'on peut exprimer $u_1, v_1$ en fonction de $u, v$ de
+%% -----File: 030.png---Folio 22-------
+telle sorte que l'on ait,
+\[
+E\, du^2 + 2F\, du·dv + G\, dv^2 = \chi(u, v)(E_1\, du^2 + 2F_1\, du·dv + G_1\, dv^2).
+\]
+Décomposons les deux $ds^2$ en facteurs du \Ord{1}{er} degré. Remarquons
+que $EG - F^2$ est la somme des carrés des déterminants déduits du
+tableau
+\[
+\begin{Vmatrix}
+\mfrac{\dd x}{\dd u} & \mfrac{\dd y}{\dd u} & \mfrac{\dd z}{\dd u} \\
+\mfrac{\dd x}{\dd v} & \mfrac{\dd y}{\dd v} & \mfrac{\dd z}{\dd v}
+\end{Vmatrix};
+\]
+$EG - F^2$ est positif pour toute surface réelle. Posons
+\[
+EG - F^2 = H^2;
+\]
+alors
+\[
+ds^2 = E\left(du + \frac{F + iH}{E}\, dv\right)
+        \left(du + \frac{F - iH}{E}\, dv\right);
+\]
+chacun des facteurs du \Ord{2}{e} membre admet un facteur intégrant.
+On a donc
+\begin{align*}
+du + \frac{F + iH}{E}\, dv &= M(u, v)\, d\alpha(u\Add{,} v)\Add{,} \\
+du + \frac{F - iH}{E}\, dv &= N(u, v)\, d\beta(u\Add{,} v)\Add{.}
+\end{align*}
+Les fonctions $\alpha\Add{,} \beta$ sont indépendantes; en effet $d\alpha$~et~$d\beta$ ne
+peuvent s'annuler en même temps si $H \neq 0$, ce que nous supposons.
+Nous pouvons donc prendre $\alpha, \beta$ comme coordonnées curvilignes
+sur la \Ord{1}{ère} surface, et nous avons
+\begin{align*}
+ds^2 &= P(u, v)\, d\alpha·d\beta
+  = \Theta(\alpha, \beta)\, d\alpha·d\beta\Add{.} \\
+\intertext{De même pour la \Ord{2}{e} surface, nous pourrons écrire}
+ds_1^2 &= P_1(u_1, v_1)\, d\alpha_1·d\beta_1
+  = \Theta_1(\alpha_1, \beta_1)\, d\alpha_1·d\beta_1.
+\end{align*}
+Nous aurons alors à satisfaire à l'équation
+\[
+\Theta(\alpha, \beta)\, d\alpha · d\beta
+  = \Omega(\alpha, \beta)\, \Theta_1(\alpha_1, \beta_1)\, d\alpha_1 · d\beta_1\Add{.}
+\]
+
+Remarquons que pour $d\alpha = 0$, on doit avoir $d\alpha_1 · d\beta_1 = 0$.
+Si nous prenons $d\alpha_1 = 0$,  $\alpha_1$~sera fonction de~$\alpha$ et de même $\beta_1$~sera
+fonction de~$\beta$
+\[
+\alpha_1(u_1, v_1) = \phi\bigl(\alpha(u, v)\bigr)\Add{,}\qquad
+\beta_1 (u_1, v_1) = \psi\bigl(\beta (u, v)\bigr)\Add{.}
+\]
+%% -----File: 031.png---Folio 23-------
+Au contraire en prenant $d\beta_1 = 0$, $\beta_1$~sera fonction de~$\alpha$ et de
+même~$\alpha_1$, de~$\beta$
+\[
+\beta_1 (u_1, v_1) = \phi\bigl(\alpha(u, v)\bigr)\qquad
+\alpha_1(u_1, v_1) = \psi\bigl(\beta (u, v)\DPtypo{}{\bigr)}\Add{.}
+\]
+On voit donc bien que l'on peut toujours établir une représentation
+conforme. Et nous avons de plus la solution \DPtypo{génerale}{générale}
+de ce problème.
+
+
+\Section{Condition pour que deux surfaces soient applicables.}
+{}{Deux surfaces données ne sont pas en général applicables
+l'une sur l'autre.}
+
+Autrement dit, étant données deux surfaces, il est impossible
+d'établir entre elles une correspondance telle que $ds^2 = ds_1^2$\Add{.}
+En effet, en reprenant le calcul précédent, il faudrait satisfaire
+à la relation
+\[
+\Theta(\alpha, \beta)\, d\alpha · d\beta = \Theta_1(\alpha_1, \beta_1)\, d\alpha_1 \Add{·} d\beta_1,
+\]
+il faudrait comme \DPtypo{précedemment}{précédemment}, prendre par exemple
+\[
+\alpha_1 = \phi(\alpha)\qquad \beta_1 = \psi(\beta);
+\]
+et la relation à satisfaire devient
+\[
+\Theta(\alpha, \beta)
+  = \Theta_1\bigl(\phi(\alpha), \psi(\beta)\bigr)\, \phi'(\alpha)\, \psi'(\beta);
+\]
+il est facile de voir que, les fonctions $\Theta, \Theta_1$\DPtypo{,}{} étant données\Add{,}
+il est impossible en général de trouver des fonctions $\phi, \psi$,
+satisfaisant à cette relation. Considérons en effet le cas
+particulier où la deuxième surface est le plan $z = 0$. Dans
+ce cas $ds_1^2 = dx^2 + dy^2 = d\alpha_1 · d\beta_1$ et on devrait avoir
+\[
+\Theta(\alpha, \beta) = \phi'(\alpha)\, \psi'(\beta)\Add{;}
+\]
+or\Add{,} la fonction $\Theta$ étant quelconque, n'est pas le produit d'une
+fonction de~$\alpha$ par une fonction de~$\beta$.
+
+Pour qu'il en soit ainsi, il faut et il suffit que l'on ait
+\[
+\log \Theta(\alpha, \beta) = \log \phi'(\alpha) + \log \psi'(\beta),
+\]
+ou
+\[
+\frac{\dd^2 \log \Theta(\alpha, \beta)}{\dd\alpha · \dd\beta} = 0.
+\]
+%% -----File: 032.png---Folio 24-------
+Nous venons ainsi de montrer qu'une surface n'est pas en
+général applicable sur un plan, et de trouver la condition
+pour qu'une surface soit applicable sur un plan.
+
+
+\Section{Les \DPtypo{Directions}{directions} conjuguées et la forme \texorpdfstring{$\sum l\, d^{2}x$}{}\Add{.}}
+{3.}{Développables circonscrites. Directions conjuguées\Add{.}}
+
+Corrélativement aux courbes tracées sur la surface,
+lieux de $\infty^{1}$~points de la surface, nous considérerons les
+développables circonscrites, enveloppes de $\infty^{1}$~plans tangents
+à la surface. Définissons le plan tangent en un point de
+la surface. Soient $l, m, n$ les coefficients directeurs de la
+normale, et supposons les coordonnées rectangulaires. Nous
+devons avoir pour toute courbe de la surface
+\[
+l\, dx + m\, dy + n\, dz = 0;
+\]
+en particulier, pour les courbes coordonnées, $u = \cte$ et $v = \cte$
+nous aurons
+\begin{align*}
+l\, \frac{\dd x}{\dd u} + m\, \frac{\dd y}{\dd u} + n\, \frac{\dd z}{\dd u} &= 0\Add{,} \\
+l\, \frac{\dd x}{\dd v} + m\, \frac{\dd y}{\dd v} + n\, \frac{\dd z}{\dd v} &= 0\Add{,}
+\end{align*}
+et ces relations montrent que $l, m, n$, sont proportionnels aux
+déterminants fonctionnels $A, B, C$,
+\[
+\Tag{(1)}
+A = \frac{\dd y}{\dd u}\, \frac{\dd z}{\dd v} - \frac{\dd z}{\dd u}\, \frac{\dd y}{\dd v}
+  = \frac{D(y, z)}{D(u, v)},\qquad
+B = \frac{D(z, x)}{D(u, v)},\qquad
+C = \frac{D(x, y)}{D(u, v)};
+\]
+nous avons vu d'ailleurs que $A^2 + B^2 + C^2 = H^2$;
+donc les \DPtypo{cosimus}{cosinus} directeurs de la normale sont
+\[
+\Tag{(2)}
+\lambda = \frac{A}{H},\qquad \mu = \frac{B}{H},\qquad \nu = \frac{C}{H}.
+\]
+
+Considérons une développable circonscrite; nous pourrons
+la définir en exprimant $u, v$ en fonction d'un paramètre~$t$,
+%% -----File: 033.png---Folio 25-------
+\[
+u = \phi(t),\qquad
+v = \psi(t);
+\]
+alors le point $(u\Add{,} v)$ décrit une courbe de la surface, soit~$(c)$,
+et les plans tangents à la surface aux divers points de~$(c)$
+enveloppent la développable considérée. Le plan tangent
+à la surface au point $(x\Add{,} y\Add{,} z)$ est, $X\Add{,} Y\Add{,} Z$ étant les coordonnées courantes,
+\[
+l·(X - x) + m·(Y - y) + n·(Z - z) = 0;
+\]
+la caractéristique est définie par l'équation précédente et
+par l'équation
+\[
+dl·(X - x) + dm·(Y - y) + dn·(Z - z) = 0
+\]
+obtenue en différentiant la précédente par rapport à~$t$, et
+remarquant que l'on~a
+\[
+l\, dx + m\, dy + n\, dz = 0\Add{.}
+\]
+
+Voyons quelle est la direction de cette caractéristique\Add{.}
+Soient $\delta x, \delta y, \delta z$ ses coefficients de direction. Elle est
+tangente à la surface, donc on peut poser
+\[
+\delta x = \frac{\dd x}{\dd u}\, \delta u + \frac{\dd x}{\dd v}\, \delta v,\quad
+\delta y = \frac{\dd y}{\dd u}\, \delta u + \frac{\dd y}{\dd v}\, \delta v,\quad
+\delta z = \frac{\dd z}{\dd u}\, \delta u + \frac{\dd z}{\dd v}\, \delta v;
+\]
+en remplaçant $X-x, Y-y, Z-z$ par les quantités proportionnelles
+$\delta x, \delta y, \delta z$, on obtient
+\[
+dl · \delta x + dm · \delta y + dn · \delta z = 0;
+\]
+or\Add{,} on a
+\[
+dl = \frac{\dd l}{\dd u}\, du + \frac{\dd l}{\dd v}\, dv,\quad
+dm = \frac{\dd m}{\dd u}\, du + \frac{\dd m}{\dd v}\, dv,\quad
+dn = \frac{\dd n}{\dd u}\, du + \frac{\dd n}{\dd v}\, dv;
+\]
+donc la relation
+\[
+\sum dl · \delta x = 0\DPtypo{.}{}
+\]
+s'écrit
+\[
+\sum \left(\frac{\dd l}{\dd u}\, du + \frac{\dd l}{\dd v}\, dv\right)
+     \left(\frac{\dd x}{\dd u}\, \delta u + \frac{\dd x}{\dd v}\, \delta v\right) = 0.
+\]
+%% -----File: 034.png---Folio 26-------
+Ordonnons par rapport à $du, dv$, $\delta u, \delta v$. Remarquons que l'on a
+\[
+\sum l\, \frac{\dd x}{\dd u} = 0;
+\]
+d'où en dérivant par rapport à~$u$
+\[
+\sum l\, \frac{\dd^2 x}{\dd u^2} + \sum \frac{\dd l}{\dd u}\, \frac{\dd x}{\dd u} = 0;
+\]
+de même, la relation
+\[
+\sum l\, \frac{\dd x}{\dd v} = 0
+\]
+donne
+\[
+\sum l\, \frac{\dd^2 x}{\dd v^2} + \sum \frac{\dd l}{\dd v}\, \frac{\dd x}{\dd v} = 0;
+\]
+et
+\[
+\sum l\, \frac{\dd^2 x}{\dd u\, \dd v} + \sum \frac{\dd l}{\dd u}\, \frac{\dd x}{\dd v} = 0;
+\]
+de sorte que la relation précédente s'écrit
+\[
+\Tag{(3)}
+\sum l\, \frac{\dd^2 x}{\dd u^2}\, du · \delta u +
+\sum l\, \frac{\dd^2 x}{\dd u\, \dd v}\, (du · \delta v + dv\Add{·} \delta u) +
+\DPtypo{}{\sum} l\, \frac{\dd^2 x}{\dd v^2}\, dv · \delta v = 0\Add{.}
+\]
+Telle est la relation qui existe entre les coefficients de direction
+de la caractéristique et de la tangente à la courbe de
+contact. Elle serait visiblement la même en coordonnées obliques,
+$l, m, n$ étant alors les coefficients de l'équation du plan tangent
+soit
+\[
+\Tag{(4)}
+E' = \sum l\, \frac{\dd^2 x}{\dd u^2},\qquad
+F' = \sum l\, \frac{\dd^2 x}{\dd u\, \dd v},\qquad
+G' = \sum l\, \frac{\dd^2 x}{\dd v^2},
+\]
+et
+\[
+\Tag{(5)}
+\Psi(du\Add{,} dv) = E'\, du^2 + 2F'\, du\, dv + G'\, dv^2.
+\]
+On a, en particulier, quand on prend $l = A$, $m = B$, $n = C$:
+\[
+E' = \begin{vmatrix}
+\mfrac{\dd^2 x}{\dd u^2} & \mfrac{\dd^2 y}{\dd u^2} & \mfrac{\dd^2 z}{\dd u^2} \\
+\mfrac{\dd x}{\dd u} & \mfrac{\dd y}{\dd u} & \mfrac{\dd z}{\dd u} \\
+\mfrac{\dd x}{\dd v} & \mfrac{\dd y}{\dd v} & \mfrac{\dd z}{\dd v}
+\end{vmatrix},\
+%
+F' = \begin{vmatrix}
+\mfrac{\dd^2 x}{\dd u\, \dd v} & \mfrac{\dd^2 y}{\dd u\, \dd v} & \mfrac{\dd^2 z}{\dd u\, \dd v} \\
+\mfrac{\dd x}{\dd u} & \mfrac{\dd y}{\dd u} & \mfrac{\dd z}{\dd u} \\
+\mfrac{\dd x}{\dd v} & \mfrac{\dd y}{\dd v} & \mfrac{\dd z}{\dd v}
+\end{vmatrix},\
+%
+G' = \begin{vmatrix}
+\mfrac{\dd^2 x}{\dd v^2} & \mfrac{\dd^2 y}{\dd v^2} & \mfrac{\dd^2 z}{\dd v^2} \\
+\mfrac{\dd x}{\dd u} & \mfrac{\dd y}{\dd u} & \mfrac{\dd z}{\dd u} \\
+\mfrac{\dd x}{\dd v} & \mfrac{\dd y}{\dd v} & \mfrac{\dd z}{\dd v}
+\end{vmatrix}\Add{.}
+\]
+La relation précédente s'écrira alors
+\[
+E' · du\, \delta u + F' · (du\, \delta v + dv\, \delta u) + G' · dv\, \delta v = 0,
+\]
+ou
+\[
+\Tag{(6)}
+\frac{\dd\Psi(du, dv)}{\dd\Add{·} du}\, \delta u + \frac{\dd\Psi(du, dv)}{\dd\Add{·} dv}\, \delta v = 0.
+\]
+%% -----File: 035.png---Folio 27-------
+Cette relation est symétrique par rapport à $d, \delta$; \emph{il y a donc
+réciprocité entre la direction de la tangente à la courbe de
+contact de la développable et la direction de la caractéristique
+du plan tangent à cette développable}. Ces deux directions
+sont dites \emph{directions conjuguées}.
+
+Cherchons en particulier la condition pour que les courbes
+$u = \cte$, $v = \cte$ forment un réseau conjugué. Alors, $dv = 0$\Add{,} $\delta u = 0$
+la condition est $F' = 0$.
+
+\Paragraph{Remarque.} On a
+\begin{gather*}
+dx = \frac{\dd x}{\dd u}\, du + \frac{\dd x}{\dd v}\, dv, \\
+d^2 x = \frac{\dd x}{\dd u}\, d^2 u
+      + \frac{\dd x}{\dd v}\, d^2 v
+      + \frac{\dd^2 x}{\DPtypo{\dd u}{\dd u^2}}\, du^2
+      + 2 \frac{\dd^2 x}{\dd u\, \dd v}\, du\, dv
+      + \frac{\dd^2 x}{\DPtypo{\dd v}{\dd v^2}}\, dv^2.
+\end{gather*}
+On en conclut, à cause de
+\[
+\sum l\, \frac{\dd x}{\dd u} = 0,\qquad
+\sum l\, \frac{\dd x}{\dd v} = 0,
+\]
+l'identité
+\[
+\sum l\, d^2x
+  = \left(\sum l\, \frac{\dd^2 x}{\dd u^2}\right) du^2
+  + 2 \left(\sum l\, \frac{\dd^2 x}{\dd u\, \dd v}\right) du\, dv
+  + \left(\sum l\, \frac{\dd^2 x}{\dd v^2}\right) dv^2\Add{,}
+\]
+c'est-à-dire
+\[
+\sum l\, d^2 x = \Psi(du, dv).
+\]
+
+
+\Section{Formules fondamentales pour une courbe de la surface.}
+{4.}{\DPchg{Eléments}{Éléments} fondamentaux d'une courbe de la surface.}
+
+Nous considérerons en un point de la courbe le trièdre
+de Serret, et un trièdre constitué par la tangente à la courbe,
+la normale~$MN$ à la surface, et la tangente~$MN'$ à la surface
+qui est normale à la courbe. Nous choisirons les directions
+positives de telle façon que le trièdre ainsi constitué
+ait même disposition que le trièdre de Serret, de sorte que
+%% -----File: 036.png---Folio 28-------
+si $l, m, n$ sont les cosinus directeurs de la normale à la surface,
+$a_1, b_1, c_1$ de la tangente à la surface normale
+à la courbe, on ait
+\[
+\begin{vmatrix}
+a   & b   & c   \\
+a_1 & b_1 & c_1 \\
+l   & m   & n
+\end{vmatrix} = 1.
+\]
+
+\Illustration[2.5in]{036a}
+Les \Card{2} trièdres considérés ont un axe
+commun et de même direction, qui est
+la tangente. Pour les définir l'un par rapport à l'autre, il
+suffira de se donner l'angle d'une des arêtes de l'un avec
+l'une des arêtes de l'autre. Nous nous donnerons l'angle
+dont il faut faire tourner la demi-normale principale~$MP$ pour
+l'amener à \DPtypo{coincider}{coïncider} avec la demi-normale à la surface~$MN$, le
+sens positif des relations étant défini par la direction positive
+$MT$ de l'axe de rotation. Cherchons les relations qui existent
+entre les cosinus directeurs des arêtes de ces trièdres.
+Quand on passe de l'un à l'autre, on fait en réalité une
+transformation de coordonnées autour de l'origine dans le
+plan normal. Considérons le point à l'unité de distance de~$M$ sur
+$MN(l, m, n)$. Rapporté au système~$PMB$ il a pour coordonnées $\cos\theta$
+et $\sin\theta$, donc
+\[
+\Tag{(1)}
+l = a' \cos \theta + a'' \sin \theta\Add{,}\qquad
+m = b' \cos \theta + b'' \sin \theta\DPtypo{;}{,}\qquad
+n = c' \cos \theta + c'' \sin \theta;
+\]
+de même le point à l'unité de distance sur $MN'(a_1, b_1, c_1)$ rapporté
+au système $PMB$ a pour coordonnées $\cos (\theta - \frac{\pi}{2}) = \sin \theta$
+et $\sin (\theta - \frac{\pi}{2}) = -\cos \theta$, donc
+%% -----File: 037.png---Folio 29-------
+\[
+\Tag{(1')}%[** TN: Renumbered duplicate equation (1)]
+a_1 = a' · \sin \theta - a'' · \cos \theta,\quad
+b_1 = b' · \sin \theta - b'' · \cos \theta,\quad
+c_1 = c' · \sin \theta - c'' · \cos \theta.
+\]
+On aura donc, en faisant la transformation de coordonnées
+inverse
+\[
+\Tag{(2)}
+\begin{aligned}
+a'  &= l \cos \theta + a_1 \sin \theta, &
+b'  &= m \cos \theta + b_1 \sin \theta, &
+c'  &= n \cos \theta + c_1 \sin \theta, \\
+a'' &= l \sin \theta - a_1 \cos \theta, &
+b'' &= m \sin \theta - b_1 \cos \theta, &
+c'' &= n \sin \theta - c_1 \cos \theta.
+\end{aligned}
+\]
+Différentions les formules~\Eq{(1)} par rapport à~$s$: il vient
+\begin{alignat*}{5}
+&\frac{dl}{ds}
+  = (-a'\Add{·}\sin \theta + a'' · \cos\theta)\, &&\frac{d\theta}{ds}
+  &&+ \cos \theta\, \frac{da'}{ds} &&+ \sin\theta\, \frac{da''}{ds},
+  \quad&&\text{et les analogues;} \\
+&\frac{da_1}{ds}
+  = (a'\Add{·}\cos\theta + a'' · \sin \theta )\, &&\frac{d\theta}{ds}
+  &&+ \sin \theta\, \frac{da'}{ds} &&- \cos\theta\, \frac{da''}{ds},
+  \quad&&\text{et les analogues;}
+\end{alignat*}
+d'où, en tenant compte des formules de Frenet et des relations \Eq{(1)}\Add{,}~\Eq{(2)}
+\begin{align*}
+\Tag{(3)}
+\frac{dl}{ds}
+  &= a_1 \left(\frac{1}{T} - \frac{d\theta}{ds}\right) - \frac{a \cos\theta}{R}
+    \qquad\text{et les analogues}; \\
+\Tag{(4)}
+\frac{da_1}{ds}
+  &= -l \left(\frac{1}{T} - \frac{d\theta}{ds}\right) \Err{}{-} \frac{a \sin\theta}{R}
+    \qquad\PadTxt{et les analogues}{(id.)}; \\
+\intertext{\DPtypo{Enfin}{enfin} nous avons}
+\Tag{(5)}
+\frac{da}{ds}
+  &= \frac{a'}{R} = l\, \frac{\cos \theta}{R} + a_1\, \frac{\sin \theta}{R}
+    \quad\qquad\PadTxt{et les analogues}{(id.)};
+\end{align*}
+\emph{les formules fondamentales \Eq{(3)}\Add{,} \Eq{(4)}\Add{,} \Eq{(5)} permettent de calculer
+$\theta, R\Add{,} T$, c'est-à-dire de déterminer le plan osculateur, la
+courbure et la torsion de la courbe considérée}.
+
+\MarginNote{Formule pour $\dfrac{\cos\theta}{R}$\Add{.}}
+En effet, les formules~\Eq{(5)} nous donnent d'abord
+\[
+\frac{\cos \theta}{R}
+  = \sum l\, \frac{da}{ds}
+  = \sum l\, \frac{d^2x}{ds^2}
+  = \frac{1}{H} \sum A\, \frac{d^2x}{ds^2},
+\]
+c'est-à-dire, d'après le calcul du paragraphe précédent, et
+%% -----File: 038.png---Folio 30-------
+et en posant:
+\begin{gather*}
+E' = \sum A\, \frac{\dd^2 x}{\dd u^2},\qquad
+F' = \sum A\, \frac{\dd^2 x}{\dd u\, \dd v},\qquad
+G' = \sum A\, \frac{\dd^2 x}{\dd v^2}, \\
+\frac{\cos\theta}{R}
+  = \frac{1}{H}\, \frac{E' · du^2 + 2F' · du\, dv + G'\Add{·} dv^2}{ds^2},
+\end{gather*}
+ou enfin
+\[
+\Tag{(6)}
+\frac{\cos \theta}{R} = \frac{1}{H} · \frac{\Psi(du, dv)}{\Phi(du, dv)}.
+\]
+
+\MarginNote{Formule pour $\dfrac{\sin\theta}{R}$\Add{.}}
+Les formules~\Eq{(5)} donnent encore
+\[
+\frac{\sin \theta}{R}
+  = \sum a_1\, \frac{da}{ds}
+  = \sum a_1\, \frac{d^2x}{ds^2}\Add{.}
+\]
+Remarquons que
+\[
+\sum a_1\, \frac{d^2x}{\Err{ds}{ds^2}}
+  = \frac{1}{ds^2}
+  \begin{vmatrix}
+    a     & b     & c     \\
+    d^2 x & d^2 y & d^2 z \\
+    l     & m     & n
+  \end{vmatrix}
+  = \frac{1}{ds^3}
+  \begin{vmatrix}
+    dx    & dy    & dz    \\
+    d^2 x & d^2 y & d^2 z \\
+    l     & m     & n
+  \end{vmatrix};
+\]
+pour calculer le déterminant, multiplions-le par
+\[
+\begin{vmatrix}
+\mfrac{\dd x}{\dd u} &  \mfrac{\dd y}{\dd u} &  \mfrac{\dd z}{\dd u} \\
+\mfrac{\dd x}{\dd v} &  \mfrac{\dd y}{\dd v} &  \mfrac{\dd z}{\dd v} \\
+l & m & n
+\end{vmatrix}
+= Al + Bm + Cn = \frac{A^2 + B^2 + C^2}{H} = H;
+\]
+le produit est
+\[
+\begin{vmatrix}
+\sum \mfrac{\dd x}{\dd u}\, dx   & \sum \mfrac{\dd x}{\dd v}\, dx   & \sum l\, dx   \\
+\sum \mfrac{\dd x}{\dd u}\, d^2x & \sum \mfrac{\dd x}{\dd v}\, d^2x & \sum l\, d^2x \\
+\sum l\, \mfrac{\dd x}{\dd u}    & \sum l\, \mfrac{\dd x}{\dd v}    & \sum l^2
+\end{vmatrix};
+\]
+or\Add{,} nous avons
+\begin{gather*}
+\sum \frac{\dd x}{\dd u} · dx  = \sum \frac{\dd x}{\dd u} · \left(\frac{\dd x}{\dd u}\, du + \frac{\dd x}{\dd v}\, dv\right) = E\, du + F\, dv, \\
+\sum \frac{\dd x}{\dd v} · dx  = \DPtypo{}{\sum} \frac{\dd x}{\dd v} · \left(\frac{\dd x}{\dd u}\, du + \frac{\dd x}{\dd v}\, dv\right)  = F\, du + G\, dv, \\
+\sum l\, dx = \sum l\, \frac{\dd x}{\dd u} = \sum l\, \frac{\dd x}{\dd v} = 0,
+\end{gather*}
+%% -----File: 039.png---Folio 31-------
+\begin{align*}
+\sum \frac{\dd x}{\dd u}\, d^2x
+  &= \sum \frac{\dd x}{\dd u} \left(\frac{\dd x}{\dd u}\, d^2u + \frac{\dd x}{\dd v}\, d^2v
+   + \frac{\dd^2 x}{\dd u^2}\, du^2 + 2\, \frac{\dd^2 x}{\dd u\, \dd v}\, du\, dv + \frac{\dd^2 x}{\dd v^2}\, dv^2\right) \\
+%
+  &= E\, d^2u + F\, d^2v + \frac{1}{2}\, \frac{\dd E}{\dd u}\, du^2 + \frac{\dd E}{\dd v}\, du·dv + \left(\frac{\dd F}{\dd v} - \frac{1}{2}\, \frac{\dd G}{\dd u}\right) dv^2, \\
+%
+\sum \frac{\dd x}{\dd v}\, d^2x
+  &= \sum \frac{\dd x}{\dd v} \left(\frac{\dd x}{\dd u}\, d^2u + \frac{\dd x}{\dd v}\, d^2v
+   + \frac{\dd^2 x}{\dd u^2}\, du^2 + 2\, \frac{\dd^2 x}{\dd u\, \dd v}\, du\, dv + \frac{\dd^2 x}{\dd v^2}\, dv^2\right) \\
+%
+  &= F\, d^2u + G\, d^2v + \left(\frac{\dd F}{\dd u} - \frac{1}{2}\, \frac{\dd E}{\dd v}\right) du^2 + \frac{\dd G}{\dd u}\, du\, dv + \frac{1}{2}\, \frac{\dd G}{\dd v}\, dv^2.
+\end{align*}
+
+Le produit précédent s'écrit donc
+\[
+\begin{vmatrix}
+E\, du + F\, dv & F\, du + G\, dv  & 0 \\
+\left[
+\begin{aligned}%[** TN: Reformatted very wide entry]
+  &E\, d^2u + F\, d^2v \\
+  &+ \mfrac{1}{2}\, \mfrac{\dd E}{\dd u}\, du^2
+  + \mfrac{\dd E}{\dd v}\, du\, dv
+  + \left(\mfrac{\dd F}{\dd v} - \mfrac{1}{2}\, \mfrac{\dd G}{\dd u}\right) dv^2
+\end{aligned}
+\right]
+  & F\, d^2u + G\, d^2v + \dots  & \sum l\, d^2x \\
+0 & 0 & 1
+\end{vmatrix}
+\]
+ou
+\[
+-\begin{vmatrix}
+E\, d^2u + F\, d^2v
+  + \mfrac{1}{2}\, \mfrac{\dd E}{\dd u}\, du^2
+  + \mfrac{\dd E}{\dd v}\, du\, dv
+  + \left(\mfrac{\dd F}{\dd v} - \mfrac{1}{2}\, \mfrac{\dd G}{\dd u}\right) dv^2
+  & E\, du + F\, dv \\
+F\, d^2u + G\, d^2v
+  + \left(\mfrac{\dd F}{\dd u} - \mfrac{1}{2}\, \mfrac{\dd E}{\dd v}\right) du^2
+  + \mfrac{\dd G}{\dd u}\, du\, dv
+  + \mfrac{1}{2}\, \mfrac{\dd G}{\dd v}\, dv^2
+  & F\, du + G\, dv
+\end{vmatrix}.
+\]
+Ce déterminant peut se décomposer en deux, dont le \Ord{1}{er} est
+\[
+-\begin{vmatrix}
+E\, d^2u + F\, d^2v & E\, du + F\, dv \\
+F\, d^2u + G\, d^2v & F\, du + G\, dv
+\end{vmatrix}
+  = H^2\, (du·d^2v - dv·d^2u),
+\]
+et on a finalement
+\begin{multline*}
+\Tag{(7)}
+\frac{\sin \theta}{R} = \frac{1}{H\, ds^3}
+\left[\MStrut
+\Err{\Omega^2}{-H^2}\, (du\Add{·}d^2v - dv·d^2u)\right. \\
+- \left.\begin{vmatrix}
+  \mfrac{1}{2}\, \mfrac{\dd E}{\dd u}\, du^2
+               + \mfrac{\dd E}{\dd v}\, du\, dv
+               + \left(\mfrac{\dd F}{\dd v}
+                 - \mfrac{1}{2}\, \mfrac{\dd G}{\dd u}\right) dv^2
+  & E\, du + F\, dv \\
+%
+  \left(\mfrac{\dd F}{\dd u} - \mfrac{1}{2}\, \mfrac{\dd E}{\dd v}\right) du^2
+  + \mfrac{\dd G}{\dd u}\, du\, dv
+  + \mfrac{1}{2}\, \mfrac{\dd G}{\dd v}\DPtypo{)}{}\, dv^2
+  & F\, du + G\, dv
+\end{vmatrix}\MStrut
+\right]\Add{.}
+\end{multline*}
+
+\MarginNote{Formule pour $\dfrac{1}{T} - \dfrac{d \theta}{ds}$\Add{.}}
+Enfin la formule~\Eq{(4)} nous donne
+\[
+\frac{1}{T} - \frac{d\theta}{ds} = \sum a_1\, \frac{dl}{ds}
+  = \frac{1}{ds}
+\begin{vmatrix}
+a  & b  & c  \\
+dl & dm & dn \\
+l  & m  & n
+\end{vmatrix}
+  = \frac{1}{ds^2}
+\begin{vmatrix}
+dx & dy & dz \\
+dl & dm & dn \\
+l  & m  & n
+\end{vmatrix};
+\]
+%% -----File: 040.png---Folio 32-------
+pour calculer le déterminant, nous le multiplierons encore
+par le même déterminant~$H$. Le produit sera,
+\[
+\begin{vmatrix}
+\sum \mfrac{\dd x}{\dd u}\, dx & \sum \mfrac{\dd x}{\dd v}\, dx            & \sum l\, dx \\
+\sum \mfrac{\dd x}{\dd u}\, dl & \sum \mfrac{\dd x}{\dd v}\, d\DPtypo{}{l} & \sum l\, dl \\
+\sum l\, \mfrac{\dd x}{\dd u}  & \sum l\, \mfrac{\dd x}{\dd v}             & \sum l^2
+\end{vmatrix}
+=
+\begin{vmatrix}
+E\, du + F\, dv                & F\, du + G\, dv                & 0 \\
+\sum \mfrac{\dd x}{\dd u}\, dl & \sum \mfrac{\dd x}{\dd v}\, dl & 0 \\
+0                              & 0                              & 1
+\end{vmatrix}.
+\]
+Nous avons d'ailleurs
+\[
+\sum l\, \frac{\dd x}{\dd u} = 0\Add{,}
+\]
+d'où en différentiant
+\[
+\sum \DPtypo{l}{dl}\, \frac{\dd x}{\dd u}
+  = -\sum \DPtypo{dl}{l}
+  \left(\frac{\dd^2 x}{\dd u^2}\, du + \frac{\dd^2 x}{\dd u\, \dd v}\, dv\right)
+  = -\frac{1}{H}\, (E'\, du + F'\, dv);
+\]
+de même
+\[
+\sum dl\, \frac{\dd x}{\dd v} = -\frac{1}{H}\, (F' · du + G' · dv)\Add{;}
+\]
+le produit est donc
+\[
+-\frac{1}{H} \begin{vmatrix}
+E\,  du + F\,  dv & F\,  du + G\,  dv \\
+E'\, du + F'\, dv & F'\, du + G'\, dv
+\end{vmatrix},
+\]
+et nous avons
+\[
+\Tag{(8)}
+\frac{1}{T} - \frac{d\theta}{ds} = \frac{1}{H^2\, ds^2}
+\begin{vmatrix}
+E'\, du + F'\, dv & E\, du + F\, dv \\
+F'\, du + G'\, dv & F\, du + G\, dv
+\end{vmatrix}.
+\]
+
+Les \Card{3} formules \Eq{(6)}\Add{,}~\Eq{(7)}\Add{,}~\Eq{(8)} permettent de calculer les
+\Card{3} éléments fondamentaux $\theta, R, T$.
+
+
+\ExSection{II}
+
+\begin{Exercises}
+\item[8.] On considère la surface~$S$ lieu des sections circulaires diamétrales
+d'une famille \DPchg{d'ellipsoides}{d'ellipsoïdes} homofocaux. Déterminer
+sur~$S$ les trajectoires orthogonales des sections circulaires
+qui l'engendrent.
+
+\item[9.] Déterminer toutes les représentations conformes d'une sphère
+sur un plan. Trouver celles qui donnent des systèmes connus
+de projections cartographiques.
+
+\item[10.] Sur une surface~$S$ on considère une courbe~$C$. Soit $M$ un de ses
+points, $MT$~la tangente à~$C$, $MN$~la normale à~$S$, et $MN'$~la normale
+à~$C$ qui \DPtypo{es}{et} tangente à~$(S)$. Montrer que les composantes
+de la rotation instantanée du trièdre~$\Tri{M}{T}{N'}{N}$ par rapport aux
+axes de ce trièdre sont les éléments fondamentaux $\dfrac{d\theta}{ds} - \dfrac{1}{T}$,
+$-\dfrac{\cos\theta}{R}$, $\dfrac{\sin\theta}{R}$.
+
+\item[11.]\phantomsection\label{exercice11}
+Si les courbes coordonnées de la surface~$S$, de l'exercice
+précédent, sont rectangulaires, soient $MU$~et~$MV$ leurs tangentes,
+et soit~$\phi$ l'angle~$(MU, MT)$. Déduire de la considération
+des mouvements des deux trièdres $\Tri{M}{T}{N'}{N}$~et~$\Tri{M}{U}{V}{N}$, lorsque $M$
+décrit~$C$, une formule de la forme
+\[
+\frac{\sin\theta}{R} - \frac{d\phi}{ds}
+  = r_{1}\frac{du}{ds} + r_{2}\frac{dv}{ds};
+\]
+et donner les expressions de $r_{1}$~et~$r_{2}$. Généraliser, en supposant
+les coordonnées $u$~et~$v$ quelconques.
+\end{Exercises}
+%% -----File: 041.png---Folio 33-------
+
+
+\Chapitre{III}{\DPchg{Etude}{Étude} des \DPtypo{Elements}{\DPchg{Eléments}{Éléments}} Fondamentaux des Courbes d'une
+Surface.}
+
+
+\Section{Courbure normale\Add{.}}
+{1.}{} Reprenons la \Ord{1}{ère} formule fondamentale
+\[
+\frac{\Cos \theta}{R}
+  = \frac{1}{H}\, \frac{E'\, du^2 + 2F'\, du\, dv + G'\, dv^2}
+                       {E\,  du^2 + 2F\,  du\, dv + G\,  dv^2},
+\]
+les différentielles secondes $d^2 u, d^2 v$ n'y figurent pas; $\dfrac{\Cos \theta}{R}$
+ne \DPtypo{depend}{dépend} que du rapport $\dfrac{dv}{du}$, c'est à dire de la direction
+de la tangente, \emph{$\dfrac{\Cos \theta}{R}$~est le même pour toutes les courbes de
+la surface tangentes à une même droite}. Considérons alors le
+\Figure{041a}
+centre de courbure~$C$ sur la normale principale~$MP$; si on prend
+pour axe polaire la normale~$MN$ à la surface,
+et pour pôle le point~$M$, $R\Add{,} \theta$ sont les
+coordonnées polaires du point~$C$. L'équation
+\[
+\frac{\Cos \theta}{R} = \cte,
+\]
+représente un cercle; le lieu du point~$C$
+est un cercle, ce qu'on peut encore voir comme il suit; considérons
+la droite polaire, elle est dans le plan normal à
+la courbe, donc elle rencontre la normale~$MN$ à la surface en
+un point~$K$, et nous avons
+\[
+R = M K \cos \theta,
+\]
+ou
+\[
+M K = \frac{R}{\Cos \theta}.
+\]
+%% -----File: 042.png---Folio 34-------
+$M K$~est constant, donc \emph{les droites polaires de toutes les
+courbes d'une surface passant par un même point~$M$ de cette
+surface et tangentes en ce point à une même droite rencontrent
+en un même point~$K$ la normale en~$M$ à la surface. Le lieu des
+centres de courbure de toutes ces courbes est le cercle de
+diamètre $M K$ \(cercle de Meusnier\)}. En particulier supposons
+$\theta = 0$, la normale principale se confond avec la normale à la
+surface, le plan osculateur passe par la normale, il est normal
+à la surface. Coupons la surface par ce plan, $K$~est le
+centre de courbure en~$M$ de la section, soit $R_n$~le rayon de
+courbure, nous avons
+\[
+\frac{\Cos \theta}{R} = \frac{1}{R_n},
+\]
+d'où
+\[
+R = R_n \Cos \theta.
+\]
+D'où le \emph{Théorème de Meusnier: Le centre de courbure en~$M$
+d'une courbe tracée sur une surface est la projection sur le
+plan osculateur en~$M$ à cette courbe du centre de courbure de
+la section normale tangente en~$M$ à la courbe}.
+
+Le Théorème est en \DPtypo{defaut}{défaut} si
+\[
+\Psi(du, dv) = E'\Add{·} du^2 + 2F'\, du · dv + G' · dv^2 = 0\Add{.}
+\]
+Alors $\dfrac{\Cos \theta}{R} = 0$, $R$~est en général infini. La formule devient
+complètement indéterminée si on a en même temps $\Cos \theta = 0$,
+alors la normale principale est perpendiculaire à la normale
+à la surface, le plan osculateur à la courbe est tangent à la
+surface. Les deux tangentes qui correspondent à ce cas d'exception
+s'appellent \emph{les deux directions asymptotiques} (correspondant
+au point~$M$ considéré).
+%% -----File: 043.png---Folio 35-------
+
+Le Théorème est également en défaut si
+\[
+\Phi(du, dv) = E\, du^2 + 2F · du · dv + G\, dv^2 = 0;
+\]
+alors $\dfrac{\cos \theta}{R}$ est infini, $R$~est nul en général. La direction de
+la tangente est telle que
+\[
+dx^2 + dy^2 + dz^2 = 0
+\]
+c'est une droite isotrope du plan tangent. Il y a donc \emph{deux
+directions isotropes correspondantes à chaque point~$M$ de la
+surface}.
+
+
+\Section{Variations de la courbure normale\Add{.}}
+{2.}{} Le Théorème de Meusnier nous montre que, pour étudier
+la courbure des diverses courbes passant par un point d'une
+surface, on peut se borner à considérer les sections normales
+passant par les différentes tangentes à la surface au point
+considéré.
+
+Nous avons
+\[
+\frac{1}{R_n} = \frac{1}{H}\,
+  \frac{E'\, du^2 + 2F'\, du · dv + G'\, dv^2}{E · du^2 + 2F\, du · dv + G · dv^2}\Add{.}
+\]
+Considérons dans la plan tangent en~$M$ les tangentes $M U, M V$
+aux courbes coordonnées $v = \Cte$ et $u = \Cte$ qui passent par~$M$,
+et considérons le trièdre constitué par $M U, M V$ et la normale
+$M N$ à la surface: les cosinus directeurs des axes sont
+\[
+\begin{array}{lr@{\,}lr@{\,}lr@{\,}l}
+MU: & \mfrac{dx}{ds}
+  = \mfrac{\dd x}{\dd u}\Add{·} \mfrac{du}{ds} = \mfrac{1}{\sqrt{E}} · \mfrac{\dd x}{\dd u} =& l', &
+\mfrac{1}{\sqrt{E}} · \mfrac{\dd y}{\dd u} =& m', &
+\mfrac{1}{\sqrt{E}}\Add{·} \mfrac{\dd z}{\dd u} =& n'\Add{,} \\
+%
+MV: & \mfrac{dx}{ds}
+  = \mfrac{\dd x}{\dd v} · \mfrac{dv}{ds} = \mfrac{1}{\sqrt{G}} · \mfrac{\dd x}{\dd v} = & l'', &
+\mfrac{1}{\sqrt{G}} · \mfrac{\dd y}{\dd v} =& m'', &
+\mfrac{1}{\sqrt{G}}\Add{·} \mfrac{\dd z}{\dd v} =& n''\Add{,} \\
+%
+MN: && \Err{1}{l}, && m, && n\Add{.}
+\end{array}
+\]
+%% -----File: 044.png---Folio 36-------
+
+Considérons alors une tangente $M T$ quelconque, \DPtypo{definie}{définie}
+par les valeurs $du, dv$ des différentielles des coordonnées.
+Les cosinus directeurs sont:
+\begin{alignat*}{4}
+\frac{dx}{ds}
+  &= \frac{\dd x}{\dd u}\, \frac{du}{ds} &&+ \frac{\dd x}{\dd v}\, \frac{dv}{ds}
+   &&= \sqrt{E}\, \frac{du}{ds} · l' &&+ \sqrt{G} · \frac{dv}{ds}\, l'', \\
+%
+\frac{dy}{ds}
+  &= \frac{\dd y}{\dd u}\, \frac{du}{ds} &&+ \frac{\dd y}{\dd v}\, \frac{dv}{ds}
+   &&= \sqrt{E}\, \frac{du}{ds} · m' &&+ \sqrt{G}\Add{·} \frac{dv}{ds}\, m'', \\
+%
+\frac{dz}{ds}
+  &= \frac{\dd z}{\dd u}\, \frac{du}{ds} &&+ \frac{\dd z}{\dd v}\, \frac{dv}{ds}
+   &&= \sqrt{E}\, \frac{du}{ds} · n' &&+ \sqrt{G}\Add{·} \frac{dv}{ds}\, n''\Add{.}
+\end{alignat*}
+
+Ces formules montrent que le segment directeur de $M T$
+est la somme géométrique de deux segments, de valeurs algébriques
+\[
+\lambda = \sqrt{E}\, \frac{du}{ds},\qquad \mu = \sqrt{G}\, \frac{dv}{ds},
+\]
+portés respectivement sur $M U$~et~$M V$. En d'autres termes:
+$\lambda, \mu$~sont les paramètres directeurs de~$M T$ dans le système
+de coordonnées $U\Add{,} M\Add{,} V$.
+
+La formule de $R_n$ devient, en y introduisant ces paramètres
+directeurs:
+\begin{align*}%[** TN: Not broken in original]
+\frac{1}{R_n}
+  &= \frac{1}{H}\left( E' \Bigl(\frac{du}{ds}\Bigr)^2 + 2 F'\, \frac{du}{ds} · \frac{dv}{ds} + G' \Bigl(\frac{dv}{ds}\Bigr)^2 \right) \\
+  &= \frac{1}{H} \left(\frac{E'}{E}\, \lambda^2 + \frac{2F'}{\sqrt{EG}}\, \lambda \mu + \frac{G'}{G}\, \mu^2 \right)\Add{.}
+\end{align*}
+Et si on considère le point~$P$ obtenu en portant sur~$M T$ un
+segment égal à $±\sqrt{R_n}$, le lieu de ce point~$P$, dont les coordonnées,
+dans le système $M\Add{,} U\Add{,} V$, sont:
+\[
+U = ±\lambda \sqrt{R_n},\qquad
+V = ±\mu \sqrt{R_n},
+\]
+aura pour équation
+\[
+\frac{E'}{E}\, U^2 + \frac{2F'}{\sqrt{EG}}\, UV + \frac{G'}{G}\, V^2 = H.
+\]
+
+C'est une conique à centre située dans le plan tangent,
+qu'on appelle \emph{indicatrice} de la surface au point~$M$. La conique
+\DPtypo{tracee}{tracée}, on a immédiatement le rayon de courbure d'une section
+%% -----File: 045.png---Folio 37-------
+normale quelconque, et on suit sans peine la variation du rayon
+de courbure, quand $M T$ varie.
+
+La nature de l'indicatrice dépend du signe de $\dfrac{E'G' - F'{}^2}{E\Del{·} G}$,
+ou puisque $E\Add{,} G$ sont positifs, de $E'G' - F'{}^2$. Dans le cas
+ou l'équation de la surface est
+\[
+Z = f(x, y)
+\]
+on a en prenant les notations habituelles
+\[
+ds^2 = (1 + p^2) · dx^2 + 2pq · dx · dy + (1 + q^2)\Add{·} dy^2
+\]
+d'où
+\[
+E = 1 + p^2\Add{,}\qquad F = p \Del{·} q\Add{,}\qquad G = 1 + q^2
+\]
+et
+\[
+H = \sqrt{E · G - F^2} = \sqrt{1 + p^2 + q^2}\Add{.}
+\]
+Maintenant
+\[
+A = -p,\qquad B = -q,\qquad C = 1,
+\]
+et
+\[
+\sum A · d^2x = -\sum dA · dx  = dp · dx + dq\Add{·} dy\Add{.}
+\]
+Mais
+\[
+dp = r\, dx + s\, dy,\qquad
+dq = s\, dx + t · dy,
+\]
+donc
+\[
+\sum A · d^2x = r · \Err{d^2x}{dx^2} + 2s · dx\, dy + t\Add{·} dy^2;
+\]
+donc
+\[
+E' = r,\qquad F' = s,\qquad G' = t,
+\]
+et
+\[
+E'G' - F'{}^2 = rt - s^2.
+\]
+
+\ParItem{\Primo} $E'G' - F'{}^2 > 0$, la conique est une \Err{e lipse}{ellipse}, tous les
+rayons de courbure sont de même signe, on dit que la surface
+est \emph{convexe} au point~$M$; elle est toute entière d'un même côté
+du plan tangent en~$M$ dans le voisinage du point~$M$.
+
+\ParItem{\Secundo} $E'G' - F'{}^2 < 0$, l'indicatrice est une hyperbole. La surface
+traverse au point~$M$ son plan tangent; elle est dite \emph{à
+courbures opposées}.
+
+\ParItem{\Tertio} $E'G' - F'{}^2 = 0$\Add{,} l'indicatrice est du genre parabole, et
+comme elle est à centre, elle se réduit à un système de deux
+droites parallèles. Le point~$M$ est dit \emph{point parabolique}.
+%% -----File: 046.png---Folio 38-------
+
+Considérons le cas particulier où $\dfrac{1}{R_n}$ est constant, quelle
+que soit la section que l'on \DPtypo{considèré}{considéré}. Il faut et il suffit
+pour cela que $\dfrac{1}{R_n}$ soit indépendant de~$\dfrac{du}{dv}$, donc que l'on ait
+\[
+\frac{E'}{E} = \frac{F'}{F} = \frac{G'}{G}\DPtypo{,}{.}
+\]
+%[** TN: Removed paragraph indentation/break.]
+Or\Add{,} l'angle de $M U, M V$ est donné par
+\[
+\cos \theta = \sum l' · l'' = \frac{F}{\sqrt{EG}};
+\]
+%[** TN: Removed paragraph indentation/break.]
+\DPtypo{Ces}{ces} conditions peuvent donc s'écrire\Del{:}
+\[
+\frac{E'}{E} =  \frac{\ \dfrac{F'}{\sqrt{EG}}\ }{\Cos \theta} = \frac{G'}{G},
+\]
+et expriment que l'indicatrice est un cercle.
+Le point~$M$ est dit alors un \emph{ombilic}.
+
+Cherchons les directions des axes de l'indicatrice. Ce
+sont des directions conjuguées par rapport aux directions asymptotiques
+de l'indicatrice, définies par
+\[
+\Psi(du, dv) = 0
+\]
+et par rapport aux directions isotropes du plan tangent, définies
+par
+\[
+\Phi(du, dv) = 0\Add{.}
+\]
+Elles sont donc définies par l'équation
+\[
+\frac{\ \dfrac{\dd \Psi}{\dd\, du}\ }{\dfrac{\dd \Phi}{\dd\, du}}
+  = \frac{\ \dfrac{\dd \Psi}{\dd\, dv}\ }{\dfrac{\dd \Phi}{\dd\, dv}}
+  = \frac{\Psi(du, dv)}{\Phi(du, dv)} = \frac{H}{R} = S,
+\]
+puisque $du, dv$ sont des coordonnées homogènes pour les directions
+$M T$ du plan tangent.
+
+Ce sont les \emph{directions principales}. Les rayons de courbure
+correspondants sont dits \emph{rayons de courbure principaux}.
+%% -----File: 047.png---Folio 39-------
+
+L'équation qui définit les directions principales est
+donc:
+\[
+\begin{vmatrix}
+E · du + F · dv & F · du + G · dv \\
+E' ·du + F'· dv & F'· du + G'· dv
+\end{vmatrix} = 0;
+\]
+le \Ord{1}{er} membre est un covariant simultané des formes $\Phi, \Psi$.
+
+L'équation aux rayons de courbure principaux s'obtiendra
+en éliminant $du, dv$ entre les équations
+\[
+\frac{\dd \Psi}{\dd\Add{·} du} = S\, \frac{\dd \Phi}{\dd\Add{·} du},\qquad
+\frac{\dd \Psi}{\dd\Add{·} dv} = S\, \frac{\dd \Phi}{\dd\Add{·} dv}\DPtypo{,}{.}
+\]
+Ce qui donne
+\[
+\begin{vmatrix}
+E' - S E & F' - S F \\
+F' - S F & G' - S G
+\end{vmatrix} = 0,
+\]
+ou
+\[
+S^2 (E · G - F^2) - S (E · G' + G · E' - 2FF') + E'G' - F'{}^2 = 0
+\]
+avec
+\[
+S = \frac{H}{R}.
+\]
+
+Supposons maintenant que les courbes coordonnées soient
+tangentes aux directions principales. Ces directions sont rectangulaires;
+donc les courbes coordonnées constituent un réseau
+orthogonal, et $F = 0$; alors l'indicatrice étant rapportée
+à ses axes on a
+\[
+F' = 0,\quad H = \sqrt{E · G},\quad\text{et}\quad
+\frac{1}{R_n} = \frac{\lambda^2 E'}{E \sqrt{EG}} + \frac{\mu^2 G'}{G \sqrt{EG}}\Add{.}
+\]
+Si nous supposons $\lambda = 1$, $\mu = 0$, nous avons un des rayons de
+courbure principaux~$R_1$
+\[
+\frac{1}{R_1} = \frac{E'}{E \sqrt{EG}}\Add{;}
+\]
+pour $\mu = 1$, $\lambda = 0$, nous avons l'autre rayon de courbure principal~$R_2$
+\[
+\frac{1}{R_2} = \frac{G'}{G \sqrt{EG}}\Add{,}
+\]
+%% -----File: 048.png---Folio 40-------
+et la formule devient
+\[
+\frac{1}{R_n} = \frac{\lambda^2}{R_1} + \frac{\mu^2}{R_2}\Add{;}
+\]
+mais ici, les coordonnées étant rectangulaires, si $\phi$ est
+l'angle de la tangente $M T$ avec l'une des directions principales,
+nous avons $\lambda = \cos \phi$, $\mu = \sin \phi$, et nous obtenons
+la \emph{formule d'Euler}
+\[
+\frac{1}{R_n} = \frac{\cos^2 \phi}{R_1} + \frac{\sin^2 \phi}{R_2}\Add{.}
+\]
+Considérons la tangente $M T'$ perpendiculaire à~$M T$, il faudra
+remplacer $\phi$ par $\phi + \frac{\pi}{2}$, et nous aurons
+\[
+\frac{1}{R'_n} = \frac{\sin^2 \phi}{R_1} + \frac{\cos^2 \phi}{R_2}
+\]
+d'où
+\[
+\frac{1}{R_n} + \frac{1}{R'_n} = \frac{1}{R_1} + \frac{1}{R_2}
+\]
+donc \emph{la somme des courbures de \Card{2} sections normales rectangulaires
+quelconques est constante et égale à la somme des courbures
+des sections normales principales}. La quantité constante
+$\dfrac{1}{2} \left(\dfrac{1}{R_1} + \dfrac{1}{R_2}\right)$ s'appelle \emph{courbure moyenne} de la surface au point
+considéré.
+
+
+\Section{Lignes minima.}
+{3.}{} En chaque point d'une surface, il y a \Card{3} couples de
+directions remarquables: les droites isotropes du plan tangent,
+définies par $\Phi(du, dv) = 0$; les directions asymptotiques
+de l'indicatrice $\Psi(du, dv) = 0$\DPtypo{;}{,} et les directions des sections
+principales.
+
+Considérons les directions isotropes, et cherchons s'il
+existe sur la surface des courbes tangentes en chacun de leurs
+points à une direction isotrope; ceci revient à intégrer l'équation
+%% -----File: 049.png---Folio 41-------
+\[
+\Phi(du, dv) = 0,
+\]
+et on obtient ainsi les \emph{courbes minima}. L'équation \DPtypo{précedente}{précédente}
+se décompose en \Card{2} équations de \Ord{1}{er} ordre, donc \emph{il y a deux
+familles de courbes minima et par tout point de la surface
+passe en général une courbe de chaque famille}. Ces courbes
+sont imaginaires; on a le long de chacune d'elles
+\[
+ds^2 = dx^2 + dy^2 + dz^2 = 0;
+\]
+c'est pourquoi on les appelle aussi lignes de longueur nulle.
+Si on les prend pour lignes coordonnées, l'équation $\Phi(du, dv) = 0$
+devant alors être vérifiée pour $du = 0$, $dv = 0$, on a
+\[
+E = 0\Add{,}\quad G = 0,\quad \text{et}\quad ds^2 = 2F\, du · dv.
+\]
+
+En général les deux systèmes de lignes minima sont distincts.
+Pour qu'ils soient confondus, il faut que
+\[
+EG - F^2 = H^2 = 0,
+\]
+dans ce cas, on a $A^2 + B^2 + C^2 = 0$, et les formules fondamentales
+ne s'appliquent plus. Pour étudier la nature d'une telle surface
+\DPtypo{Considérons}{considérons} le plan tangent:
+\[
+A(X - x) + B(Y - y) + C(Z - z) = 0;
+\]
+ce plan est alors tangent à un cône isotrope, c'est un \emph{plan isotrope}.
+\emph{Tous les plans tangents à la surface sont isotropes.}
+Cherchons l'équation générale des plans isotropes. Soit
+\[
+ax + by + cz + d = 0
+\]
+nous avons la condition
+\[
+a^2 + b^2 + c^2 = 0
+\]
+ou
+\begin{gather*}
+a^2 + b^2 = -c^2\Add{,} \\
+(a + ib) · (a - ib) = \Err{-c}{-c^2}\Add{.}
+\end{gather*}
+%% -----File: 050.png---Folio 42-------
+Posons
+\[
+a + ib = tc,\qquad
+a - ib = -\frac{1}{t}\, c,
+\]
+ou
+\[
+a + ib - tc = 0,\qquad ta - ibt + c = 0;
+\]
+de ces deux relations nous tirons
+\[
+\frac{a}{1 - t^2} = \frac{ib}{-(1 + t^2)} = \frac{c}{-2t},
+\]
+ou
+\[
+\frac{a}{1 - t^2} = \frac{b}{i(1 + t^2)} = \frac{c}{-2t};
+\]
+d'où l'équation générale des plans isotropes
+\[
+\Tag{(1)}
+(1 - t^2)x + i(1 + t^2)y - 2tz + 2w = 0.
+\]
+Un plan isotrope dépend de deux paramètres. La surface considérée
+est l'enveloppe de plans isotropes; si ces plans dépendent
+de deux paramètres, elle se réduit au cercle imaginaire
+à l'infini. Supposons alors que $w$~soit fonction de~$t$ par exemple;
+le plan tangent ne dépendant que d'un paramètre, la surface
+est développable, c'est une \emph{développable isotrope}. Cherchons
+son arête de rebroussement. Différentions l'équation~\Eq{(1)}
+\Card{2} fois par rapport à~$t$. Nous avons
+\begin{gather*}
+\Tag{(2)}
+-tx + ity - z + w' = 0 \\
+\Tag{(3)}
+-x + iy + w'' = 0
+\end{gather*}
+les équations \Eq{(1)}\Add{,}~\Eq{(2)}\Add{,}~\Eq{(3)} définissent l'arête de rebroussement;
+\Eq{(3)}~donne
+\[
+x - iy = w''\Add{,}
+\]
+\Eq{(2)}~s'écrit
+\[
+z = -t(x - iy) + w' = w' - tw''\Add{,}
+\]
+et~\Eq{(1)}
+\[
+x + iy = t^2(x - iy) + 2tz - 2w = t^2w'' + 2t(w' - tw'') - 2w
+\]
+d'où, pour les équations de l'arête de rebroussement:
+\[
+\Tag{(4)}
+x - iy = w'',\qquad
+x + iy = -2w + 2tw' - t^2w'',\qquad
+z = w' - tw''.
+\]
+Nous en tirons
+\[
+d(x - iy) = w'''\, dt,\qquad
+d(x + iy) = -t^2w'''\, dt,\qquad
+dz = -tw'''\, dt;
+\]
+%% -----File: 051.png---Folio 43-------
+d'où
+\[
+d(x - i y) · d(x + i y)
+  = - t^2 \DPtypo{w'''}{(w''')^2}\, dt^2
+  = \DPtypo{}{-}dz^2\Add{,}
+\]
+ou
+\begin{gather*}
+d(x - i y) · d(x + i y) + dz^2 = 0, \\
+dx^2 + dy^2 + dz^2 = 0;
+\end{gather*}
+c'est une courbe minima. \emph{L'arête de rebroussement d'une développable
+isotrope est une courbe minima.}
+
+Réciproquement, considérons une courbe minima. Nous avons
+la relation
+\[
+dx^2 + dy^2 + dz^2 = 0\Add{.}
+\]
+Différentions
+\[
+dx · d^2 x + dy · d^2 y + dz · d^2 z = 0\Add{,}
+\]
+mais l'identité de Lagrange nous donne
+\[
+\sum dx^2 \sum (d^2 x)^2 - \DPtypo{\sum (dx·d^2 x)}{\left(\sum dx·d^2 x\right)^2}
+  = \sum (dy·d^2 z - dz·d^2 y)^2 = 0\Add{,}
+\]
+ou\DPtypo{,}{} $A\Add{,} B\Add{,} C$ désignant les coefficients du plan osculateur
+\[
+A^2 + B^2 + C^2 = 0\Add{.}
+\]
+\emph{Le plan osculateur en un point d'une courbe minima est isotrope.
+Toute courbe minima peut être considérée comme l'arête
+de rebroussement d'une développable isotrope.}
+
+Il en résulte que cette arête de rebroussement est la
+courbe minima la plus générale, et que les coordonnées d'un
+point d'une courbe minima quelconque sont données par les
+formules~\Eq{(4)}, ou $w$~est une fonction arbitraire de~$t$.
+
+%[** TN: Renumber 3 -> 4]
+\Section{Lignes asymptotiques.}
+{4.}{} Si nous cherchons maintenant les courbes d'une surface
+tangentes en chacun de leurs points à une asymptote de
+l'indicatrice, nous sommes ramenés à intégrer l'équation
+\[
+\Psi(du , dv) = 0\Add{,}
+\]
+et nous obtenons les \emph{lignes asymptotiques}. Comme précédemment,
+%% -----File: 052.png---Folio 44-------
+nous voyons qu'\emph{il y a deux familles de lignes asymptotiques,
+et par tout point de la surface passe en général une asymptotique
+de chaque famille}.
+
+L'équation différentielle précédente s'écrit
+\begin{align*}
+\sum A\, d^2 x &= 0\Add{,} \\
+\intertext{on a d'ailleurs}
+\sum A\, dx &= 0;
+\end{align*}
+mais $A\Add{,} B\Add{,} C$ sont les coefficients du plan tangent à la surface;
+les équations précédentes montrent qu'il contient les directions
+$dx, dy, dz$ et $d^2 x, d^2 y, d^2 z$, donc \DPtypo{coincide}{coïncide} avec le plan osculateur
+à la courbe; donc \emph{les lignes asymptotiques sont telles
+que le plan osculateur en chacun de leurs points soit tangent
+à la surface}. En particulier, \emph{toute génératrice rectiligne
+d'une surface est une ligne asymptotique}, car le plan osculateur
+en un point d'une droite étant indéterminé, peut être
+considéré comme \DPtypo{coincidant}{coïncident} avec le plan tangent en ce point
+à la surface. \emph{Si donc une surface est réglée, un des systèmes
+de lignes asymptotiques est constitué par les génératrices
+rectilignes.}
+
+Si nous prenons les lignes asymptotiques pour courbes
+coordonnées, nous aurons
+\[
+E' = G' = 0
+\]
+et
+\[
+\Psi(du, dv) = 2F'\, du · dv.
+\]
+
+Les lignes asymptotiques sont réelles aux points où la
+surface est à courbures opposées, imaginaires aux points où
+elle est convexe. Elles sont en général distinctes, et distinctes
+aussi des lignes minima. Nous allons examiner les cas
+d'exception.
+
+\ParItem{\Primo.} \emph{Les lignes asymptotiques sont confondues.} Prenons
+%% -----File: 053.png---Folio 45-------
+l'équation de la surface sous la \DPtypo{formé}{forme}
+\[
+Z = f(x, y):
+\]
+nous avons $E'G' - F^2 = 0$, condition qui se réduit ici à
+\[
+rt - s^2 = 0;
+\]
+tous les points de la surface doivent être paraboliques. L'équation
+différentielle précédente peut s'écrire
+\[
+dp \DPtypo{,}{·} dx + dq · dy = 0.
+\]
+Elle montre que si l'une des différentielles $dp, dq$ est nulle,
+l'autre est aussi nulle, donc $p, q$ sont fonctions l'un de l'autre.
+Le plan tangent en un point s'écrit
+\[
+p(X - x) \DPtypo{,}{+} q(Y - y) - (Z - z) = 0 ,
+\]
+ou
+\[
+pX + qY - Z = px + qy - z.
+\]
+Mais
+\[
+d(px + qy - z) = x·dp + y·dq
+\]
+et nous voyons que si $dp = 0$, puisque $dq = 0$, on a en même
+temps $d(px + qy - z) = 0$, donc $px + qy - z$ est fonction
+de~$p$, de même que~$q$, et alors le plan tangent ne dépend que
+d'un seul paramètre, et la surface est développable. La \DPtypo{reciproque}{réciproque}
+est évidente, car si l'équation $pX + qY - Z = px + qy - z$
+ne dépend que d'un paramètre~$\theta$, $dp$~et~$dq$ sont proportionnels
+à~$d\theta$, et les deux formes linéaires $dp = r·dx + s·dy$\Add{,}
+$dq = s·dx + t·dy$ ne sont pas indépendantes. On a donc bien
+\[
+\begin{vmatrix}
+r & s \\
+s & t
+\end{vmatrix} = rt - s^2 = 0.
+\]
+
+Donc \emph{les surfaces à lignes asymptotiques doubles sont
+les surfaces développables, et les lignes asymptotiques doubles
+sont les génératrices rectilignes. Pour les développables
+isotropes, les lignes asymptotiques doubles sont confondues
+avec les lignes minima doubles, qui sont les génératrices
+%% -----File: 054.png---Folio 46-------
+rectilignes isotropes}.
+
+\Paragraph{Remarque.} Pour les surfaces développables, l'arête de
+rebroussement ayant son plan osculateur tangent à la surface
+doit être considérée comme une ligne asymptotique. Son équation
+est en effet une \DPtypo{integrale}{intégrale} singulière de l'\DPtypo{equation}{équation}
+différentielle des lignes asymptotiques.
+
+\ParItem{\Secundo.} \emph{Une famille de lignes asymptotiques est confondue
+avec une famille de lignes minima.} \DPchg{Ecartons}{Écartons} le cas des développables
+isotropes, qui vient d'être examiné. Prenons les
+lignes minima comme courbes coordonnées, $E = 0$, $G = 0$, et si
+nous supposons que la famille $v = \cte$ constitue une famille
+d'asymptotiques, $dv = 0$ doit être solution de $\Psi(du, dv) = 0$,
+donc \DPtypo{$E' = 0$}{}
+\[
+E' =
+\begin{vmatrix}
+\mfrac{\dd x}{\dd u} & \mfrac{\dd y}{\dd u} & \mfrac{\dd z}{\dd u} \\
+\mfrac{\dd x}{\dd v} & \mfrac{\dd y}{\dd v} & \mfrac{\dd z}{\dd v} \\
+\mfrac{\dd^2 x}{\dd u^2} & \mfrac{\dd^2 y}{\dd u^2} & \mfrac{\dd^2 z}{\dd u^2}
+\end{vmatrix} = 0\Add{.}
+\]
+Il existe donc entre les éléments des lignes de ce déterminant
+une même relation linéaire et homogène.
+On a
+\[
+\left\{%[** Moved brace to left-hand side]
+\begin{aligned}
+\frac{\dd^2 x}{\dd u^2} &= M\, \frac{\dd x}{\dd u} + N\, \frac{\dd x}{\dd v}\Add{,} \\
+\frac{\dd^2 y}{\dd u^2} &= M\, \frac{\dd y}{\dd u} + N\, \frac{\dd y}{\dd v}\Add{,} \\
+\frac{\dd^2 z}{\dd u^2} &= M\, \frac{\dd z}{\dd u} + N\, \frac{\dd z}{\dd v}\Add{.}
+\end{aligned}
+\right.
+\]
+Multiplions respectivement par $\dfrac{\dd x}{\dd u}, \dfrac{\dd y}{\dd u}, \dfrac{\dd z}{\dd u}$ et ajoutons. Le
+coefficient de~$M$ est $E = 0$, le \Ord{1}{e} membre est $\dfrac{1}{2}\, \dfrac{\dd E}{\dd u} = 0$, donc
+$NF = 0$, et comme $F \neq 0$, $N = 0$, et nous avons:
+%% -----File: 055.png---Folio 47-------
+\[
+\frac{\ \dfrac{\dd^2 x}{\dd u^2}\ }{\dfrac{\dd x}{\dd u}} =
+\frac{\ \dfrac{\dd^2 y}{\dd u^2}\ }{\dfrac{\dd y}{\dd u}} =
+\frac{\ \dfrac{\dd^2 z}{\dd u^2}\ }{\dfrac{\dd z}{\dd u}} = M\Add{,}
+\]
+les courbes $v = \cte$ sont des droites, et comme ce sont des
+lignes minima, ce sont des droites isotropes. Et réciproquement
+si les courbes $v = \cte$ sont des droites, on a
+\[
+\frac{\dd^2 x}{\dd u^2} = M\, \frac{\dd x}{\dd u},\qquad
+\frac{\dd^2 y}{\dd u^2} = M\, \frac{\dd y}{\dd u},\qquad
+\frac{\dd^2 z}{\dd u^2} = M\, \frac{\dd z}{\dd u};
+\]
+et par suite
+\[
+\sum A\, \frac{\dd^2 x}{\dd u^2} = M \sum A\, \frac{\dd x}{\dd u} = 0
+\]
+les courbes $v = \cte$ qui sont des droites minima sont des lignes
+asymptotiques. Donc \emph{les surfaces qui ont une famille
+d'asymptotiques confondue avec une famille de lignes minima
+sont des surfaces \DPtypo{réglees}{réglées} à génératrices isotropes, et ces
+génératrices sont les asymptotiques confondues avec les courbes
+minima}.
+
+\ParItem{\Tertio.} \emph{Les deux systèmes d'asymptotiques sont des courbes
+minima.} En prenant toujours les lignes minima comme courbes
+coordonnées, il faut que l'équation $\Psi(du, dv) = 0$ soit satisfaite
+pour $du = 0$, $dv = 0$, il faut donc que $E' = G' = 0$.
+Alors les formes quadratiques $\Phi$~et~$\Psi$ sont proportionnelles.
+Il en est de même avec un système de coordonnées quelconques
+et on~a
+\[
+\frac{E'}{E} = \frac{F'}{F} = \frac{G'}{G}\Add{.}
+\]
+L'indicatrice en un point quelconque est un cercle, \emph{tous les
+points de la surface sont des ombilics}. En reprenant le calcul
+comme précédemment, on verra que la surface admet deux
+systèmes de génératrices rectilignes isotropes. \emph{C'est une
+sphère.}
+%% -----File: 056.png---Folio 48-------
+
+%[** TN: Renumber 4 -> 5]
+\Section{Surfaces minima\Add{.}}
+{5.}{} Ce dernier cas nous a conduit à étudier la surface
+telle que l'indicatrice soit toujours un cercle. Examinons
+maintenant \emph{le cas où cette indicatrice est toujours une hyperbole
+équilatère}. Ceci revient à chercher les surfaces
+pour lesquelles les lignes asymptotiques sont orthogonales.
+Il faut pour cela que l'on ait
+\[
+EG' + GE' - 2FF' = 0,
+\]
+ou
+\[
+\frac{1}{R_1} + \frac{1}{R_2} = 0.
+\]
+Les rayons de courbure en chaque point sont égaux et de signes
+contraires; la surface est dite une \emph{surface minima}.
+
+Prenons pour coordonnées les lignes minima. Alors $E = 0$\Add{,}
+$G = 0$, et
+\[
+ds^2 = 2F · du · dv;
+\]
+la condition précédente donne $F' = 0$, et
+\[
+\Psi(du, dv) = E'\, du^2 + G'\, dv^2.
+\]
+Mais on a
+\[
+F' =
+\begin{vmatrix}
+\mfrac{\dd x}{\dd u} & \mfrac{\dd y}{\dd u} & \mfrac{\dd z}{\dd u} \\
+\mfrac{\dd x}{\dd v} & \mfrac{\dd y}{\dd v} & \mfrac{\dd z}{\dd v} \\
+\mfrac{\dd^2 x}{\dd u\, \dd v} & \mfrac{\dd^2 y}{\dd u\, \dd v} & \mfrac{\dd^2 z}{\dd u\, \dd v}
+\end{vmatrix} = 0.
+\]
+Il existe donc une même relation linéaire et homogène entre
+les éléments des lignes. On a
+\[
+\left\{%[** TN: Moved brace to left]
+\begin{aligned}
+\frac{\dd^2 x}{\dd u\, \dd v} = M\, \frac{\dd x}{\dd u} + N\, \frac{\dd x}{\dd v}\Add{,} \\
+\frac{\dd^2 y}{\dd u\, \dd v} = M\, \frac{\dd y}{\dd u} + N\, \frac{\dd y}{\dd v}\Add{,} \\
+\frac{\dd^2 z}{\dd u\, \dd v} = M\, \frac{\dd z}{\dd u} + N\, \frac{\dd z}{\dd v}\Add{.}
+\end{aligned}
+\right.
+\]
+Multiplions respectivement par $\dfrac{\dd x}{\dd u}$, $\dfrac{\dd y}{\dd u}$, $\dfrac{\dd z}{\dd u}$ et ajoutons. Le
+\Ord{1}{er} membre est $\dfrac{1}{2}\, \dfrac{\dd E}{\dd u} = 0$; le coefficient de~$M$ est $E = 0$;
+%% -----File: 057.png---Folio 49-------
+nous avons donc $N F = 0$, donc $N = 0$. De même en multipliant
+par $\dfrac{\dd x}{\dd v}$, $\dfrac{\dd y}{\dd v}$, $\dfrac{\dd z}{\dd v}$ et ajoutant, on trouvera $M = 0$; donc on a
+\[
+\frac{\dd^2 x}{\dd u\, \dd v} = 0,\qquad
+\frac{\dd^2 y}{\dd u\, \dd v} = 0,\qquad
+\frac{\dd^2 z}{\dd u\, \dd v} = 0.
+\]
+Ce qui donne
+\[
+x = f(u) + \phi(v),\qquad
+y = g(u) + \psi(v),\qquad
+z = h(u) + \chi(v);
+\]
+les surfaces représentées par des équations de cette forme
+sont dites \emph{surfaces de translation. Elles peuvent être engendrées
+de deux façons différentes par la translation d'une
+courbe de forme invariable dont un point décrit une autre
+courbe}. Considérons en effet sur la surface les \Card{4} points
+$M_0(u_0\Add{,} v_0)$, $M_1(u\Add{,} v_0)$, $M_2(u_0\Add{,} v)$, $M(u\Add{,} v)$. D'après les formules précédentes
+ces points sont les sommets d'un parallélogramme. Si,
+laissant $v_0$~fixe, on fait varier~$u$, le point~$M_1$ décrit une
+courbe~$\Gamma$ de la surface; de même si, laissant $u_0$~fixe, on fait
+varier~$v$, le point~$M_2$ décrit une autre courbe~$\Gamma'$ de la surface.
+On peut donc considérer la surface comme engendrée par la
+courbe~$\Gamma$ animée d'un mouvement de translation dans lequel le
+point~$M_2$ décrit la courbe~$\Gamma'$, ou par la courbe~$\Gamma'$ animée d'un
+mouvement de translation dans lequel le point~$M_1$ décrit la
+courbe~$\Gamma$.
+
+Pour les surfaces minima, les \Card{6} fonctions $f\Add{,} g\Add{,} h\Add{,} \phi\Add{,} \psi\Add{,} \chi$ ne
+sont pas quelconques. Elles doivent satisfaire aux relations
+\[
+E = f'{}^2 + g'{}^2 + h'{}^2 = 0,\qquad G = \phi'{}^2 + \psi'{}^2 + \chi'{}^2 = 0;
+\]
+il en résulte que la courbe
+\[
+x = f(u),\qquad y = g(u),\qquad  z = \Err{h()}{h(u)}
+\]
+est une courbe minima, et si nous nous reportons aux équations
+générales d'une courbe minima, nous voyons que nous pouvons
+%% -----File: 058.png---Folio 50-------
+écrire, $F$~étant une fonction quelconque de~$u$
+\begin{align*}%[** TN: Unaligned in original]
+f(u) - ig(u) &= F''(u), \\
+f(u) + ig(u) &= -2F(u) + 2uF'(u) - u^2F''(u), \\
+h(u) &= F'(u) - uF''(u).
+\end{align*}
+De même la courbe
+\[
+x = \phi(v),\qquad
+y = \psi(v),\qquad
+z = \chi(v)
+\]
+étant une courbe minima, on aura, $\Phi$~étant une fonction quelconque
+de~$v$,
+\begin{align*}%[** TN: Unaligned in original]
+\phi(v) - i \psi(v) &= \Phi''(v)\Add{,} \\
+\phi(v) + i \psi(v) &= -2 \Phi(v) + 2v \Phi'(v) - v^2 \Phi''(v)\Add{,} \\
+\chi(v) &= \Phi'(v) - v \Phi''(v);
+\end{align*}
+d'où les coordonnées d'un point de la surface minima la plus
+générale
+\begin{align*}
+x + iy &= - 2F(u) + 2u F'(u) - u^2 F''(u) - 2 \Phi(v) + 2v \Phi'(v) + v^2 \Phi''(v)\Add{,} \\
+x - iy &= F''(u) + \Phi''(v), \\
+z &= F'(u) - u F''(u) + \Phi'(v) - v \Phi''(v).
+\end{align*}
+
+Dans le cas où l'équation de la surface est mise sous la
+forme
+\[
+z = f(x, y),
+\]
+l'équation aux dérivées partielles des surfaces minima est
+\[
+(1 + p^2)·t + (1 + q^2)·r + 2pq\, s = 0.
+\]
+
+%[** TN: Renumber 5 -> 6]
+\Section{Lignes courbure.}
+{6.}{} Les \emph{lignes de courbure} sont les lignes tangentes en
+chacun de leurs points aux directions principales ou axes de
+l'indicatrice. Ce sont les intégrales de l'équation
+\[
+\frac{\dd\Phi}{\dd·du} · \frac{\dd\Psi}{\dd·dv} -
+\frac{\dd\Phi}{\dd·dv} · \frac{\dd\Psi}{\dd·du} = 0,
+\]
+les directions principales étant conjuguées par rapport aux
+directions isotropes et aux directions asymptotiques. Si ces
+\Card{2} couples constituent \Card{4} directions distinctes, les directions
+%% -----File: 059.png---Folio 51-------
+principales seront aussi distinctes et distinctes des précédentes.
+Il en \DPtypo{resulte}{résulte} qu'il n'y aura pas d'autres cas singuliers
+pour les lignes de courbure que ceux déjà rencontrés
+pour les lignes minima et les lignes asymptotiques.
+
+\ParItem{\Primo.} \emph{Surfaces réglées non développables à génératrices
+isotropes \(la sphère exceptée\).} Une famille de lignes minima
+est constituée par des lignes asymptotiques. Prenant les lignes
+minima comme coordonnées, nous avons
+\[
+\Phi = 2 F·du·dv;
+\]
+prenons les lignes $u = \cte$ confondues avec les asymptotiques,
+$du = 0$ doit annuler~$\Psi$; donc
+\[
+\Psi = E' · du^2 + 2F' · du·dv;
+\]
+l'équation différentielle des lignes de courbure est
+\[
+F·dv · F'·du - F·du (E'·du + F'·dv) = 0,
+\]
+ou
+\[
+E'\Del{·}F · du^2 = 0.
+\]
+\emph{Les lignes de courbure sont doubles, ce sont les génératrices
+rectilignes isotropes qui sont déjà lignes minima et asymptotiques.}
+
+\ParItem{\Secundo.} \emph{Sphère.} $\Phi, \Psi$ sont proportionnels, l'équation différentielle
+est identiquement vérifiée. \emph{Sur la sphère toutes
+les lignes sont lignes de courbure.}
+
+\ParItem{\Tertio.} \emph{Surfaces développables non isotropes.} Prenons les
+génératrices rectilignes comme courbes $u = \cte$, ce sont des
+lignes asymptotiques doubles, nous avons
+\begin{align*}
+ds^2 &= E·du^2 + 2F·du·dv + G·dv^2, \\
+\Psi &= E'·du^2;
+\end{align*}
+l'équation différentielle des lignes de courbure est
+\[
+(F·du + G·dv)\, E'·du = 0.
+\]
+%% -----File: 060.png---Folio 52-------
+\emph{Les lignes de courbure sont les génératrices rectilignes, qui
+sont déjà lignes asymptotiques, et leurs trajectoires orthogonales.}
+
+\ParItem{\Quarto.} \emph{Surfaces développables isotropes.} Prenant pour courbe
+$v = \cte$ les lignes minima doubles confondues avec les lignes
+asymptotiques doubles, nous avons
+\[
+\Phi = E·du^2\Add{,}\qquad
+\Psi = E'·du^2\Add{.}
+\]
+L'équation aux lignes de courbure est identiquement vérifiée.
+\emph{Sur les développables isotropes toutes les lignes sont lignes
+de courbure.}
+
+\Paragraph{Remarque.} Pour un plan, les courbes minima sont des
+droites; et toute ligne du plan est asymptotique et ligne de
+courbure.
+
+%[** TN: Renumber 6 -> 7]
+\Section{Courbure géodésique.}
+{7.}{} Examinons maintenant la \Ord{2}{e} formule fondamentale
+\begin{multline*}
+\frac{\sin \theta}{R}
+  = \frac{1}{H\, ds^2} \left[\MStrut
+    H^2\, (du\, d^2v - dv\, d^2u) \right. \\
+  - \left.\MStrut
+    \begin{vmatrix}
+      \mfrac{1}{2}\, \mfrac{\dd E}{\dd u}\, du^2
+      + \mfrac{\dd E}{\dd v}\, du\, dv
+      + \left(\mfrac{\dd F}{\dd v}
+        - \mfrac{1}{2}\, \mfrac{\dd G}{\dd u}\right) dv^2 & E\, du + F\, dv \\
+      %
+      \left(\mfrac{\dd F}{\dd u} - \mfrac{1}{2}\, \mfrac{\dd E}{\dd v}\right) du^2
+      + \mfrac{\dd G}{\dd u}\, du\, dv
+      + \mfrac{1}{2}\, \mfrac{\dd G}{\dd v}\, dv^2 & F\, du + G\, dv
+    \end{vmatrix}
+  \right]\Add{;}
+\end{multline*}
+\Illustration[1.625in]{060a}
+$\theta$~est l'angle de la normale principale avec la
+normale à la surface. Soit $C$~le centre de courbure.
+Considérons la droite polaire, qui rencontre
+le plan tangent en~$G$, nous avons
+\[
+\DPtypo{MO}{MC} = MG \cos \left(\theta - \tfrac{\pi}{2}\right) = MG \sin\theta.
+\]
+$MG$ est ce qu'on appelle le \emph{rayon de courbure
+géodésique~$R_g$}. On a
+\[
+R = R_g \sin\theta.
+\]
+%% -----File: 061.png---Folio 53-------
+
+Le point~$G$ est le \emph{centre de courbure géodésique. La projection
+du centre de courbure géodésique sur la normale principale
+est le centre de courbure}. L'inverse du rayon de courbure
+géodésique s'appelle \emph{courbure géodésique}. Son expression
+ne dépend que de $E, F, G$ et de leurs dérivées; \emph{la courbure géodésique
+se conserve quand on déforme la surface}.
+
+Cherchons s'il existe des courbes de la surface dont le
+rayon de courbure géodésique soit constamment infini. De telles
+courbes sont appelées lignes géodésiques. Alors $\dfrac{\sin \theta}{R}$ est
+constamment nul, et comme $R$~n'est pas constamment infini, $\sin \theta = 0$.
+\emph{Le plan osculateur est normal à la surface en chaque
+point de la courbe.} Les lignes géodésiques sont définies par
+une équation différentielle de la forme
+\[
+v'' = \Phi (u, v, v').
+\]
+De l'étude des équations de cette forme il résulte qu'\emph{il y a
+une ligne géodésique et une seule passant par chaque point de
+la surface et tangente en ce point à une direction donnée du
+plan tangent. Il y en a une et une seule joignant deux points
+donnés dans un domaine suffisamment petit.}
+
+Prenons pour \Err{}{lignes }coordonnées les lignes minima. Alors $E = G = 0$
+et $H^2 = - F^2$. L'équation différentielle des lignes géodésiques
+devient
+\[
+- F^2 (du·d^2 v - dv·d^2 u)
+  - \begin{vmatrix}
+  \mfrac{\dd F}{\dd v}\, dv^2 & F\, dv \\
+  \mfrac{\dd F}{\dd u}\, du^2 & F\, du
+  \end{vmatrix} = 0,
+\]
+%% -----File: 062.png---Folio 54-------
+ou
+\[
+du · d^2v - dv · d^2u
+  + \frac{\dd · \log F}{\dd v}\, du · dv^2
+  - \frac{\dd · \log F}{\dd u}\, du^2\Add{·} dv = 0
+\]
+on voit qu'elle est vérifiée pour $du = 0$, $dv = 0$. Ainsi \emph{les
+lignes minima sont des lignes géodésiques}.
+
+\Paragraph{Remarque.} Si le plan osculateur se confond avec le plan
+tangent, le centre de courbure se confond avec le centre de
+courbure géodésique; et si en particulier on considère un plan
+\emph{dans ce plan la courbure géodésique n'est autre que la courbure}.
+Il en résulte que \emph{les lignes géodésiques du plan sont les
+droites de ce plan}, ce qu'on vérifie facilement par le calcul.
+
+\Illustration{062a}
+\Paragraph{Définition directe de la courbure géodésique.} Considérons
+sur la surface une courbe~$(C)$ et une famille
+de courbes~$(K)$ orthogonales à~$(C)$. Sur chaque
+courbe~$(K)$ portons à partir du point~$M$ où elle
+rencontre la courbe~$(C)$ une longueur d'arc
+constante~$M N$. Pour chaque valeur de cette
+constante nous obtenons une courbe~$(C')$ lieu
+du point~$N$. Prenons comme courbes coordonnées
+les courbes $(C)\Add{,} (C')\Add{,} \dots\Add{,} (v = \cte)$, la courbe~$(C)$ étant $v = 0$,
+et les courbes~$(K)\Add{,} (u = \cte)$. Alors $v$~ne sera autre que la
+longueur d'arc~$MN$. Nous avons
+\[
+ds^2 = E · du^2 + 2F \Add{·} du · dv + G · dv^2\Add{.}
+\]
+La courbe $v = 0$ est orthogonale à toutes les courbes~$(K)$, donc
+on a, quel que soit~$u$
+\[
+F(u, \DPtypo{o}{0}) = 0;
+\]
+$v$~représentant l'arc~$M N$, on a $ds^2 = dv^2$, d'où $G = 1$, et alors
+\[
+ds^2 = E · du^2 + 2F \Add{·} du · dv + dv^2.
+\]
+Nous pouvons de même supposer que sur la courbe~$(C)$\Add{,} $u$~\DPtypo{represente}{représente}
+%% -----File: 063.png---Folio 55-------
+l'arc. Alors pour $v = 0$, on a $ds = du$, donc
+\[
+E(u, 0) = 1,
+\]
+et pour cette courbe~$(C)$ on a
+\[
+H^2 = E · G - F^2 = 1,
+\]
+d'où~$H = 1$. Nous avons alors
+\[
+\frac{\sin \theta}{ R} = - \frac{1}{ds^2}
+  \begin{vmatrix}
+    \mfrac{1}{2}\, \mfrac{\dd E}{\dd u}\, du^2  & E\, du \\
+    \left(\mfrac{\dd F}{\dd u} - \mfrac{1}{2}\, \mfrac{\dd E}{\dd v} \right) du^2 & F\, du
+  \end{vmatrix}
+  = - \frac{1}{2}\, \dfrac{\dd E}{\dd v}.
+\]
+Pour la courbe~$(C')$ nous aurons
+\[
+ds^2 = E\, du^2,
+\]
+d'où
+\[
+ds = \sqrt{E}\, du;
+\]
+prenons la dérivée logarithmique par rapport à $v$
+\[
+\frac{\dd · \log ds}{\dd v}
+  = \frac{\dd · \log \DPtypo{E}{\sqrt{E}}}{\dd v}
+%[** TN: Explicit \cdot; Latin-1 char converts to thinspace]
+  = \frac{1}{\sqrt{E}} \cdot \frac{1}{2}\, \frac{\dfrac{\dd E}{\dd v}}{\sqrt{E}}
+  = \frac{1}{2E} · \frac{\dd E}{\dd v}\Add{.}
+\]
+Si on considère la courbe~$(C)$, $E = 1$, et on a pour cette courbe
+\[
+\frac{\dd · \log ds}{\dd v}
+  = \frac{1}{2}\, \frac{\dd E}{\dd v},
+\]
+d'où
+\[
+\frac{1}{R_{g}}
+  = \frac{\sin \theta}{R}
+  = -\frac{\dd \log · ds}{\dd v}\Add{,}
+\]
+ce qui donne une \DPtypo{definition}{définition} de la courbure géodésique n'empruntant
+aucun élément extérieur à la surface.
+
+\MarginNote{Propriétés des
+géodésiques.}
+Supposons en particulier que toutes les courbes~$(K)$ soient
+des géodésiques. Avec les mêmes conventions que précédemment,
+$du = 0$~doit être une solution de \DPtypo{l'equation differentielle}{l'équation différentielle} des
+lignes géodésiques, ce qui donne
+\[
+\begin{vmatrix}
+  \mfrac{\dd F}{\dd v} & F \\
+  0 & 1
+\end{vmatrix}
+= \dfrac{\dd F}{\dd v} = 0;
+\]
+%% -----File: 064.png---Folio 56-------
+donc $F$~est une fonction de $u$~seulement, et comme $F = 0$ pour
+$v = 0$, $F$~est identiquement nul, et on a
+\[
+ds^2 = E\, du^2 + dv^2;
+\]
+et alors toutes les courbes~$(C)$ coupent orthogonalement les
+géodésiques~$(K)$. Donc \emph{si nous considérons une courbe~$(c)$, si
+nous menons en chaque point de~$(c)$ la géodésique qui lui est
+orthogonale, et si nous portons sur chacune de ces géodésiques
+un arc constant, le lieu des extrémités de ces arcs est une
+courbe~$(c')$ normale aux géodésiques}. Nous obtenons ainsi les
+\emph{courbes parallèles} sur une surface quelconque.
+
+\emph{Réciproquement, si nous considérons une famille de géodésiques
+et leurs trajectoires orthogonales, ces trajectoires
+déterminent sur les géodésiques des longueurs d'arc égales.}
+Toujours avec les mêmes hypothèses, les courbes $u = \cte$ et
+$v = \cte$ étant orthogonales, on a $F = 0$. Les $u = \cte$ étant des
+géodésiques, nous avons
+\[
+\begin{vmatrix}
+-\mfrac{1}{2}\, \mfrac{\dd G}{\dd u}\, dv^2 & 0 \\
+\phantom{-} \mfrac{1}{2}\, \mfrac{\dd G}{\dd v}\, dv^2 & G
+\end{vmatrix}
+  = -\frac{1}{2}\, G \frac{\dd G}{\dd u}\, dv^2 = 0.
+\]
+$G \neq 0$, sans quoi les courbes $u = \cte$ seraient des courbes minima,
+donc $\dfrac{\dd G}{\dd u} = 0$ et $G = \phi(v)$. Calculons alors l'arc d'une
+courbe~$(K)$ compris entre la courbe $v = v_{0}$ et la courbe $v = v_{1}$\Add{.}
+Nous avons
+\begin{align*}
+ds^2 &= G\, dv^2 = \phi(v)\, dv^2, \\
+\intertext{et}
+s &= \int_{v_{0}}^{v_{1}} \sqrt{\phi(v)} · dv;
+\end{align*}
+$s$~est indépendant de~$u$, l'arc est le même sur toutes les géodésiques.
+%% -----File: 065.png---Folio 57-------
+
+Si on prend encore pour~$v$ l'arc sur les courbes $u = \cte$
+on a
+\[
+ds^{2} = E\, du^{2} + dv^{2},
+\]
+et \emph{cette forme est caractéristique du système de coordonnées}
+employé, \emph{constitué par une famille de géodésiques et leurs
+trajectoires orthogonales}.
+
+Prenons alors sur la surface deux points voisins~$A\Add{,} B$.
+Il existe une ligne géodésique et une seule dans le domaine
+de ces deux points et joignant ces deux points. Considérons
+une famille de géodésiques voisines ne se coupant pas dans
+le domaine, et leurs trajectoires orthogonales. Prenons-les
+comme courbes coordonnées. Considérons une ligne quelconque
+de la surface allant de $A$ à~$B$, soit
+\[
+u = f(v)\Add{.}
+\]
+Si $A$ a pour coordonnées $u_{0}, v_{0}$ et~$B$, $u_{0}\Add{,} v_{1}$, la longueur de l'arc~$AB$
+de cette ligne est
+\[
+\int_{v_{0}}^{v_{1}} \sqrt{E\, du^{2} + dv^{2}}
+  = \int_{v_{0}}^{v_{1}} \sqrt{E\bigl(f(v), v\bigr)\, f'{}^{2}(v) + 1}\, dv.
+\]
+Cette intégrale est visiblement minima si $f'(v) = 0$, c'est-à-dire
+si la courbe joignant~$A\Add{,} B$ est la géodésique. Ainsi
+donc, \emph{dans un domaine suffisamment petit entourant deux points
+d'une surface, la géodésique est le plus court chemin entre
+ces deux points}.
+
+%[** TN: Renumber 7 -> 8]
+\Section{Torsion géodésique.}
+{8.}{} Voyons enfin la \Ord{3}{e} formule fondamentale
+\[
+\frac{1}{T} - \frac{d\theta}{ds}
+  = \frac{1}{H^{2}\, ds^{2}}
+\begin{vmatrix}
+\Err{E}{E'}\, du + F'\, dv & F'\, du + G'\, dv \\
+E\, du + F\, dv & F\, du + G\, dv
+\end{vmatrix}.
+\]
+Si $\theta$ est constant, et en particulier constamment nul, la formule
+%% -----File: 066.png---Folio 58-------
+précédente donne la torsion; elle donne donc en particulier
+la torsion d'une géodésique. L'expression précédente ne
+dépend que de~$\dfrac{du}{dv}$, c'est-à-dire de la direction de la tangente\Add{.}
+Considérons alors sur la surface une courbe~$(c)$ et un point~$M$.
+Il existe une géodésique tangente à~$(c)$ au point~$M$ et $\dfrac{1}{T} - \dfrac{d \theta}{ds}$
+est la torsion de cette géodésique. C'est pourquoi $\dfrac{1}{T} - \dfrac{d \theta}{ds}$ s'appelle
+\emph{torsion géodésique}. On voit ainsi que \emph{la torsion géodésique
+en un point d'une courbe est la torsion de la géodésique
+tangente en ce point à la courbe \DPtypo{donnee}{donnée}}. Posons
+\[
+\frac{1}{T_{g}} = \frac{1}{T} - \frac{d \theta}{ds}\Add{;}
+\]
+$T_{g}$~est le \emph{rayon de torsion géodésique}. Contrairement au rayon
+de courbure géodésique, il change dans la déformation des surfaces.
+
+La formule précédente montre que la torsion géodésique
+est nulle si la direction $du, dv$ est une direction principale;
+\emph{la torsion géodésique est nulle pour toute courbe tangente à
+une ligne de courbure}. Il en résulte que \emph{les lignes de courbure
+ont une torsion géodésique constamment nulle \(Théorème
+de Lancret\)}.
+
+$\dfrac{1}{T_{g}}$ est le quotient de deux trinômes du \Ord{2}{e} degré en $du,
+dv$, on peut donc étudier sa variation. Prenons pour courbes
+coordonnées les lignes de courbure, elles sont conjuguées et
+rectangulaires, donc $F - F' = 0$, et
+\[
+\frac{1}{T_{g}} = \frac{1}{H^{2}\, ds^{2}} (E'G - G'E)\, du\, dv
+  = \left( \frac{E'}{E} - \frac{G'}{G} \right) \frac{du}{ds}\, \frac{dv}{ds}\Add{.}
+\]
+Si nous revenons aux notations employées au §1 pour l'étude
+de la courbure normale, les coefficients de direction de la
+%% -----File: 067.png---Folio 59-------
+tangente dans le plan tangent sont
+\[
+\lambda = \sqrt{E}\, \frac{du}{ds}\Add{,} \qquad
+\mu     = \sqrt{G}\, \frac{dv}{ds}\Add{,}
+\]
+et alors
+\[
+\frac{1}{T_{g}}
+  = \frac{1}{\sqrt{EG}} \left(\frac{E'}{E} - \frac{G'}{G}\right) \lambda \mu;
+\]
+les rayons de courbure principaux sont
+\[
+\frac{1}{R_{1}} = \frac{1}{\sqrt{EG}} · \frac{E'}{E}\Add{,} \qquad
+\frac{1}{R_{2}} = \frac{1}{\sqrt{EG}} \Add{·} \frac{G'}{G}\Add{,}
+\]
+d'où
+\[
+\frac{1}{T_{g}} = \left(\frac{1}{R_{1}} - \frac{1}{R_{2}} \right) \lambda\mu\Add{,}
+\]
+d'où la \emph{formule d'Ossian Bonnet}, analogue à la formule d'Euler
+\[
+\frac{1}{T_{g}}
+  = \left(\frac{1}{R_{1}} - \frac{1}{R_{2}} \right) \sin\phi · \cos\phi\Add{.}
+\]
+
+\MarginNote{Théorèmes de
+Joachimsthal.}
+Considérons une courbe~$(c)$ intersection de deux surfaces;
+le plan normal à~$(c)$ en l'un de ses points~$M$ contient la normale
+principale à la courbe et les normales aux deux surfaces. Soit
+$V$~l'angle des normales $MN, MN'$; $\theta, \theta'$ leurs angles avec~$MP$.
+
+Nous avons
+\[
+V = \theta' - \theta;
+\]
+mais
+\[
+\frac{1}{T} - \frac{d \theta}{ds} = \frac{1}{T_{g}} \qquad
+\frac{1}{T} - \frac{d \theta'}{ds} = \frac{1}{T'_{g}}
+\]
+d'où en retranchant
+\[
+\frac{dV}{ds} = \frac{1}{T_{g}} - \frac{1}{T'_{g}}.
+\]
+Supposons alors que $(c)$ soit ligne de courbure des deux surfaces;
+$\dfrac{1}{T_{g}}$~et~$\dfrac{1}{T'_{g}}$ sont tous deux nuls, $\dfrac{dV}{ds} = 0$, $V$ est constant.
+D'où les \emph{Théorèmes de Joachimsthal: Si 2 surfaces se coupent
+suivant une ligne de courbure, leur angle est constant tout
+le long de cette ligne}, et la même formule montre immédiatement
+que réciproquement: \emph{si deux surfaces se coupent sous un
+angle constant, et si l'intersection est ligne de courbure
+pour l'une des surfaces, elle est aussi ligne de courbure pour
+l'autre.} Sur un plan ou sur une sphère, toutes les lignes sont
+%% -----File: 068.png---Folio 60-------
+lignes de courbure; donc \emph{si une ligne de courbure d'une surface
+est plane ou sphérique, le plan ou la sphère qui la contient
+coupe la surface sous un angle constant, et réciproquement, si
+un plan ou une sphère coupe une surface sous un angle constant
+l'intersection est une ligne de courbure de la surface}. Enfin
+si un cercle est ligne de courbure d'une surface, il y a une
+sphère passant par ce cercle qui est tangente à la surface en
+un point du cercle; et, par suite, en tous les points du cercle.
+Donc toute \emph{ligne de courbure circulaire est la courbe de
+contact d'une sphère inscrite ou circonscrite à la surface}.
+
+
+\ExSection{III}
+
+\begin{Exercises}
+\item[12.] On considère la surface
+\[
+x = \frac{c^{2} - b^{2}}{bc} · \frac{uv}{u + v},\quad
+y = \frac{\sqrt{c^{2} - b^{2}}}{b} · \frac{v \sqrt{b^{2} - u^{2}}}{u + v},\quad
+z = \frac{\sqrt{c^{2} - b^{2}}}{c} · \frac{u \sqrt{v^{2} - c^{2}}}{u + v};
+\]
+déterminer ses lignes de courbure, et calculer les rayons de
+courbure principaux.
+
+\item[13.] On considère la surface
+\begin{align*}
+x &= \frac{1}{2} \int (1 - u^{2})\, f(u)\, du
+   + \frac{1}{2} \int (1 - v^{2})\, \phi(v)\, dv, \\
+y &= \frac{1}{2} \int (1 + u^{2})\, f(u)\, du
+   - \frac{1}{2} \int (1 + v^{2})\, \phi(v)\, dv, \\
+z &= \int uf(u)\, du + \int v \phi(v)\, dv.
+\end{align*}
+Calculer les rayons de courbures principaux et les coordonnées
+des centres de courbures principaux. Former l'équation
+différentielle des lignes de courbure et des lignes asymptotiques.
+\DPchg{Etudier}{Étudier} les lignes de courbure en prenant
+\[
+f(u)    = \frac{2m^{2}}{(m^{2} + u^{2})^{2}},\qquad
+\phi(v) = \frac{2m^{2}}{(m^{2} + v^{2})^{2}},
+\]
+et en introduisant de nouvelles coordonnées par les formules
+\[
+u = m \tg \frac{\lambda + i \mu}{2},\qquad
+v = m \tg \frac{\lambda - i \mu}{2}.
+\]
+
+\item[14.] Soient, en coordonnées rectangulaires, les équations
+\begin{align*}
+x &= \frac{1}{2} e^{u} \cos(v - \alpha) + \frac{1}{2} e^{-u} \cos(v + \alpha),\\
+y &= \frac{1}{2} e^{u} \sin(v - \alpha) + \frac{1}{2} e^{-u} \sin(v + \alpha),\\
+z &= u \cos\alpha + v \sin\alpha.
+\end{align*}
+
+\Primo. Pour chaque valeur de $\alpha$, ces formules définissent une
+surface $S_{\alpha}$. Indiquer un mode de génération de cette surface.
+Que sent en particulier $S_{0}$ et $S_{\frac{\pi}{2}}$?
+
+\Secundo. On considère deux de ces surfaces $S_{\alpha}$~et~$S_{\beta}$, et on les fait
+correspondre point par point de manière que les plans tangents
+aux points correspondants soient parallèles. Démontrer que les
+tangentes à deux courbes correspondantes, menées en deux
+points homologues, font un angle constant.
+
+\Tertio. Chercher les lignes de courbure et les lignes asymptotiques
+de~$S_{\alpha}$ et trouver une propriété géométrique des courbes
+auxquelles elles correspondent sur~$S_{\beta}$, dans la transformation
+précédente. Qu'arrive-t-il pour $\alpha = \frac{\pi}{2}$?
+
+\item[15.] Chercher les surfaces dont les lignes de courbure d'un système
+sont les courbes de contact des cônes circonscrits ayant
+leurs sommets sur~$Oz$. Quelles sont les autres lignes de courbure?
+
+\item[16.] \DPchg{Etudier}{Étudier} les surfaces dont les lignes de courbure d'un système
+sont situées sur des sphères concentriques. Que peut-on dire
+des lignes de courbure de l'autre système?
+
+\item[17.] Si les courbes coordonnées $u = \const.$, $v = \const.$ sur une
+surface~$S$ sont les lignes asymptotiques de cette surface, et
+si $l, m, n$ sont les cosinus directeurs de la normale à~$S$, en un
+point quelconque de~$S$, montrer qu'il existe une fonction~$\theta$
+telle que l'on ait
+\begin{alignat*}{5}
+dx &= \theta \Biggl[
+  && m &&\left(\frac{\dd n}{\dd u}\, du - \frac{\dd n}{\dd v}\, dv\right)
+  &&-n &&\left(\frac{\dd m}{\dd u}\, du - \frac{\dd m}{\dd v}\, dv\right)
+  \Biggr], \\
+dy &= \theta \Biggl[
+  && n &&\left(\frac{\dd l}{\dd u}\, du - \frac{\dd l}{\dd v}\, dv\right)
+  &&-l &&\left(\frac{\dd n}{\dd u}\, du - \frac{\dd n}{\dd v}\, dv\right)
+  \Biggr], \\
+dz &= \theta \Biggl[
+  && l &&\left(\frac{\dd m}{\dd u}\, du - \frac{\dd m}{\dd v}\, dv\right)
+  &&-m &&\left(\frac{\dd l}{\dd u}\, du - \frac{\dd l}{\dd v}\, dv\right)
+  \Biggr].
+\end{alignat*}
+Calculer, en partant de ces formules, le~$ds^2$ de la surface,
+l'équation des lignes de courbure, l'équation aux rayons de
+courbure principaux. Calculer la torsion des lignes asymptotiques,
+et montrer qu'elle s'exprime au moyen des rayons de
+courbure principaux seulement.
+\end{Exercises}
+%% -----File: 069.png---Folio 61-------
+
+
+\Chapitre{IV}{Les Six Invariants --- La Courbure Totale.}
+
+
+\Section{Les six \DPtypo{Invariants}{invariants}.}
+{1.}{} Dans l'étude des courbes tracées sur une surface~$(S)$
+ne sont intervenus que les coefficients des deux formes quadratiques
+fondamentales:
+\begin{align*}
+\Phi (du, dv) &= ds^{2} = E\, du^{2} + 2F · du · dv + G\, dv^{2}, \\
+\Psi (du, dv) &= \sum A · d^{2}x = E'\, du^{2} + 2F'du · dv + G'\, dv^{2},
+\end{align*}
+et les différentielles de $u, v$, considérées comme les fonctions
+d'une variable indépendante~$t$ qui correspondent à chaque courbe
+particulière considérée.
+
+Si l'on déplace la surface~$(S)$ dans l'espace, sans la déformer,
+et sans changer les coordonnées superficielles $u, v$ employées,
+ces formes quadratiques demeureront les mêmes, de
+sorte que \emph{leurs six coefficients} $E, F, G$, $E', F', G'$ \emph{sont six invariants
+différentiels, pour le groupe des mouvements dans l'espace}\Add{.}
+
+Cela résulte, pour la forme $ds^{2} = \Phi (du, dv)$, de ce qu'elle
+représente le carré de la différentielle d'un arc qui reste
+le même dans les conditions énoncées.
+
+Dès lors $H = \sqrt{EG - F^{2}}$ est un invariant, et la formule
+\[
+\Psi(du, dv) = H · \Phi(du, dv) · \frac{\cos\theta}{R},
+\]
+dont tous les facteurs du second membre sont invariants, montre
+que $\Psi$~possède aussi la propriété d'invariance.
+
+Il n'y a du reste aucune difficulté à vérifier, par un
+calcul direct, l'invariance des six coefficients sur les formules
+%% -----File: 070.png---Folio 62-------
+qui les définissent:
+\begin{alignat*}{3}
+\Tag{(1)}
+\sum \left(\frac{\dd x}{\dd u} \right)^{2} &= E,\qquad
+&\sum \frac{\dd x}{\dd u}\, \frac{\dd x}{\dd v} &= F,\qquad
+&\sum \left(\frac{\dd x}{\dd v} \right)^{2} &= G, \\
+\Tag{(2)}
+\sum A\, \frac{\dd^{2} x}{\dd u^{2}} &= E',\qquad
+&\sum A\, \frac{\dd^{2} x}{\dd u\, \dd v} &= F',\qquad
+&\sum A\, \frac{\dd^{2} x}{\dd v^{2}} &= G',
+\end{alignat*}
+\DPtypo{où}{ou} l'on se rappelle que $A, B, C$ sont les trois déterminants
+fonctionnels
+\[
+A = \frac{D(y, z)}{D(u, v)}, \quad
+B = \frac{D(z, x)}{D(u, v)}, \quad
+C = \frac{D(x, y)}{D(u, v)}.
+\]
+Rappelons enfin l'identité
+\[
+H = \sqrt{A^{2} + B^{2} + C^{2}} = \sqrt{EG - F^{2}} \Add{.}
+\]
+
+\MarginNote{La forme de la
+surface définie
+par les six invariants.}
+Supposons maintenant que $E, F, G$, $E', F', G'$ aient été calculés,
+en fonction de $u, v$, pour une surface~$(S)$ particulière
+\[
+\Tag{(3)}
+x = f(u, v),\quad
+y = g(u, v),\quad
+z = h(u, v);
+\]
+et considérons les équations \Eq{(1)},~\Eq{(2)} comme un système d'équations
+aux dérivées partielles, où $x, y, z$ sont les fonctions inconnues,
+$u, v$ les variables indépendantes, et $E, F, G$, $E', F', G'$ des
+fonctions données. En vertu de l'invariance que nous venons
+d'expliquer, ce système \DPtypo{differentiel}{différentiel} admettra comme intégrales,
+non seulement les fonctions~\Eq{(3)}, qui \DPtypo{definissent}{définissent}~$(S)$,
+mais encore toutes les fonctions
+\[
+\Tag{(4)}
+\left\{%[** TN: Added brace]
+\begin{alignedat}{4}
+x &= x_{0} &&+ \alpha f &&+ \alpha' g &&+ \alpha'' h,\\
+y &= y_{0} &&+ \beta f &&+ \beta' g &&+ \beta'' h, \\
+z &= z_{0} &&+ \gamma f &&+ \gamma' g &&+ \gamma'' h,
+\end{alignedat}
+\right.
+\]
+qui \DPtypo{definissent}{définissent} les surfaces obtenues en déplaçant~$(S)$ de toutes
+les manières possibles, lorsqu'on donne à $x_{0}, y_{0}, z_{0}$, toutes
+les valeurs constantes possibles, et à $\alpha, \beta, \gamma$, $\alpha', \beta', \gamma'$,
+$\alpha'', \beta'', \gamma''$ toutes les valeurs constantes compatibles avec les
+conditions d'orthogonalité bien connues.
+%% -----File: 071.png---Folio 63-------
+
+Cela donne donc des intégrales dépendant de six constantes
+arbitraires. Nous prouverons que le système \Eq{(1)},~\Eq{(2)} n'en
+a pas d'autres; ce que l'on pourra exprimer en disant que \emph{la
+forme de la surface est entièrement définie par les six invariants
+$E, F, G$, $E', F', G'$}.
+
+On démontre dans la théorie des équations aux dérivées
+partielles que, dans tout système dont l'intégrale générale
+ne dépend que de constantes arbitraires, toutes les dérivées
+partielles d'un certain ordre peuvent s'exprimer en fonction
+des variables indépendantes et dépendantes et des dérivées
+d'ordre inférieur. Nous devons donc essayer de constater que
+cela a lieu pour notre système; et commencer par différentier
+les équations~\Eq{(1)}. Les résultats obtenus peuvent s'écrire:
+\[%[** TN: Rearranged in pairs, added brace]
+\Tag{(5)}
+\left\{
+\begin{aligned}
+&\sum \frac{\dd x}{\dd u} · \frac{\dd^{2}x}{\dd u^{2}}
+  = \frac{1}{2}\, \frac{\dd E}{\dd u},
+&&\sum \frac{\dd x}{\dd v}\Add{·} \frac{\dd^{2}x}{\dd u^{2}}
+  = \frac{\dd F}{\dd u} - \frac{1}{2}\, \frac{\dd E}{\dd v}, \\
+%
+&\sum \frac{\dd x}{\dd u}\Add{·} \frac{\dd^{2}x}{\dd u\, \dd v}
+  = \frac{1}{2}\, \frac{\dd E}{\dd v},
+&&\sum \frac{\dd x}{\dd v}\Add{·} \frac{\dd^{2}x}{\dd u\, \dd v}
+  = \frac{1}{2}\, \frac{\dd G}{\dd u}, \\
+%
+&\sum \frac{\dd x}{\dd u}\Add{·} \frac{\dd^{2}x}{\dd v^{2}}
+  = \frac{\dd F}{\dd v} - \frac{1}{2}\, \frac{\dd G}{\dd u};
+&&\DPtypo{}{\sum} \frac{\dd x}{\dd v} · \frac{\dd^{2}x}{\dd v^{2}}
+  = \frac{1}{2}\, \frac{\dd G}{\dd v};
+\end{aligned}
+\right.
+\]
+et l'on voit qu'en associant ces équations aux équations~\Eq{(2)},
+on obtiendra effectivement les expressions de toutes les dérivées
+du second ordre.
+
+Pour faciliter ce calcul, nous introduirons les cosinus
+directeurs de la normale:
+\[
+\Tag{(6)}
+l = \frac{A}{H},\qquad
+m = \frac{B}{H},\qquad
+n = \frac{C}{H};
+\]
+et nous substituerons à la forme $\sum A\, d^{2}x$ la forme
+\[
+\Tag{(7)}
+\sum l · d^{2}x
+  = \frac{1}{H} \sum A · d^{2}x
+  = L · du^{2} + 2 · M · du · dv + N · dv^{2}
+\]
+de sorte qu'on aura
+\[
+\Tag{(8)}
+L = \frac{E'}{H},\qquad
+M = \frac{F'}{H},\qquad
+N = \frac{G'}{H};
+\]
+et que les équations~\Eq{(2)} seront remplacées par
+%% -----File: 072.png---Folio 64-------
+\[
+\Tag{(9)}
+\sum l\, \frac{\dd^{2} x}{\dd u^{2}} = L,\qquad
+\sum l\, \frac{\dd^{2} x}{\dd u\, \dd v} = M,\qquad
+\sum l\, \frac{\dd^{2} x}{\dd v^{2}} = N.
+\]
+Cela fait, si on pose:
+\begin{alignat*}{3}
+\frac{\dd^{2} x}{\dd u^{2}}
+  &= L'\, \frac{\dd x}{\dd u} &&+ L''\, \frac{\dd x}{\dd v} &&+ L''' l, \\
+\frac{\dd^{2} y}{\dd u^{2}}
+  &= L'\, \frac{\dd y}{\dd u} &&+ L''\, \frac{\dd y}{\dd v} &&+ L''' m, \\
+\frac{\dd^{2} z}{\dd u^{2}}
+  &= L'\, \frac{\dd z}{\dd u} &&+ L''\, \frac{\dd z}{\dd v} &&+ L''' n,
+\end{alignat*}
+$L', L'', L'''$ étant des coefficients à déterminer, on aura pour
+les calculer les conditions
+\[
+\sum \frac{\dd x}{\dd u}\, \frac{\dd^{2} x}{\dd u^{2}} = E L' + F L'',\quad
+\sum \frac{\dd x}{\dd v}\, \frac{\dd^{2} x}{\dd u^{2}} = F L' + G \DPtypo{L'}{L''},\quad
+\sum l\, \DPtypo{\frac{dx}{du}}{\frac{\dd^2 u}{\dd x^2}} = L''';
+\]
+d'où on conclut d'abord $L''' = L$; et ensuite, en se servant
+des formules~\Eq{(5)}, des équations qui donneront $L'$~et~$L''$.
+
+En opérant de même pour les autres dérivées, on obtient
+les résultats suivants
+\[
+\Tag{(10)}
+\left\{
+\begin{alignedat}{5}
+\frac{\dd^{2} x}{\dd u^{2}}
+  &= L'\, &&\frac{\dd x}{\dd u} &&+ L''\, &&\frac{\dd x}{\dd v} &&+ L·l, \\
+\frac{\dd^{2} x}{\dd u\, \dd v}
+  &= M'\, &&\frac{\dd x}{\dd u} &&+ M''\, &&\frac{\dd x}{\dd v} &&+ M·l, \\
+\frac{\dd^{2} x}{\dd v^{2}}
+  &= N'\, &&\frac{\dd x}{\dd u} &&+ N''\, &&\frac{\dd x}{\dd v} &&+ N·l,
+\end{alignedat}
+\right.
+\]
+avec les équations auxiliaires:
+\[
+\Tag{(11)}
+\left\{
+\begin{alignedat}{4}
+E L' &+ F L'' &&= \frac{1}{2}\, \frac{\dd E}{\dd u}, &
+F L' &+ G L'' &&= \frac{\dd F}{\dd u} - \frac{1}{2}\, \frac{\dd E}{\dd v}, \\
+E M' &+ F M'' &&= \frac{1}{2}\, \frac{\dd E}{\dd v}, &
+F M' &+ G M'' &&= \frac{\dd G}{\dd u}, \\
+E N' &+ F N'' &&= \frac{\dd F}{\dd v} - \frac{1}{2}\, \frac{\dd G}{\dd u},
+\quad &
+F N' &+ G N'' &&= \frac{1}{2}\, \frac{\dd G}{\dd v},
+\end{alignedat}
+\right.
+\]
+d'où on déduirait les valeurs des coefficients $L', L''$, $M', M''$, $N',
+N''$. On remarquera qu'elles ne dépendent que des coefficients
+$E, F, G$ de l'élément linéaire $ds^{2} = \Phi(du, dv)$, et des dérivées
+premières de ces coefficients.
+%% -----File: 073.png---Folio 65-------
+
+Enfin, les mêmes équations~\Eq{(10)} subsisteront pour les
+autres coordonnées $y, z$; il n'y aura qu'à y laisser les mêmes
+coefficients, et à y remplacer la lettre~$x$ par la lettre~$y$ ou
+la lettre~$z$, en même temps qu'on changera~$l$ en~$m$ ou en~$n$.
+
+Nous concluons de là que si on \DPchg{connait}{connaît}, pour un système
+de valeurs de $u, v$, les valeurs de $x, y, z$ et de leurs dérivées
+premières, on pourra calculer les valeurs de leurs dérivées
+secondes; et, par des différentiations nouvelles, celles
+de toutes leurs dérivées d'ordre supérieur. Et par suite les
+développements en séries de Taylor d'une intégrale quelconque
+ne peuvent contenir d'autres arbitraires que les valeurs initiales
+de
+\[
+x,\ y,\ z,\quad
+\frac{\dd x}{\dd u},\
+\frac{\dd x}{\dd v},\quad
+\frac{\dd y}{\dd u},\
+\frac{\dd y}{\dd v},\quad
+\frac{\dd z}{\dd u},\
+\frac{\dd z}{\dd v};
+\]
+et encore celles-ci doivent être liées par les équations~\Eq{(1)};
+et, lorsque ces valeurs initiales sont données, l'intégrale
+est entièrement déterminée.
+
+Donc, pour prouver que \Eq{(4)}~donne l'intégrale générale, il
+reste seulement à montrer que \Eq{(4)}~peut satisfaire aux conditions
+initiales énoncées. Or, si nous introduisons les cosinus
+directeurs $l', m', n'$; $l'', m'', n''$ des tangentes $MU, MV$ aux deux
+courbes coordonnées qui passent par un point quelconque~$M$ de
+la surface, nous aurons
+\begin{alignat*}{3}
+\frac{\dd x}{\dd u} &= l' \sqrt{E},\qquad &
+\frac{\dd y}{\dd u} &= m' \sqrt{E},\qquad &
+\frac{\dd z}{\dd u} &= n' \sqrt{E}, \\
+\frac{\dd x}{\dd v} &= l'' \sqrt{G},\qquad &
+\frac{\dd y}{\dd v} &= m'' \sqrt{G},\qquad &
+\frac{\dd z}{\dd v} &= n'' \sqrt{G};
+\end{alignat*}
+et les conditions~\Eq{(1)} se réduiront à
+\[
+\sum l'{}^{2} = 1,\qquad
+\sum l''{}^{2} = 1,\qquad
+\sum l'l'' = \frac{F}{\sqrt{EG}} = \cos \omega,
+\]
+%% -----File: 074.png---Folio 66-------
+$\omega$ étant l'angle $\DPchg{\widehat{UMV}}{(MU, MV)}$.
+
+Les conditions initiales signifient donc que l'on se donne
+arbitrairement la position du point~$M$, et la direction des
+tangentes $M U, M V$, sous la réserve que ces directions fassent
+entre elles le même angle qu'elles font au point correspondant
+de~$(S)$. Il y a donc bien une des positions de~$(S)$ qui satisfait
+à ces conditions, et notre résultat se trouve définitivement
+établi.
+
+\Paragraph{Remarque.} Le raisonnement précédent serait en défaut,
+si les courbes coordonnées étaient les lignes minima (à cause
+de $E = G = 0$). Mais il suffit de remarquer que si $\Phi$~et~$\Psi$\DPtypo{,}{}
+sont connues, pour un système de coordonnées $u, v$, on en déduit
+leurs expressions pour un autre système de coordonnées $u, v$, en
+y effectuant directement le changement de variables correspondant.
+Notre théorème est donc vrai pour tout système de coordonnées
+superficielles, dès qu'il est vrai pour un seul.
+
+
+\Section{Les \DPtypo{Conditions}{conditions} d'\DPtypo{Intégrabilité}{intégrabilité}.}
+{2.}{} Les coefficients des formules~\Eq{(10)} satisfont à certaines
+conditions, dites \emph{conditions d'intégrabilité} qu'on
+obtient en écrivant que les dérivées du troisième ordre $\dfrac{\dd^{3} x}{\dd u^{2}\, \dd v}$,
+$\dfrac{\dd^{3} x}{\dd u\, \dd v^{2}}$ ont la même valeur, qu'on les obtienne en différentiant
+l'une ou l'autre des formules~\Eq{(10)}.
+
+Pour pouvoir calculer ces conditions, il est commode d'avoir
+des formules qui donnent les dérivées des cosinus directeurs
+$l, m, n$ de la normale. Ils sont définis par les équations
+\[
+\sum l\, \frac{\dd x}{\dd u} = 0,\qquad
+\sum l\, \frac{\dd x}{\dd v} = 0,\qquad
+\sum l^{2} = 1,
+\]
+qui donnent, par différentiation:
+%% -----File: 075.png---Folio 67-------
+\[
+\Tag{(12)}
+\begin{aligned}
+&\sum \frac{\dd l}{\dd u}\, \frac{\dd x}{\dd u}
+  = - \sum l\, \frac{\dd^{2} x}{\dd u^{2}} = -L, &&
+\sum \frac{\dd l}{\dd v}\, \frac{\dd x}{\dd u}
+  = - \sum l\, \frac{\dd^{2} x}{\dd u\, \dd v} \rlap{${} = -M$,} \\
+&\sum \frac{\dd l}{\dd u}\, \frac{\dd x}{\dd v}
+  = - \sum l\, \frac{\dd^{2} x}{\dd u\, \dd v} = -M, &&
+\sum \frac{\dd l}{\dd v}\, \frac{\dd x}{\dd v}
+  = - \sum l\, \frac{\dd^{2} x}{\dd v^{2}} = -N\Add{,} \\
+&\sum \frac{\dd l}{\dd u}\, l = 0, &&
+\sum \frac{\dd l}{\dd v}\, \DPtypo{v}{l} = 0.
+\end{aligned}
+\]
+Si donc on pose, en suivant la même méthode qu'au paragraphe
+précédent,
+\begin{align*}
+\frac{\dd l}{\dd u} &= P'\, \frac{\dd x}{\dd u} + P''\, \frac{\dd x}{\dd v} + Pl, \\
+\frac{\dd \DPtypo{n}{m}}{\dd u} &= P'\, \frac{\dd y}{\dd u} + P''\, \frac{\dd y}{\dd v} + Pm, \\
+\frac{\dd n}{\dd \DPtypo{v}{u}} &= P'\, \frac{\dd z}{\dd u} + P''\, \frac{\dd \DPtypo{x}{z}}{\dd v} + Pn,
+\end{align*}
+on trouvera:
+\[
+\sum \frac{\dd x}{\dd u}\, \frac{\dd l}{\dd u} = EP' + FP'',\quad
+\sum \frac{\dd x}{\dd v}\, \frac{\dd l}{\dd u} = FP' + GP'',\quad
+\sum l\, \frac{\dd l}{\dd u} = P;
+\]
+c'est-à-dire qu'on peut écrire, en tenant compte des formules~\Eq{(12)},
+\[
+\Tag{(13)}
+\left\{
+\begin{aligned}
+\frac{\dd l}{\dd u} &= P'\, \frac{\dd x}{\dd u} + P''\, \frac{\dd x}{\dd v},
+\quad\text{et de même:} \\
+\frac{\dd l}{\dd v} &= Q'\, \frac{\dd x}{\dd u} + Q''\, \frac{\dd x}{\dd v},
+\end{aligned}
+\right.
+\]
+les coefficients $P', P'', Q', Q''$ étant définis par:
+\[
+\Tag{(14)}
+\left\{
+\begin{alignedat}{2}
+EP' + FP'' &= -L,\qquad & FP' + GP'' &= -M, \\
+EQ' + FQ'' &= -M,\qquad & FQ' + GQ'' &= -N,
+\end{alignedat}
+\right.
+\]
+et qu'on aura les mêmes formules pour $m, n$ en changeant~$x$ en~$y$,
+et en~$z$, respectivement.
+
+Nous achèverons le calcul, en supposant la surface rapportée
+à ses lignes minima. Les calculs précédents se simplifient
+alors beaucoup. Si nous appliquons directement les formules
+trouvées, nous obtenons:
+\begin{gather*}%[** TN: Rearranged]
+E = 0,\qquad G = 0, \\
+L'' = 0,\quad L' = \frac{\dd \log F}{\dd u},\quad
+M'' = 0,\quad M' = 0, \quad
+N'' = \frac{\dd \log F}{\dd v}, \quad N' = 0; \\
+P'' = - \frac{L}{F},\quad P' = - \frac{M}{F},\qquad
+Q'' = - \frac{M}{F},\quad Q' = - \frac{N}{F};
+\end{gather*}
+%% -----File: 076.png---Folio 68-------
+c'est-à-dire
+\begin{align*}
+\Tag{(15)}
+&\left\{
+\begin{aligned}
+\frac{\dd^{2} x}{\dd u^{2}}
+  &= \frac{\dd \log F}{\dd u} · \frac{\dd x}{\dd u} + L · l, \\
+\frac{\dd^{2} x}{\dd u\, \dd v}
+  &= M · l, \\
+\frac{\dd^{2} x}{\dd v\DPtypo{}{^{2}}}
+  &= \frac{\dd \log F}{\dd v} · \frac{\dd x}{\dd v} + N · l,
+\end{aligned}
+\right. \\
+\Tag{(16)}
+&\left\{
+\begin{aligned}
+\frac{\dd l}{\dd u}
+  &= -\frac{1}{F} \left(M\, \frac{\dd x}{\dd u}
+                      + L\, \frac{\dd x}{\dd v}\right), \\
+\frac{\dd l}{\dd v}
+  &= -\frac{1}{F} \left(N\, \frac{\dd x}{\dd u}
+                      + M\, \frac{\dd x}{\dd v}\right).
+\end{aligned}
+\right.
+\end{align*}
+Alors, en différentiant la première équation~\Eq{(15)}, il vient:
+\begin{align*}
+\frac{\dd^{3} x}{\dd u^{2}\, dv}
+  &= \left(\frac{\dd^{2} \log F}{\dd u\, \dd v} - \frac{NL}{F} \right)
+       \frac{\dd x}{\dd u}
+   - \frac{LM}{F}\, \frac{\dd x}{\dd v}
+   + \left(\frac{\dd \log F}{\dd u}\, M + \frac{\dd L}{\dd v} \right)l, \\
+\intertext{en différentiant la deuxième équation~\Eq{(15)}, il vient}
+\frac{\dd^{3} x}{\dd u^{2}\, \dd v}
+  &= \frac{-M^{2}}{F} · \frac{\dd x}{\dd u} - \frac{LM}{F}\, \frac{\dd x}{\dd v} + \frac{\dd M}{\dd u}\, l;
+\end{align*}
+et en égalant, on obtient:
+\[
+\Tag{(17)}
+\left( \frac{\dd^{2} \log F}{\dd u\, \dd v} - \frac{LN - M^{2}}{F} \right) \frac{\dd x}{\dd u}
+  + \left( \frac{\dd \log F}{\dd u}\, M + \frac{\dd L}{\dd v} - \frac{\dd M}{\dd u} \right) l = 0.
+\]
+C'est là une condition de la forme
+\begin{alignat*}{3}
+S'\, \frac{\dd x}{\dd u} &+ S''\, \frac{\dd x}{\dd v} &&+ Sl &&= 0, \\
+\intertext{et en reprenant le même calcul, pour $y$~et~$z$, on obtiendrait
+les conditions analogues}
+S'\, \frac{\dd y}{\dd u} &+ S''\, \frac{\dd y}{\dd v} &&+ Sm &&= 0, \\
+S'\, \frac{\dd z}{\dd u} &+ S''\, \frac{\dd z}{\dd v} &&+ Sn &&= 0.
+\end{alignat*}
+On en conclut qu'on a nécessairement $\DPtypo{S'}{S} = S' = S'' = 0$, c'est-à-dire
+ici
+\[
+\Tag{(18)}
+\frac{\dd^{2} \log F}{\dd u\, \dd v} - \frac{LN - M^{2}}{F} = 0,\qquad
+M\, \frac{\dd \log F}{\dd u} + \frac{\dd L}{\dd v} - \frac{\dd M}{\dd u} = 0;
+\]
+et cela est suffisant pour que~\Eq{(17)} ait lieu.
+
+En égalant de même les deux valeurs de $\dfrac{\dd^{3} x}{\dd u\, \dd v^{2}}$, on obtiendra
+%% -----File: 077.png---Folio 69-------
+les conditions qui se déduisent de~\Eq{(18)} en échangeant les
+rôles des variables $u, v$; cela ne modifie que la seconde de ces
+conditions.
+
+Les conditions d'intégrabilité cherchées sont donc:
+\[
+\Tag{(19)}
+\left\{
+\begin{aligned}
+M\, \frac{\dd \log F}{\dd u} &= \frac{\dd M}{\dd u} - \frac{\dd L}{\dd v}\Add{,} \\
+\frac{\dd^{2} \log F}{\dd u\, \dd v} &= \frac{LN - M^{2}}{F}\Add{,} \\
+M\, \frac{\dd \log F}{\dd v} &= \frac{\dd M}{\dd v} - \frac{\dd N}{\dd u}\Add{,}
+\end{aligned}
+\right.
+\]
+et ce sont là, d'après la théorie des équations différentielles,
+les seules conditions d'intégrabilité du système considéré.
+
+
+\Section{Courbure totale.}
+{3.}{} La \Ord{2}{e} des formules précédentes, due à Gauss
+\[
+\frac{\dd^{2} \log F}{\dd u\, \dd v} = \frac{LN - M^2}{F}
+\]
+conduit à une conséquence importante. Reprenons en effet l'équation
+aux rayons de courbure principaux qui est ici
+\[
+H^{2}(LN - M^{2}) + 2SFHM - S^{2}F^{2} =0,
+\]
+où
+\[
+S = \frac{H}{R}.
+\]
+
+On peut l'écrire
+\[
+LN - M^{2} + 2FM · \frac{1}{R} - \frac{F^{2}}{R^{2}} = 0,
+\]
+d'où
+\[
+\frac{1}{R_{1}R_{2}} = - \frac{LN - M}{F},
+\]
+c'est-à-dire
+\[
+\frac{1}{R_{1}R_{2}} = - \frac{1}{F}\, \frac{\dd^{2} \log F}{\dd u\, \dd v}\Add{;}
+\]
+\emph{le produit des rayons de courbure principaux ne dépend que de
+l'élément linéaire; il se conserve donc dans la déformation
+des surfaces}. On donne à $\dfrac{1}{R_{1}R_{2}}$ le nom de \emph{\DPtypo{Courbure}{courbure} totale}.
+
+\Paragraph{Représentation sphérique.} De même que l'on a fait correspondre
+%% -----File: 078.png---Folio 70-------
+à une courbe son indicatrice sphérique, on peut imaginer
+une correspondance entre une surface quelconque et la
+sphère de rayon~$1$, l'homologue d'un point $(u\Add{,} v)$ de la surface
+étant le point $(l, m, n)$. A une aire de la surface correspond une
+aire sphérique. La considération de la limite du rapport de
+ces aires lorsqu'elles deviennent infiniment petites dans toutes
+leurs dimensions va nous conduire à une définition directe
+de la courbure totale.
+
+L'aire sur la surface a pour expression
+\[
+\Area = \iint \sqrt{A^{2} + B^{2} + C^{2}}\, du\, dv = \iint H\, du\, dv,
+\]
+Pour avoir l'aire homologue sur la sphère, il faut d'abord
+calculer l'élément linéaire $dl^{2} + dm^{2} + dn^{2}$. D'après les formules~\Eq{(16)}
+du \Numero~précédent, nous avons
+\begin{align*}
+dl &= \frac{\dd l}{\dd u}\, du + \frac{\dd l}{\dd v}\, dv
+  = - \frac{du}{F} \left(L\, \frac{\dd x}{\dd v} + M\, \frac{\dd x}{\dd u} \right)
+    - \frac{dv}{F} \left(N\, \frac{\dd x}{\dd u} + M\, \frac{\dd x}{\dd v} \right) \\
+  &= -\frac{1}{F}
+    \left[L\, \frac{\dd x}{\DPtypo{d}{\dd v}}\, du
+        + M\, dx + N\, \frac{\dd x}{\dd u}\, dv \right];
+\end{align*}
+d'où
+\begin{gather*}
+\sum dl^{2} = \frac{1}{F^{2}} \left[M^{2}·2F\, du\, dv + 2LMF·du^{2} + 2MNF·dv^{2} + 2LNF·du\, dv \right], \\
+\sum dl^{2} = \frac{2LM}{F}\, du^{2} + 2\, \frac{LN + M^2}{F}\, du\, dv + \frac{2MN}{F}\, dv^{2}.
+\end{gather*}
+Pour la sphère la fonction~$H$ sera donc
+\[
+\sqrt{4\, \frac{LM^{2}N}{F^{2}} - \frac{(LN + M^{2})^{2}}{F^{2}}}
+  = \frac{LN - M^{2}}{iF} = \frac{LN - M^{2}}{H},
+\]
+et l'aire sphérique a pour expression
+\[
+\Area' = \iint \frac{LN - M^{2}}{H}\, du\, dv;
+\]
+ce qui peut s'écrire, en remarquant que
+\begin{gather*}%[** TN: Set first two equations on a single line]
+d\Area = H·du\, dv,\qquad
+\Area' = \iint \frac{LN - M^{2}}{H^{2}} · d\Area
+  = \iint \frac{1}{R_{1} R_{2}}\, d\Area\Add{,} \\
+\intertext{donc}
+d\Area' = \frac{1}{R_{1} R_{2}}\, d\Area;
+\end{gather*}
+%% -----File: 079.png---Folio 71-------
+\emph{le rapport des aires homologues sur la surface et sur la sphère
+a donc pour limite la courbure totale, lorsque ces aires
+deviennent infiniment petites dans toutes leurs dimensions}.
+
+
+\Section{Coordonnées orthogonales et isothermes.}
+{4.}{} Pour éviter l'emploi des imaginaires dans les considérations
+qui \DPtypo{précèdent}{précédent}, nous introduirons un nouveau système
+de coordonnées curvilignes. La surface étant supposée réelle,
+nous choisirons les coordonnées minima de façon que $u, v$ soient
+imaginaires conjugués. Nous poserons donc
+\[
+u = u' + iv',\qquad
+v = u' - iv',
+\]
+$u'\Add{,} v'$ étant des quantités réelles. Nous en tirons
+\[
+du = du' + i\, dv',\qquad
+dv = du' - i\, dv',
+\]
+d'où
+\[
+du\, dv = du'{}^{2} + dv'{}^{2}.
+\]
+L'élément linéaire prend la forme
+\[
+ds^{2} = 2F·du\, dv = 2F (du'{}^{2} + dv'{}^{2});
+\]
+les coordonnées $u'\Add{,} v'$ sont orthogonales; on leur donne le nom
+de \emph{coordonnées orthogonales et isothermes}. On peut dire que
+\emph{ces coordonnées divisent la surface en un réseau de carrés infiniment
+petits}. Considérons en effet les courbes coordonnées
+$u', u' + h, u' + 2h\Add{,} \dots$ et $v', v' + h, v'+ 2h\Add{,} \dots$; si on prend l'un
+des quadrilatères curvilignes obtenus, ses \Card{4} angles sont droits,
+ses côtés sont $\sqrt{2F}\Add{·}du'$~et~$\sqrt{2F}·dv'$, c'est-à-dire~$\sqrt{2F}·h$, aux
+infiniment petits d'ordre supérieur près; ces arcs sont égaux.
+
+Avec ce système de coordonnées particulières, en désignant
+%[** TN: Reworded to follow the typeset edition]
+par $\overline{E}, \overline{F}, \overline{G}, \overline{H}$ les valeurs des fonctions \DPtypo{$\overline{E}, \overline{F}, \overline{G}, \overline{H}$}{analogues à $E, F, G, H$}, nous avons
+\[%[** TN: Not displayed in manuscript, but displayed in typeset edition]
+\overline{E} = 2F,\quad
+\overline{G} = 2F,\quad
+\overline{F} = 0,\quad
+\overline{H}^{2} = \overline{E}\overline{G} - \overline{F}^{2} = 4 F^{2},\quad
+\overline{H} = 2F,
+\]
+donc
+\[
+ds^{2} = \overline{H} (du'{}^{2} + dv'{}^{2}).
+\]
+%% -----File: 080.png---Folio 72-------
+Mais nous avons
+\[
+\frac{\dd \Phi}{\dd u'} = \frac{\dd \Phi}{\dd u} + \frac{\dd \Phi}{\dd v},\qquad
+\frac{\dd \Phi}{\dd v'} = i \left(\frac{\dd \Phi}{\dd u} - \frac{\dd \Phi}{\dd v} \right);
+\]
+d'où
+\[
+\frac{\dd^{2} \Phi}{\dd u'{}^{2}}
+  = \frac{\dd^{2} \Phi}{\dd u^{2}}
+  + 2\, \frac{\dd^{2} \Phi}{\dd u\, \dd v}
+  + \frac{\dd^{2} \Phi}{\dd v^{2}},\qquad
+\frac{\dd^{2} \Phi}{\dd v'{}^{2}}
+  = - \left[ \frac{\dd^{2} \Phi}{\dd u^{2}}
+    - 2\, \frac{\dd^{2} \Phi}{\dd u\, \dd v}
+    + \frac{\dd^{2} \Phi}{\dd v^{2}} \right],
+\]
+et
+\[
+\frac{\dd^{2} \Phi}{\dd u'{}^{2}} + \frac{\dd^{2} \Phi}{\dd v'{}^{2}}
+  = 4\, \frac{\dd^{2} \Phi}{\dd u\, \dd v}\Add{,}
+\]
+\DPtypo{D'où}{d'où} par conséquent
+\[
+4\, \frac{\dd^{2} \log F}{\dd u\, \dd v}
+  = \DPtypo{}{4\,}\frac{\dd^{2} \log \overline{H}}{\dd u\, \dd v}
+  = \frac{\dd^{2} \log \overline{H}}{\dd u'{}^{2}}
+  + \frac{\dd^{2} \log \overline{H}}{\dd v'{}^{2}}.
+\]
+En supprimant les accents, nous avons donc les formules suivantes,
+en coordonnées orthogonales et isothermes:
+\begin{gather*}
+ds^{2} = H (du^{2} + dv^{2}), \\
+\frac{1}{R_{1} R_{2}}
+  = -\frac{1}{2H} \left( \frac{\dd^{2} \log H}{\dd u^{2}} + \frac{\dd^{2} \log H}{\dd v^{2}} \right).
+\end{gather*}
+Nous poserons encore
+\[
+\sum l\, d^{2} x = L\, du^{2} + 2M\, du\, dv + N\, dv^{2}.
+\]
+L'équation aux rayons de courbure principaux sera
+\[
+(LN -M^{2}) - \frac{H}{R} (L + N) + \frac{H^{2}}{R^{2}} = 0,
+\]
+et on aura
+\[
+\frac{1}{R_{1}R_{2}} = \frac{LN - M^{2}}{H^{2}}.
+\]
+
+Calculons la représentation sphérique. Posons
+\begin{alignat*}{3}
+l' &= \frac{1}{\sqrt{H}}\Add{·} \frac{\dd x}{\dd u},\qquad &
+m' &= \frac{1}{\sqrt{H}} · \frac{\dd y}{\dd u},\qquad &
+n' &= \frac{1}{\sqrt{H}} · \frac{\dd z}{\dd u}, \\
+l'' &= \frac{1}{\sqrt{H}}\Add{·} \frac{\dd x}{\dd v},\qquad &
+m'' &= \frac{1}{\sqrt{H}} · \frac{\dd y}{\dd v},\qquad &
+n'' &= \frac{1}{\sqrt{H}} · \frac{\dd z}{\dd v}.
+\end{alignat*}
+De la relation
+\[
+\sum  l^{2} = 1,
+\]
+nous tirons
+\[
+\sum l · \frac{\dd l}{\dd u} = 0.
+\]
+Maintenant
+\[
+L = \sum l\, \frac{\dd^{2} x}{\dd u^{2}}
+  = - \sum \frac{\dd l}{\dd u} · \frac{\dd x}{\dd u}
+  = - \sqrt{H} · \sum l'\, \frac{\dd l}{\dd u};
+\]
+d'où
+\[
+\sum l'\, \frac{\dd l}{\dd u} = - \frac{L}{\sqrt{H}};
+\]
+de même
+\[%[** TN: Set on two lines in original]
+M = \sum l\, \frac{\dd^{2} x}{\dd u\, \dd v}
+  = - \sum \frac{\dd l}{\dd u} · \frac{\dd x}{\dd v}
+  = - \sqrt{H} \sum l''\, \frac{\dd l}{\dd u}, \qquad
+\sum l''\, \frac{\dd l}{\dd u} = - \frac{M}{\sqrt{H}}.
+\]
+D'où \Card{3} équations en $\dfrac{\dd l}{\dd u}, \dfrac{\dd m}{\dd u}, \dfrac{\dd n}{\dd u}$. Multiplions respectivement par
+%% -----File: 081.png---Folio 73-------
+$l'\Add{,} l''\Add{,} l'''$ et ajoutons, il vient
+\begin{align*}
+\frac{\dd l}{\dd u}
+  &= - \frac{L}{H} · \frac{\dd x}{\dd u}
+     - \frac{M}{H} · \frac{\dd x}{\dd v}; \\
+\intertext{de même:}
+\frac{\dd m}{\dd u}
+  &= - \frac{L}{H} · \frac{\dd y}{\dd u}
+     - \frac{M}{H}\Add{·} \frac{\dd y}{\dd v}, \\
+\frac{\dd n}{\dd u}
+  &= - \frac{L}{H}\Add{·} \frac{\dd z}{\dd u}
+     - \frac{M}{H} · \frac{\dd z}{\dd v}.
+\end{align*}
+On obtiendra de même
+\begin{align*}
+\frac{\dd l}{\dd v}
+  &= - \frac{1}{H} \left(M\, \frac{\dd x}{\dd u}
+                       + N\, \frac{\dd x}{\dd v}\right), \\
+\frac{\dd m}{\dd v}
+  &= - \frac{1}{H} \left(M\, \frac{\dd y}{\dd u}
+                       + N\, \frac{\dd y}{\dd v}\right), \\
+\frac{\dd n}{\dd v}
+  &= - \frac{1}{H} \left(M\, \frac{\dd z}{\dd u}
+                       + N\, \frac{\dd z}{\dd v}\right).
+\end{align*}
+Alors, sur la sphère, les \Card{3} fonctions $E\Add{,} F\Add{,} G$ seront
+\begin{alignat*}{2}
+\scrE &= \sum \left(\frac{\dd l}{\dd u}\right)^{2}\!\!
+  &&= \frac{1}{H^{2}} \sum \left(L\, \frac{\dd x}{\dd u}
+                               + M\, \frac{\dd x}{\dd v}\right)^{2}
+  = \frac{L^{2} + M^{2}}{H}, \\
+%
+\scrF &= \sum \frac{\dd l}{\dd u} · \frac{\dd l}{\dd v}
+  &&= \frac{1}{H^{2}}
+      \sum \left(L\, \frac{\dd x}{\dd u} + M\, \frac{\dd x}{\dd v}\right)
+     \!·\! \left(M\, \frac{\dd x}{\dd u} + N\, \frac{\dd x}{\dd v}\right)
+  = \frac{M (L + N)}{H}, \\
+%
+\scrG &= \sum \left(\frac{\dd l}{\dd v}\right)^{2}\!\!
+  &&= \frac{1}{H^{2}} \sum \left(M\, \frac{\dd x}{\dd u}
+                               + N\, \frac{\dd x}{\dd v}\right)^{2}
+  = \frac{M^{2} + N^{2}}{H};
+\end{alignat*}
+et
+\[
+\scrH^{2} = \scrE\scrG - \scrF^{2}
+  = \frac{(L^{2} + M^{2}) (M^{2} + N^{2})- M^{2} (L + N)^{2}}{H^{2}}
+  = \left( \frac{LN - M^{2}}{H}\right)^{2},
+\]
+et alors l'aire sur la sphère a pour expression
+\[
+\Area' = \iint \frac{LN - M^{2}}{H}\, du\, dv.
+\]
+On retrouve la même expression que précédemment, et on arriverait
+de même à la définition directe de la courbure totale.
+
+\Paragraph{Remarque.} Dans l'expression précédente, $\Area$ a un signe,
+qui est celui de $LN - M^{2}$. Considérons le déterminant des cosinus
+$l, m, n$; $l', m', n'$; $l'', m'', n''$: il est égal, à un facteur positif près
+à
+\[
+\begin{vmatrix}
+l & m & n \\
+\mfrac{\dd l}{\dd u} & \mfrac{\dd m}{\dd u} & \mfrac{\dd n}{\dd u} \\
+\mfrac{\dd l}{\dd v} & \mfrac{\dd m}{\dd v} & \mfrac{\dd n}{\dd v}
+\end{vmatrix}
+= \frac{LN  - M^{2}}{H^{2}}
+\begin{vmatrix}
+l & m & n \\
+\mfrac{\dd x}{\dd u} & \mfrac{\dd y}{\dd u} & \mfrac{\dd z}{\dd u} \\
+\mfrac{\dd x}{\dd v} & \mfrac{\dd y}{\dd v} & \mfrac{\dd z}{\dd v}
+\end{vmatrix}\Add{.}
+\]
+Il résulte de cette formule que, si $\Area\Area' > 0$, le point mobile $x\Add{,}y\Add{,}z$
+%% -----File: 082.png---Folio 74-------
+décrivant le contour qui limite l'aire sur la surface dans le
+sens direct le point $l, m, n$ décrira le contour qui limite l'aire
+homologue sur la sphère aussi dans le sens direct. Si $\Err{\Area'}{\Area\Area'} < 0$,
+les conclusions sont inverses.
+
+
+\Section{Relations entre la courbure totale et la courbure géodésique.}
+{5.}{} La courbure totale est un élément qui reste invariant
+dans la déformation des surfaces. Cherchons s'il y a des relations
+entre elle et les autres éléments invariants dans la
+déformation. Considérons la courbure géodésique. En coordonnées
+orthogonales et isothermes, son expression est
+\[
+\llap{$\dfrac{1}{R_{g}} = $}\dfrac{1}{H\, ds^{3}}
+  \left[ H^{2} (du\, d^{2} v - dv\, d^{2} u)
+  -
+  \begin{vmatrix}
+  \phantom{-}  \mfrac{1}{2}\, \mfrac{\dd H}{\dd u}\, du^{2}
+                 + \mfrac{\dd H}{\dd v}\, du\, dv
+  - \mfrac{1}{2}\, \mfrac{\dd H}{\dd u}\, dv^{2} & H\, du \\
+  - \mfrac{1}{2}\, \mfrac{\dd H}{\dd v}\, du^{2}
+                 + \mfrac{\dd H}{\dd u}\, du\, dv
+  + \mfrac{1}{2}\, \mfrac{\dd H}{\dd v}\, dv^{2} & H\, dv
+\end{vmatrix}
+\right],
+\]
+ou
+\[
+\frac{1}{R_{g}} = \frac{1}{ds^{3}} \left[
+  H (du\, d^{2} v - dv\, d^{2} u)
+  + \frac{1}{2} \left(\frac{\dd H}{\dd u}\, dv - \frac{\dd H}{\dd v}\, du\right) (du^{2} + dv^{2})\right];
+\]
+mais on a
+\[
+ds^{2} = (du^{2} + dv^{2})\, H,
+\]
+et la formule précédente s'écrit
+\begin{align*}
+\frac{ds}{R_{g}}
+  &= \frac{du\, d^{2} v - dv\, d^{2} u}{du^{2} + dv^{2}}
+   + \frac{1}{2}\, \frac{\dd \log H}{\dd u}\, dv - \frac{1}{2}\, \frac{\dd \log H}{\dd v}\, du; \\
+\intertext{ou encore}
+\frac{ds}{R_{g}}
+  &= d \left(\arctg \frac{dv}{du}\right)
+   + \frac{1}{2}\, \frac{\dd \log H}{\dd u}\, dv - \frac{1}{2}\, \frac{\dd \log H}{\dd v}\, du.
+\end{align*}
+Imaginons alors dans le plan tangent les tangentes $MU, MV$ aux
+courbes coordonnées dans le sens des $u, v$ croissants; considérons
+la tangente à une courbe quelconque $MT$ de la surface, et
+soit $(MU, MT) = \phi$. Nous avons
+%[** TN: Set each of next two aligned pairs on a single line]
+\[
+\Cos \phi = \sqrt{H} · \frac{du}{ds}, \qquad
+\sin \phi = \sqrt{H}\Add{·} \frac{dv}{ds};
+\]
+%% -----File: 083.png---Folio 75-------
+d'où
+\[
+\tg \phi = \frac{dv}{du}, \qquad
+\phi = \arctg\DPtypo{.} \frac{dv}{du};
+\]
+et la formule précédente devient
+\[
+\frac{ds}{R_{g}}
+  = d \phi
+  + \frac{1}{2}\, \frac{\dd \log H}{\dd u}\,  dv
+  - \frac{1}{2}\, \frac{\dd \log H}{\dd v}\,  du.
+\]
+Prenons alors sur~$S$ un contour fermé et intégrons le long de
+ce contour dans le sens direct
+\[
+\int \frac{ds}{R_{g}}
+  = \int d \phi
+  + \frac{1}{2} \int \frac{\dd \log H}{\dd u}\,  dv
+  - \frac{1}{2} \int \frac{\dd \log H}{\dd v}\,  du.
+\]
+\Figure[2.75in]{083a}
+Rappelons le \emph{Théorème de Green}. Dans le plan des $u, v$, le
+point $(u\Add{,}v)$ décrit un contour fermé, aussi
+dans le sens direct. Menons les tangentes
+parallèles à l'axe des~$u$; soient $A\Add{,} B$~les
+points de contact. Nous avons ainsi \Card{2} arcs
+$AMB$~et~$ANB$, et si nous désignons par~$\DPtypo{C}{c}$ le
+contour, nous avons
+\[
+\int_c \frac{\dd f}{\dd u}\, dv
+  = \int_{AMB} \frac{\dd f}{\dd u}\, dv
+  + \int_{BNA} \frac{\dd f}{\dd u}\, dv\Add{.}
+\]
+Menons une parallèle à $OU$ qui coupe le contour en deux points
+$M (u_{2})$~et~$N (u_{1})$\Add{.}
+
+Soient $a, b$ les valeurs de~$U$ qui correspondent aux deux points
+$A, B$. Nous avons
+\[
+\int_c \frac{\dd f}{\dd u}\, dv
+  = \int_a^b \left(\frac{\dd f}{\dd u}\right)_{u_{1}u_{2}} \kern -12pt \Err{dv\, dv}{d\gamma}
+  - \int_a^b \left(\frac{\dd f}{\dd u}\right)_{u_{1}u_{2}} \kern -12pt \Err{dv\, du}{d\gamma}
+  = \int_a^b \left[
+    \left(\frac{\dd f}{\dd u}\right)_{2}
+  - \left(\frac{\dd f}{\dd u}\right)_{1}\right] dv.
+\]
+Mais
+\[
+\left(\frac{\dd f}{\dd u}\right)_{2} - \left(\frac{\dd f}{\dd u}\right)_{1}
+  = \int_{u_{1}}^{u_{2}} \frac{\dd^{2} f}{\dd u^{2}} · du,
+\]
+et alors
+\[
+\int_c \frac{\dd f}{\dd u} · dv
+  = \int_a^b dv \int_{u_{1}}^{u_{2}} \frac{\dd^{2} f}{\dd u^{2}} · du
+  = \iint \frac{\dd^{2} f}{\dd u^{2}}\, du\, dv,
+\]
+%% -----File: 084.png---Folio 76-------
+l'intégrale double étant étendue à toute l'aire limitée par le
+contour. Cette formule subsiste pour un contour simple quelconque.
+De même
+\[
+\int_c \frac{\dd f}{\dd v}\, du
+  = - \iint \frac{\dd^{2} f}{\dd v^{2}} · du\, dv\Add{.}
+\]
+Alors nous aurons
+\[
+\int \frac{ds}{R_{g}}
+  = \int d \phi
+  + \frac{1}{2} \iint \left[\frac{\dd^{2} \log H}{\dd u^{2}}
+                           + \frac{\dd^{2} \log H}{\dd v^{2}}\right] du\, dv,
+\]
+ou
+\[
+\int \frac{ds}{R_{g}}
+  = \int d \phi - \iint \frac{H}{R_{1} R_{2}} · du\, dv
+  = \int d \phi - \iint \frac{d\Area}{R_{1} R_{2}},
+\]
+d'où la \emph{formule d'Ossian Bonnet}
+\[
+\Area' = \iint \frac{d\Area}{R_{1} R_{2}}
+  = \int d \phi - \int \frac{ds}{R_{g}}.
+\]
+
+\Paragraph{Remarque.} L'angle~$\phi$ est l'angle de~$MU$ avec la tangente
+à la courbe. Supposons qu'en chaque point de la surface on
+détermine une direction~$MO$, dont les cosinus directeurs sont
+des fonctions bien déterminées de~$u\Add{,}v$. Soit $\psi = (MO, MU)$ et
+$\psi' = (MO, MT)$.
+
+Nous avons
+\[
+\phi' = \psi + \phi,
+\]
+d'où
+\[
+d \phi' = d \psi + d \phi.
+\]
+Intégrons le long d'un contour fermé quelconque
+\[
+\int d \phi' = \int d \psi + \int d \phi;
+\]
+or\Add{,} $\psi$~est une fonction de~$u\Add{,}v$, et le long d'un contour fermé,
+on a
+\[
+\int d \psi (u, v) = 0;
+\]
+donc
+\[
+\int d \phi' = \int d \phi,
+\]
+et l'on peut substituer à l'angle~$\phi$ l'angle~$\phi'$ précédemment
+défini.
+%% -----File: 085.png---Folio 77-------
+
+
+\Section{Triangles géodésiques.}{}{}
+Nous appellerons \emph{triangle géodésique} la figure formée par
+\Card{3} lignes géodésiques. Le long de chacun des côtés on a
+\[
+\int \frac{ds}{R_{g}} = \int \frac{\sin \theta}{R}\, ds = 0,
+\]
+et la formule d'O.~Bonnet nous donne
+\[
+\Area' = \int d \phi,
+\]
+c'est-à-dire
+\[
+\Area' = \int_{AB} d \phi + \int_{BC} d \phi + \int_{CA} d \phi.
+\]
+
+Les coordonnées
+orthogonales et isothermes
+constituent
+une représentation
+conforme de la surface
+\Figure[5in]{085a}
+sur le plan des~$u\Add{,}v$. Considérons donc sur ce plan la représentation~$a\Add{,} b\Add{,} c$ du triangle~$ABC$. Menons aux extrémités $a, b, c$ les
+tangentes aux côtés dans le sens direct: Soient $T_{1}, T_{2}, T_{3}$, $T'_{1},
+T'_{2}, T'_{3}$ ces tangentes. Si par un point du plan nous menons des
+parallèles à ces tangentes, nous aurons
+\[
+\int_{AB} d \phi = (T'_{1}, T_{2}),\qquad
+\int_{BC} d \phi = (T'_{2}, T_{3}),\qquad
+\int_{CA} d \phi = (T'_{3}, T_{1});
+\]
+or\Add{,} si nous appelons $a, b, c$ les \Card{3} angles du triangle géodésique,
+nous avons
+\begin{multline*}%[** TN: Re-breaking]
+(T'_{1}, T_{2}) + (T'_{2}, T_{3}) + (T'_{3}, T_{1}) \\
+\begin{aligned}
+  &= - \bigl[(T_{1}, T'_{1}) + (T_{2}, T'_{2}) + (T_{3}, T'_{3})\bigr]
+    + \bigl[(T_{1}, T_{2}) + (T_{2}, T_{3}) + (T_{3}, T_{1})\bigr] \\
+  &= 2 \pi - \bigl[(\pi - a) + (\pi - b) + (\pi - c)\DPtypo{}{\bigr]}
+  = a + b + c - \pi,
+\end{aligned}
+\end{multline*}
+d'où la \emph{formule de Gauss}
+\[
+a + b + c - \pi = \Area'.
+\]
+Si en particulier la surface est la sphère de rayon~$1$, on a
+%% -----File: 086.png---Folio 78-------
+la formule qui donne l'aire d'un triangle sphérique.
+
+\Section{Nouvelle expression de la courbure géodésique.}{}{}%
+Considérons un arc de courbe~$AB$; menons en $AB$ les géodésiques
+tangentes à cette courbe, qui se
+coupent en~$C$ sous un angle~$\epsilon$ que nous appellerons
+\emph{angle de contingence géodésique}.
+Le long du contour de ce triangle on a
+\[
+\int d \phi = - \epsilon,
+\]
+et la formule d'O.~Bonnet nous donne
+\[
+- \epsilon - \int_{AB} \frac{ds}{R_{g}} = \iint d\Area'.
+\]
+Supposons que $A$~corresponde au paramètre~$t$, $B$~à~$t + \Delta t$, et que
+$\Delta t$ tende vers~$0$; soit $\Delta s$~l'arc~$AB$. Nous avons
+\[
+-\frac{\epsilon}{\Delta s} - \frac{1}{\Delta s} \int_{AB} \frac{ds }{R_{g}}
+  = \frac{1}{\Delta s} \iint d\Area'.
+\]
+Soit $\left(\dfrac{1}{R_{g}}\right)_{m}$ la valeur moyenne de la courbure géodésique sur l'arc~$AB$;
+nous avons,
+\[
+\frac{1}{\Delta s} \int_{AB} \frac{ds}{R_{g}}
+  = \left(\frac{1}{R_{g}}\right)_{m},
+\]
+et par suite
+\[
+-\frac{\epsilon}{\Delta s} - \left(\frac{1}{R_{g}}\right)_{m}
+  = \frac{1}{\Delta s} \iint dA'\Add{.}
+\]
+\begin{figure}[hbt]
+\centering
+\Input[1.75in]{086a}\hfil\hfil
+\Input[2.25in]{086b}
+\end{figure}
+Si $\Delta s$ tend vers~$0$, $\left(\dfrac{1}{R_{g}}\right)_{m}$ a pour limite la courbure géodésique
+au point~$A$. Je dis que le \Ord{2}{e} membre a pour limite~$0$; il suffit
+de montrer que $\ds\iint d\Area'$ est infiniment petit
+du \Ord{2}{e} ordre au moins. Considérons la représentation
+$a, b, c$ du triangle~$ABC$ sur le
+plan~$U\Add{,}V$.
+
+Nous avons
+%% -----File: 087.png---Folio 79-------
+\[
+\iint d\Area' = \iint \psi(u, v)\, du\, dv
+  = \bigl[\psi (u, v)\bigr]_{m} \iint du\, dv
+\]
+et au signe près $\ds\iint du\, dv$ est égale à l'intégrale curviligne $\ds\int v\, du$.
+Soient $v_{2}\Add{,} v_{1}$ les fonctions~$v$ sur les arcs $bc$~et~$bk$. La partie de
+$\smash{\ds\int v\, du}$ donnée par ces arcs est $\ds\int_{u_{0}}^{u'} (v_{2} - v_{1})\, du$. Or\Add{,} les courbes $ab$~et~$bc$
+étant tangentes en~$b$, $v_{2} - v_{1}$ est infiniment petit du \Ord{2}{e} ordre au
+moins par rapport à $u'- u$ et \textit{à~fortiori} par rapport à $(u'- u_{0})$.
+L'intégrale $\ds\int_{u_{0}}^{u'} (v_2 - v_1)\, du$, qui est égale au produit de $(u' - u_{0})$ par
+la valeur moyenne de $(v_{2} - v_{1})$ sera donc du troisième ordre au moins
+par rapport à $(u'- u_{0})$, et, par suite, par rapport à~$\Delta s$. Le même
+raisonnement s'appliquant aux autres arcs $ac$~et~$ak$, on voit que
+$\ds\iint d\Area'$ est du troisième ordre au moins.
+
+
+\ExSection{IV}
+
+\begin{Exercises}
+\item[18.] \DPchg{Etablir}{Établir} les conditions d'intégrabilité qui lient les invariants
+fondamentaux, en supposant la surface rapportée à ses
+lignes de courbure.
+
+\item[19.] Même question, en supposant la surface \DPtypo{rapportee}{rapportée} à une famille
+de géodésiques et à leurs trajectoires orthogonales. Exprimer,
+en fonction de la quantité~$H$, la courbure totale, et la forme
+différentielle~$\dfrac{ds}{R_{g}} - d\phi$ (voir \hyperref[exercice11]{exercice~11}); et retrouver ainsi
+la formule d'Ossian Bonnet.
+
+\item[20.] En supposant les coordonnées quelconques, trouver celle des
+conditions d'intégrabilité qui donne l'expression de la courbure
+totale.
+\end{Exercises}
+%% -----File: 088.png---Folio 80-------
+
+
+\Chapitre{V}{Surfaces Réglées.}
+
+\Section{Surfaces développables.}
+{1.}{} Pour définir la variation de la droite qui engendre
+la surface réglée, nous nous donnerons la trajectoire d'un
+point~$M$ de cette droite, et la direction de cette droite pour
+\Figure{088a}
+chaque position du point~$M$. Les coordonnées
+d'un point de la surface sont ainsi exprimées
+en fonction de deux paramètres, l'un
+définissant la position du point~$M$ sur sa
+trajectoire~$(K)$, l'autre définissant la position
+du point~$P$ considéré sur la droite~$D$. Soit
+\[
+x = f(v),\qquad y = g(v),\qquad z = h(v),
+\]
+la courbe~$K$. Soient $l(v), m(v), n(v)$ les coefficients de direction
+de la génératrice, et $u$~le rapport du segment~$MP$ au
+segment de direction de la génératrice. Les coordonnées de~$P$
+sont
+\[
+\Tag{(1)}
+x = f(v) + u·l(v),\qquad
+y = g(v) + u·m(v),\qquad
+z = h(v) + u·n(v).
+\]
+
+Cherchons la condition pour que la surface définie par
+les équations précédentes soit développable. Si nous exceptons
+les cas du cylindre et du cône, la condition nécessaire et
+suffisante est que les génératrices soient tangentes à une même
+courbe gauche. On peut donc trouver sur la génératrice~$G$ un
+point~$P$ tel que sa trajectoire soit constamment tangente à~$G$;
+on doit donc avoir, en appelant $x, y, z$ les coordonnées de ce
+point
+\[
+\frac{dx}{l} = \frac{dy}{m} = \frac{dz}{n} = d \rho;
+\]
+%% -----File: 089.png---Folio 81-------
+d'où
+\[
+\Tag{(2)}
+dx = l\, d \rho,\qquad
+dy = m\, d \rho,\qquad
+dz = n\, d \rho,
+\]
+Mais les équations~\Eq{(1)} donnent
+\[
+dx = df + u\, dl + l\, du,\quad
+dy = dg + u\, dm + m\, du,\quad
+dz = dh + u\, dn + n\, du
+\]
+et les équations~\Eq{(2)} s'écrivent
+\begin{alignat*}{5}
+&df &&+ u\, dl &&+ l\,&&(du - d \rho) &&= 0, \\
+&dg &&+ u\, dm &&+ m\,&&(du - d \rho) &&= 0, \\
+&dh &&+ u\, dn &&+ n\,&&(du - d \rho) &&= 0;
+\end{alignat*}
+ou, en posant
+\begin{gather*}%[** TN: Set second group on separate lines, added brace]
+\Tag{(3)}
+d \sigma = du - d \rho, \\
+\Tag{(4)}
+\left\{
+\begin{alignedat}{4}
+&df &&+ u\, dl &&+ l\, &&d \sigma = 0, \\
+&dg &&+ u\, dm &&+ m\, &&d \sigma = 0, \\
+&dh &&+ u\, dn &&+ n\, &&d \sigma = 0,
+\end{alignedat}
+\right.
+\end{gather*}
+$d \sigma$~et~$u$ doivent satisfaire à ces \Card{3} équations linéaires; donc
+on doit avoir
+\[
+\Tag{(5)}
+\begin{vmatrix}
+df & dl & l \\
+dg & dm & m \\
+dh & dn & n
+\end{vmatrix}
+= 0\Add{.}
+\]
+
+Si les \Card{3} déterminants déduits du tableau
+\[
+\begin{Vmatrix}
+dl & dm & dn \\
+l  &  m &  n
+\end{Vmatrix}
+\]
+ne sont pas tous nuls, la condition~\Eq{(5)} est suffisante. Si ces
+\Card{3} déterminants sont nuls, on a
+\[
+\frac{dl}{l} = \frac{dm}{m} = \frac{dn}{n}\Add{,}
+\]
+et l'intégration de ces équations nous montre que $l, m, n$ sont
+proportionnels à des quantités fixes; la surface est alors un
+cylindre. En écartant ce cas, la condition~\Eq{(5)} est nécessaire
+et suffisante.
+
+\Paragraph{Remarque \1.} Pour que le point~$P$ décrive effectivement
+une courbe, il faut que $dx, dy, dz$, et par suite~$d\rho$, ne soient
+%% -----File: 090.png---Folio 82-------
+pas identiquement \DPtypo{nu}{nul}. Si on avait $d \rho = 0$, toutes les génératrices
+passeraient par un point fixe, la surface serait un
+cône. La condition~\Eq{(5)} s'applique donc au cas du cône.
+
+\Paragraph{Remarque \2.} On emploie souvent les équations de la génératrice
+sous la forme
+\[
+x = Mz + P,\qquad
+y = Nz + Q,
+\]
+$M, N, P, Q$, étant fonctions d'un paramètre arbitraire. C'est un
+cas particulier de la représentation générale~\Eq{(1)} dans laquelle
+on fait $h(v) = 0$ et $n(v) = 1$; alors $z = u$, et on peut écrire
+\[
+\Tag{(6)}
+x = f(v) + z·l(v)\Add{,}\qquad
+y = g(v) + z·m(v)\Add{,}
+\]
+les coefficients de direction sont $l, m, 1$. La courbe~$(K)$ est
+alors la section par le plan $z = 0$; dans ce cas la condition
+\Eq{(5)}~prend la forme simple
+\[
+\Tag{(7)}
+\begin{vmatrix}
+df & dl \\
+dg & dm
+\end{vmatrix} = 0,\quad\text{c'est-à-dire}\quad
+\begin{vmatrix}
+dM & dP \\
+dN & dQ
+\end{vmatrix} = 0.
+\]
+
+
+\Section{Propriétés des développables\Add{.}}{}{}%
+Revenons au cas général; supposons que $l, m, n$ soient les
+cosinus directeurs de la génératrice; on a
+\[
+l^{2} + m^{2} + n^{2} = 1,
+\]
+d'où
+\[
+l\, dl + m\, dm + n\, dn = 0.
+\]
+Multiplions alors les équations~\Eq{(4)} respectivement par $dl, dm,
+dn$, et ajoutons, il vient
+\[
+u = - \frac{\sum dl\, df}{\sum dl^{2}}.
+\]
+Supposons en outre que $MG$ soit normale à la courbe~$(K)$. Il
+est toujours possible de trouver sur une surface réglée des
+trajectoires orthogonales des génératrices. Il suffit que
+%% -----File: 091.png---Folio 83-------
+l'on ait
+\[
+\sum l\, dx =0,
+\]
+ou
+\[
+\sum l\, df + u \sum l\, dl + \sum l^{2}\, du =0;
+\]
+ou, comme ici $\sum l^{2} = 1$, $\sum l\,dl = 0$\Add{,}
+\[
+\sum l\, df + du = 0;
+\]
+la détermination des trajectoires orthogonales se fait donc
+au moyen d'une quadrature. Si donc nous supposons $(K)$~normale
+à la génératrice, nous avons
+\[
+\sum l\, df = 0.
+\]
+Multiplions alors les équations~\Eq{(4)} respectivement par $l, m, n$
+et ajoutons, il vient $d\sigma = 0$, d'où $d \rho = du$, et les équations~\Eq{(2)}
+deviennent
+\[
+\Tag{(3')}
+dx = l\, du,\qquad dy = m\, du,\qquad dz = n\, du.
+\]
+\DPtypo{mais}{Mais}, $l, m, n$ étant les cosinus directeurs de la tangente à
+l'arête de rebroussement~$(R)$, $u$~représente l'arc de cette
+courbe compté dans le sens positif choisi sur la génératrice
+à partir d'une origine arbitraire~$I$; et comme $u$~représente
+aussi la longueur~$MP$, on a
+\[
+d·MP = d·(\arc IP);
+\]
+d'où
+\[
+MP = \arc IP + \cte[].
+\]
+On peut toujours choisir l'origine des arcs telle que la constante
+soit nulle. Alors $MP = \arc IP$. La courbe~$(K)$ est une
+développante de la courbe~$(R)$. \emph{Sur une surface développable,
+les trajectoires orthogonales des génératrices sont des développantes
+de l'arête de rebroussement.}
+
+Les formules~\Eq{(4)} donnent alors
+%% -----File: 092.png---Folio 84-------
+\[
+\Tag{(4')}
+df + u\, dl = 0,\qquad dg + u\, dm = 0,\qquad dh + u\, dn = 0.
+\]
+
+\Section{Développées des courbes gauches.}
+{2.}{} Supposons qu'on se donne la courbe~$(K)$, et cherchons
+à mener à cette courbe une normale en chacun de ses points
+de façon à obtenir une surface développable. Nous prendrons
+pour variable~$v$ l'arc~$s$ de la courbe~$(K)$. Considérons le trièdre
+de Serret au point~$M$ de la courbe. Soit $MG$ la normale
+%[** Original diagram uses \alpha, \beta, \gamma; changed to match text]
+\Figure{092a}
+cherchée; elle est dans le plan normal à
+la courbe, pour la définir, il suffira
+donc de se donner l'angle $(MN, MG) = \chi$. Le
+point à l'unité de distance sur~$MG$ a pour
+coordonnées par rapport au trièdre de Serret
+$0, \cos \chi, \sin \chi$; si donc $l, m, n$ sont les cosinus directeurs
+de~$MG$, nous avons
+\begin{align*}%[** TN: Set on one line in original]
+l &= a' \cos \chi + a'' \sin \chi, \\
+m &= b' \cos \chi + b'' \sin \chi, \\
+n &= c' \cos \chi + c'' \sin \chi.
+\end{align*}
+Or\Add{,} $v$~étant l'arc de la courbe~$(K)$, on a
+\[
+df = a\, dv,\qquad dg = b\, dv,\qquad dh = c\, dv;
+\]
+les formules~\Eq{(4')} donnent
+\[
+a\, dv + u\left[(-a'\sin \chi + a''\cos \chi)\, d \chi
+  - \cos \chi \left(\frac{a}{R} + \frac{a''}{T}\right) dv
+  + \sin \chi\, \frac{a'}{T}\, dv\right] = 0;
+\]
+ou
+\[
+a \left(1 - \frac{u \cos \chi}{R}\right)
+  + a' u \sin \chi \left(-\frac{\DPtypo{dx}{d\chi}}{dv} + \frac{1}{T}\right)
+  + a''u \cos \chi \left( \frac{\DPtypo{dx}{d\chi}}{dv} - \frac{1}{T}\right)
+  = 0\Add{.}
+\]
+On a \Card{2} équations analogues avec $b, b', b''$ et $c, c', c''$; nous avons
+ainsi \Card{3} équations linéaires et homogènes par rapport aux coefficients
+de $a, a', a''$. Le déterminant est~$1$, donc les inconnues
+sont toutes nulles; et comme $u$~n'est pas constamment nul, on a
+%% -----File: 093.png---Folio 85-------
+\[
+1 - \frac{u \cos \chi}{R} = 0,\qquad
+\sin \chi \left( - \frac{d \chi}{dv} + \frac{1}{T}\right) = 0,\qquad
+\cos \chi \left(\frac{d \chi}{dv} - \frac{1}{T}\right) = 0\Add{.}
+\]
+Les \Card{2} dernières donnent, en remplaçant $v$ par l'arc~$\DPtypo{S}{s}$
+\[
+\Tag{(1)}
+\frac{d \chi}{ds} = \frac{1}{T},
+\]
+et la \Ord[f]{1}{e} donne
+\[
+\Tag{(2)}
+u = \frac{R}{\cos \chi}\Add{.}
+\]
+Il y a donc une infinité de solutions: $\chi$~se détermine par une
+quadrature.
+
+\Illustration[2.25in]{093a}
+La formule~\Eq{(2)} nous montre que
+\[
+R = u \cos \chi;
+\]
+donc la projection du point~$P$, où la normale~$MG$
+rencontre son enveloppe, sur la
+normale principale, est le centre de courbure.
+\emph{Le point de contact de la normale
+avec son enveloppe est sur la droite polaire.
+Les développées d'une courbe sont
+sur la surface polaire.}
+
+Considérons \Card{2} solutions $\chi\Add{,} \chi'$ de l'équation~\Eq{(1)}, la différence
+$\chi - \chi'$ est constante; les deux normales $MG, MG'$ se coupent
+sous un angle constant. Donc, \emph{lorsque une normale à une courbe
+décrit une surface développable, si on la fait tourner dans
+chacune de ses positions d'un angle constant autour de la tangente,
+la droite obtenue décrit encore une développable}.
+
+Le plan osculateur à une développée est le plan tangent
+à la développable correspondante: c'est le plan~$GMT$, ce plan
+est normal au plan~$BMC$, plan tangent à la surface polaire.
+%% -----File: 094.png---Folio 86-------
+\emph{Donc les développées sont des géodésiques de la surface polaire.}
+
+Considérons la normale principale~$P''$ en $P$ à la développée,
+elle est dans le plan osculateur $GMT$, elle est perpendiculaire
+à la tangente $MP$, donc parallèle à~$MT$. \emph{Les normales principales
+aux développées d'une courbe sont parallèles aux tangentes à la
+courbe. Le plan normal à la courbe est le plan rectifiant de
+toutes ses développées.}
+
+En partant d'une courbe~$(G)$, et remarquant que la courbe
+donnée~$(K)$ en est la développante, on pourra énoncer les propriétés
+précédentes de façon à obtenir les propriétés des développantes
+d'une courbe.
+
+
+\Section{Lignes de courbure.}
+{3.}{} Considérons sur une surface~$(S)$ une ligne de courbure~$(K)$,
+et la développable circonscrite à~$(S)$ le long de~$(K)$. La
+direction d'une génératrice $MG$ de cette développable est conjuguée
+de la tangente~$MT$ à la ligne de courbure, et par conséquent
+est perpendiculaire à~$MT$, c'est-à-dire normale à~$(K)$. Cette génératrice~$MG$
+est donc constamment tangente à la développée d'une
+ligne de courbure, et nous voyons que \emph{les normales à une ligne
+de courbure tangentes à la surface engendrent une développable}.
+
+\begin{wrapfigure}[10]{O}{2.25in}
+\Input[2.25in]{094a}
+\end{wrapfigure}
+Faisons tourner $\DPtypo{MG'}{MG}$ d'un angle droit
+autour de la tangente, nous obtenons une
+droite~$MG'$ qui, étant perpendiculaire aux
+\Card{2} tangentes à la surface $MT, MG$, sera la
+normale à la surface. Donc \emph{les normales à
+la surface en tous les points d'une ligne
+%% -----File: 095.png---Folio 87-------
+de courbure engendrent une développable}.
+
+Considérons le point~$P'$ où la droite $MG'$ touche son enveloppe;
+c'est le point où la droite polaire de la ligne de
+courbure rencontre la normale à la surface. Or, d'après le
+Théorème de Meusnier, les droites polaires de toutes les courbes
+de la surface tangentes en~$M$ rencontrent la normale en~$M$
+en un même point, qui est le centre de courbure de la section
+normale correspondante: $P'$~est donc le centre de courbure de
+la section principale $G'MT$, c'est l'un des centres de courbure
+principaux de la surface au point~$M$.
+
+Reprenons alors les formules~\Eq{(4')} du §1, que nous écrirons
+\[
+dx + u\, dl = 0,\quad dy + u\, dm = 0,\quad dz + u\, dn = 0;
+\]
+$l, m, n$ sont ici les cosinus directeurs de la normale, $u$~est le
+rayon de courbure principal~$R$; et pour un déplacement sur une
+ligne de courbure, nous avons les \emph{formules d'Olinde Rodrigues}
+\[
+dx + R\, dl =0,\qquad dy + R\, dm = 0,\qquad dz + R\, dn = 0.
+\]
+
+Les \emph{Théorèmes de Joachimsthal} se déduisent aisément de ce
+qui précède. Supposons que l'intersection~$(K)$ de \Card{2} surfaces $(S)\Add{,}
+(S_{1})$ soit une ligne de courbure pour chacune d'elles. Soient
+$MG, MG_{1}$ les normales aux \Card{2} surfaces en un point~$M$ de~$(K)$. Elles
+engendrent deux développables, donc enveloppent deux développées
+de~$(K)$, et par suite leur angle est constant. \emph{Réciproquement},
+si l'intersection $(K)$ de $(S)\Add{,} (S_{1})$ est ligne de courbure
+de~$(S_{1})$, et si l'angle des \Card{2} surfaces est constant tout
+le long de~$(K)$, la normale $MG_{1}$\DPtypo{,}{} à~$(S_{1})$ engendre une développable,
+et comme $\Err{MG}{MG_{1}}$~fait avec~$MG$ un angle constant, elle engendre
+aussi une développable, donc $(K)$~est une ligne de courbure
+sur~$(S)$.
+%% -----File: 096.png---Folio 88-------
+
+La condition~\Eq{(5)} pour qu'une droite engendre une surface
+développable est ici
+\[
+\begin{vmatrix}
+dx & dl & l \\
+dy & dm & m \\
+dz & dn & n
+\end{vmatrix}
+= 0,
+\]
+ou
+\[%[** TN: Added elided entries]
+\begin{vmatrix}
+\mfrac{\dd x}{\dd u} · du + \mfrac{\dd x}{\dd v} · dv &
+\mfrac{\dd l}{\dd u}\Add{·} du + \mfrac{\dd l}{\dd v}\Add{·} dv & l \\
+\mfrac{\dd y}{\dd u} · du + \mfrac{\dd y}{\dd v} · dv &
+\mfrac{\dd m}{\dd u}\Add{·} du + \mfrac{\dd m}{\dd v}\Add{·} dv & m \\
+\mfrac{\dd z}{\dd u} · du + \mfrac{\dd z}{\dd v} · dv &
+\mfrac{\dd n}{\dd u}\Add{·} du + \mfrac{\dd n}{\dd v}\Add{·} dv & n
+\end{vmatrix}
+= 0.
+\]
+Multiplions par
+\[
+\begin{vmatrix}
+\mfrac{\dd x}{\dd u} & \mfrac{\dd x}{\dd v} & l \\
+\mfrac{\dd y}{\dd u} & \mfrac{\dd y}{\dd v} & m \\
+\mfrac{\dd z}{\dd u} & \mfrac{\dd z}{\dd v} & n
+\end{vmatrix}
+\neq 0;
+\]
+Nous obtenons
+\[
+\begin{vmatrix}
+E\, du + F\, dv & -L\, du - M\, dv & 0 \\
+F\, du + G\, dv & -M\, du - N\, dv & 0 \\
+0 & 0 & 1
+\end{vmatrix}
+= 0;
+\]
+et nous retrouvons ainsi \emph{l'équation différentielle des lignes
+de courbure}
+\[
+\begin{vmatrix}
+E\, du + F\, dv & L\, du + M\, dv \\
+F\, du + G\, dv & M\, du + N\, dv
+\end{vmatrix}
+= 0.
+\]
+
+La même méthode, appliquée à l'équation~\Eq{(6)}
+\[
+\begin{vmatrix}
+dx & dl \\
+dy & dm
+\end{vmatrix}
+= 0
+\]
+donne facilement l'équation différentielle
+\[
+\begin{vmatrix}
+dx + p\, dz & dp \\
+dy + q\, dz & dq
+\end{vmatrix}
+= 0.
+\]
+
+%% -----File: 097.png---Folio 89-------
+
+\Section{Développement d'une surface développable sur un plan.}
+{4.}{Toute surface développable est applicable sur un plan\Add{.}}
+
+Considérons d'abord le cas du cylindre, dont les équations
+sont
+\begin{align*}
+x &= f(v) + u·l, & y &= g(v) + u·m, &  z &= h(v) + u·n; \\
+dx &= f'(v)\, dv + l·du, &
+dy &= g'(v)\, dv + m·du, &
+dz &= h'(v)\, dv + n·du.
+\end{align*}
+Nous avons
+\[
+ds^{2} = \sum f'{}^{2}(v)·dv^{2} + 2 \sum lf'(v)·du\, dv + \sum l^{2}·du^{2}.
+\]
+Nous pouvons supposer que la directrice: $x = f(v)$, $y = g(v)$, $z = h(v)$
+est une section droite, ce qui donne $\sum lf' = 0$; que $l, m, n$
+sont cosinus directeurs: $\sum l^{2} = 1$; enfin que $v$~est l'arc sur
+la section droite: $\sum f'{}^{2} = 1$. Alors on a
+\[
+\Tag{(1)}
+ds^{2} = du^{2} + dv^{2};
+\]
+\DPtypo{On}{on} a l'élément linéaire d'un plan. \emph{Un cylindre est applicable
+sur un plan}, $\Phi$~et~\Eq{(1)} donne la loi du développement.
+
+Voyons maintenant le cas du cône
+\[
+x = u·l(v),\qquad y = u·m(v),\qquad z = u·n(v);
+\]
+$u$~est la longueur prise sur la génératrice à partir du sommet;
+supposons que $l, m, n$ soient cosinus directeurs de la génératrice,
+$v$~étant l'arc de courbe sphérique intersection du cône
+avec la sphère $u = 1$. Alors
+\begin{alignat*}{3}%[** TN: Set on one line in original]
+dx &= ul'(v)\, &&dv + l(v)\, &&du, \\
+dy &= um'(v)\, &&dv + m(v)\, &&du, \\
+dz &= u\DPtypo{n}{n'}(v)\, &&dv + n(v)\, &&du;
+\end{alignat*}
+et
+\[
+\Tag{(2)}
+ds^{2} = u^{2}\, dv^{2} + du^{2}.
+\]
+C'est l'élément linéaire d'un plan en coordonnées polaires. Un
+\emph{cône est applicable sur un plan}, $\Phi$~\Add{et}~\Eq{(2)} donne la loi du développement.
+
+Passons enfin au cas général
+\[
+x = f(v) + u·l(v),\qquad y = g(v) + u·m(v),\qquad z = h(v) + u·n(v)\Add{.}
+\]
+%% -----File: 098.png---Folio 90-------
+Nous supposerons que la courbe $x = f(v)$, $y = g(v)$, $z = h(v)$
+soit l'arête de rebroussement, $v$~l'arc sur cette courbe, $l, m, n$
+les cosinus directeurs de la tangente en un point, et $u$~la
+distance comptée sur cette tangente à partir du point de contact.
+Alors $l = f'= a$; $m = g'= b$; $n = h'= c$;
+et
+\begin{gather*}%[** TN: Aligned last three equations]
+l' = \frac{da}{dv} = \frac{a'}{R},\qquad
+m' = \frac{db}{dv} = \frac{b'}{R},\qquad
+n' = \frac{dc}{dv} = \frac{c'}{R}; \\
+\begin{alignedat}{3}
+dx &= a\, dv &&+ u\, \frac{a'}{R}\, dv &&+ a\, du, \\
+dy &= b\, dv &&+ u\, \frac{b'}{R}\, dv &&+ b\, du, \\
+dz &= c\, dv &&+ u\, \frac{c'}{R}\, dv &&+ c\, du\Add{;}
+\end{alignedat}
+\end{gather*}
+et
+\[
+ds^{2} = \bigl[d(u + v)\bigr]^{2} + \frac{u^{2}}{R^{2}}\, dv^{2}.
+\]
+Cet élément reste le même si $R$~garde la même expression en
+fonction de~$v$. Donc \emph{l'élément linéaire est le même pour toutes
+les surfaces développables dont les arêtes de rebroussement
+sont des courbes dont le rayon de courbure a la même expression
+en fonction de l'arc}:
+\[
+R = \Phi (v).
+\]
+Nous pouvons déterminer une courbe plane dont le rayon de courbure
+s'exprime en fonction de l'arc par l'équation précédente.
+Nous prendrons pour coordonnées dans le plan de cette courbe
+l'arc~$\DPtypo{S}{s}$ de la courbe, et la distance comptée sur la tangente
+à partir du point de contact et on aura pour l'élément linéaire
+du plan la forme précédente. La développable sera donc applicable
+sur ce plan. Quand la développable est donnée, on détermine
+par des opérations algébriques son arête de rebroussement,
+et par une quadrature l'arc de cette arête de rebroussement.
+On a alors
+\[
+R = \Phi(s)\Add{.}
+\]
+%% -----File: 099.png---Folio 91-------
+Il faut construire une courbe plane satisfaisant à cette condition.
+Si $\alpha$~est l'angle de la tangente avec~$Ox$, on a
+\[
+R = \frac{ds}{d \alpha};
+\]
+d'où
+\[%[** TN: Set on two lines in original]
+\frac{ds}{d \alpha} = \Phi (s), \qquad
+\alpha = \int \frac{ds}{\Phi (s)};
+\]
+et alors
+\[
+dx = \cos \alpha · ds,\qquad dy = \sin \alpha · ds;
+\]
+$x, y$ se déterminent au moyen de \Card{3} quadratures. La courbe que
+l'on obtient est le développement de l'arête de rebroussement.
+
+
+\Section{Réciproque.}
+{}{Réciproquement toute surface applicable sur un plan est
+une surface développable.}
+
+Soit la surface
+\[
+x = f(u,v),\qquad y = g(u,v),\qquad z = h(u,v),
+\]
+que nous supposons applicable sur un plan. Nous avons, en
+choisissant convenablement les coordonnées $u, v$:
+\[
+ds^{2} = E\Add{·} du^{2} + 2 F·du\, dv + G\Add{·} dv^{2} = du^{2} + dv^{2};
+\]
+d'où
+\[
+\sum \left(\frac{\dd x}{\dd u}\right)^{2} = 1,\qquad
+\sum \frac{\dd x}{\dd u} · \frac{\dd x}{\dd v} = 0,\qquad
+\sum \left(\frac{\dd x}{\dd v}\right)^{2} = 1.
+\]
+Différentions ces relations successivement par rapport à $u, v$,
+nous avons
+\begin{align*}
+&\sum \frac{\dd x}{\dd u} · \frac{\dd^{2} x}{\dd u^{2}} = 0, &
+&\sum \frac{\dd^{2} x}{\dd u^{2}} · \frac{\dd x}{\dd v}
+  + \sum \frac{\dd x}{\dd u} · \frac{\dd^{2} x}{\dd u\, \dd v} = 0, &
+&\sum \frac{\dd x}{\dd v} · \frac{\dd^{2} x}{\dd u\, \dd v} = 0, \\
+%
+&\sum \frac{\dd x}{\dd u}\Add{·} \frac{\dd^{2} x}{\dd u\, \dd v} = 0, &
+&\sum \frac{\dd^{2} x}{\dd u\, \dd v} · \frac{\dd x}{\dd v}
+  + \DPtypo{}{\sum} \frac{\dd x}{\dd u} · \frac{\dd^{2} x}{\dd v^{2}} = 0, &
+&\sum \frac{\dd x}{\dd v} · \frac{\dd^{2} x}{\dd v^{2}} = 0;
+\end{align*}
+d'où nous tirons:
+\[
+\sum \frac{\dd^{2} x}{\dd \DPtypo{u}{u^{2}}} · \frac{\dd x}{\dd v} = 0,\qquad
+\sum \frac{\dd x}{\dd u} · \frac{\dd^{2} x}{\dd \DPtypo{v}{v^{2}}} = 0.
+\]
+Considérons les \Card{2} fonctions $\dfrac{\dd x}{\dd u}$ et $\dfrac{\dd y}{\dd u}$. Leur déterminant fonctionnel
+est:
+%% -----File: 100.png---Folio 92-------
+\[
+\begin{vmatrix}
+\mfrac{\dd^{2} x}{\dd u^{2}} & \mfrac{\dd^{2} x}{\dd u\, \dd v} \\
+\mfrac{\dd^{2} y}{\dd u^{2}} & \mfrac{\dd^{2} y}{\dd u\, \dd v}
+\end{vmatrix}\Add{.}
+\]
+Or, considérons les équations
+\begin{alignat*}{4}
+&X\, \frac{\dd^{2} x}{\dd u^{2}} &&+ Y\, \frac{\dd^{2} y}{\dd u^{2}} &&+ Z\, \frac{\dd^{2} z}{\dd u^{2}} &&= 0, \\
+&X\, \frac{\dd^{2} x}{\dd u\, \dd v} &&+ Y\, \frac{\dd^{2} y}{\dd u\, \dd v} &&+ Z\, \frac{\dd^{2} z}{\dd u\, \dd v} &&= 0;
+\end{alignat*}
+d'après les relations précédemment écrites, ce système admet
+\Card{2} solutions distinctes
+\begin{alignat*}{3}
+X &= \frac{\dd x}{\dd u},\qquad &
+Y &= \frac{\dd y}{\dd u},\qquad &
+Z &= \frac{\dd z}{\dd u}; \\
+X &= \frac{\dd x}{\dd v}, &
+Y &= \frac{\dd y}{\dd v}, &
+Z &= \frac{\dd z}{\dd v}.
+\end{alignat*}
+Ces solutions ne sont pas proportionnelles, sans quoi les
+courbes $u = \cte$ et $v = \cte$ seraient constamment tangentes.
+Donc les \Card{3} déterminants déduits du tableau
+\[
+\begin{Vmatrix}
+\mfrac{\dd^{2} x}{\dd u^{2}}     & \mfrac{\dd^{2} y}{\dd u^{2}}     & \mfrac{\dd^{2} z}{\dd u^{2}} \\
+\mfrac{\dd^{2} x}{\dd u\, \dd v} & \mfrac{\dd^{2} y}{\dd u\, \dd v} & \mfrac{\dd^{2} z}{\dd u\, \dd v}
+\end{Vmatrix}
+\]
+\DPtypo{Sont}{sont} nuls; or\Add{,} ce sont les déterminants fonctionnels des \Card{3} quantités
+$\dfrac{\dd x}{\dd u}$, $\dfrac{\dd y}{\dd u}$, $\dfrac{\dd z}{\dd u}$ prises \Card{2} à \Card{2}, donc ces \Card{3} quantités sont
+fonctions de l'une d'entre elles, c'est-à-dire d'une seule
+variable~$t$. De même $\dfrac{\dd x}{\dd v}$, $\dfrac{\dd y}{\dd v}$, $\dfrac{\dd z}{\dd v}$ sont fonctions d'une même variable~$\theta$.
+De plus la relation
+\[
+\sum \frac{\dd x}{\dd u} · \frac{\dd x}{\dd v} = 0
+\]
+montre que $\theta$ par exemple s'exprime en fonction de~$t$.
+Les \Card{6} dérivées partielles sont donc fonctions d'une même variable;
+le plan tangent à la surface dépend d'un seul paramètre.
+La surface est développable.
+
+%[** TN: Roman numerals in original]
+\Paragraph{Remarque \1.} Dans le développement les lignes géodésiques
+%% -----File: 101.png---Folio 93-------
+se conservent; or\Add{,} les géodésiques du plan sont des droites.
+\emph{Les lignes géodésiques de la surface développable sont donc
+les lignes qui dans le développement de cette surface sur un
+plan, correspondent aux droites de ce plan.}
+
+En particulier, considérons la surface rectifiante d'une
+courbe, enveloppe du plan rectifiant. Cette courbe est une
+géodésique de sa surface rectifiante, puisque son plan osculateur
+est perpendiculaire au plan tangent; elle se développe
+donc suivant une droite lorsqu'on effectue le développement
+de la surface rectifiante sur un plan. \emph{De là le nom de plan
+rectifiant.}
+
+\Paragraph{Remarque \2.} Il résulte de là que la recherche des géodésiques
+d'une surface développable se ramène à son développement,
+et exige par conséquent \Card{4} quadratures.
+
+\Paragraph{Remarque \3.} La détermination des lignes de courbure,
+développantes de l'arête de rebroussement, revient à une quadrature.
+
+
+\Section{Lignes géodésiques d'une surface développable.}
+{5.}{} Nous avons trouvé les lignes géodésiques d'une surface
+développable en considérant le développement de cette
+surface sur un plan. On peut les chercher directement.
+Soit l'arête de rebroussement
+\[
+\Tag{(1)}
+x = f(s),\qquad y = g(s),\qquad z = h(s),
+\]
+$s$~désignant l'arc. Si $a, b, c$ sont les cosinus directeurs de la
+tangente, et $u$~une longueur comptée sur cette tangente à partir
+du point de contact, la surface est représentée par
+\[
+x = f + u · a,\qquad y = g + u · b,\qquad z = h + u · c;
+\]
+%% -----File: 102.png---Folio 94-------
+%[** TN: Seeming error in erratum, changed in accord with surrounding text]
+en désignant par $\Err{u, v, w}{\DPtypo{u', v', w'}{a', b', c'}}$,
+les dérivées de $\DPtypo{u v w}{a\Add{,} b\Add{,} c}$ par rapport \Err{$a, s$}{à~$s$},
+on a
+\[%[** TN: Added elided equations]
+\frac{dx}{ds} = a + u\, \frac{a'}{R} + au',\qquad
+\frac{dy}{ds} = b + u\, \frac{b'}{R} + bu',\qquad
+\frac{dz}{ds} = c + u\, \frac{c'}{R} + cu';
+\]
+ou
+\begin{gather*}%[** TN: Added elided equations in first set]
+\frac{dx}{ds} = a(1 + u') + a'\, \frac{u}{R},\quad
+\frac{dy}{ds} = b(1 + u') + b'\, \frac{u}{R},\quad
+\frac{dz}{ds} = c(1 + u') + c'\, \frac{u}{R}; \\
+%
+\frac{d^{2}x}{ds^{2}} = a\left(u'' - \frac{u}{R^{2}}\right)
+  + a'\, \frac{1}{R} \left(1 + 2u' - u\, \frac{R'}{R}\right)
+  + a''\, \frac{-u}{RT},
+\end{gather*}
+et les analogues.
+
+L'équation des lignes géodésiques est, en remarquant que la
+normale à la surface n'est autre que la binormale à l'arête de
+rebroussement
+\[
+\begin{vmatrix}
+\mfrac{d^{2}x}{ds^{2}} & \mfrac{d^{2}y}{ds^{2}} & \mfrac{d^{2}z}{ds^{2}} \\
+\mfrac{dx}{ds}         & \mfrac{dy}{ds}         & \mfrac{dz}{ds} \\
+a''                   & b''                   & c''
+\end{vmatrix}
+= 0,
+\]
+ou
+\[
+\begin{vmatrix}
+a \left(u'' - \mfrac{u}{R^{2}}\right) + \mfrac{a'}{R} \left(1 + 2u' - u\, \mfrac{R'}{R}\right) + a''\, \mfrac{- u}{RT} & \dots & \dots \\
+a (1 + u') + a'\, \mfrac{u}{R}  & \dots & \dots \\
+a''                             & \dots & \dots \\
+\end{vmatrix}
+= 0,
+\]
+ou, en décomposant en déterminants simples,
+\[
+\frac{1}{R} \left(1 + 2u' - u\, \frac{R'}{R}\right) (1 + u')
+\begin{vmatrix}
+a'  & b'  & c' \\
+a   & b   & c  \\
+a'' & b'' & c''
+\end{vmatrix}
++ \frac{u}{R} \left(u'' - \frac{u}{R^{2}}\right)
+\begin{vmatrix}
+a   & b   & c   \\
+a'  & b'  & c'  \\
+a'' & b'' & c''
+\end{vmatrix}
+= 0;
+\]
+ou enfin
+\[
+\frac{u}{R} \left(u'' - \frac{u}{R^{2}}\right)
+  - \frac{1}{R} (1 + u') \left(1 + 2u' - u\, \frac{R'}{R}\right) = 0,
+\]
+c'est-à-dire
+\[
+\Tag{(2)}
+u · u'' - 2u'{}^{2}
+  - u'\left(3 - u\, \frac{R'}{R}\right)
+  - \frac{u^{2}}{R^{2}} + u · \frac{R'}{R} - 1 =0.
+\]
+Telle est l'équation différentielle qui détermine~$u$.
+%% -----File: 103.png---Folio 95-------
+
+Cherchons la nature de l'intégrale générale. Si nous développons
+la surface sur un plan, la courbe~\Eq{(1)} sera représentée
+par une courbe
+\[
+x = F(s),\qquad y = G(s),
+\]
+dont le rayon de courbure sera encore~$R$. Le point homologue du
+point $(u,s)$ de la surface sera
+\[
+x = F + uF',\qquad y = G + uG'.
+\]
+Les droites du plan sont
+\[
+A(F + uF') + B(G + uG') + C = 0,
+\]
+d'où
+\[
+u = - \frac{AF + BG + C}{AF' + BG'};
+\]
+en remarquant que le dénominateur est la dérivée du numérateur
+nous sommes donc conduits à poser
+\[
+u = -\frac{w}{w'}\DPtypo{.}{,}
+\]
+et à prévoir que l'équation en~$w$ sera linéaire, homogène du
+\Ord{3}{e} ordre. Effectivement
+\begin{align*}
+u' &= -1 + \frac{ww''}{w'{}^{2}}, \\
+\intertext{et}
+u'' &= \frac{ww'''}{w'{}^{2}} + \frac{w''}{w'} - \frac{2ww''{}^{2}}{w'{}^{3}};
+\end{align*}
+\Eq{(2)}~devient alors
+\begin{multline*}
+\DPtypo{}{-}\frac{w}{w'} \left(\frac{ww'''}{w'{}^{2}} + \frac{w''}{w'} - \frac{2ww''{}^{2}}{w'{}^{3}}\right)
+  - 2\left(-1 + \frac{ww''}{w'{}^{2}}\right)^{2}
+  - 3\left(-1 + \frac{ww''}{w'{}^{2}}\right) \\
+  - \frac{R'}{R}\, \frac{w}{w'} \left(-1 + \frac{ww''}{w'{}^{2}}\right)
+  \DPtypo{- \frac{1}{R} · \frac{w}{w} - - \frac{w}{w} - 1}
+         {- \frac{1}{R^{2}} · \frac{w^{2}}{w'{}^{2}} - \frac{R'}{R} · \frac{w}{w'} - 1}
+  = 0;
+\end{multline*}
+ou, après simplification
+\[
+w''' + \frac{R'}{R}\, w'' + \frac{1}{R^{2}}\, w' = 0.
+\]
+Posons
+\[
+w' = \theta\Add{,}
+\]
+il vient
+\[
+\Tag{(3)}
+\theta'' + \frac{R'}{R}\, \theta' + \frac{1}{R^{2}}\, \theta = 0\Add{,}
+\]
+%% -----File: 104.png---Folio 96-------
+équation linéaire du \Ord{2}{e} ordre en~$\theta$. Faisons \DPchg{disparaitre}{disparaître} le \Ord{2}{e} terme
+par le changement de variable
+\[
+\alpha = \phi (s),
+\]
+d'où
+\begin{align*}
+\theta' &= \frac{d \theta}{d \alpha} · \phi', \\
+\theta''
+  &= \frac{d^{2} \theta}{d \DPtypo{\alpha}{\alpha^{2}}} · \phi'{}^{2}
+   + \frac{d \theta}{d \alpha}\, \phi'';
+\end{align*}
+\Eq{(3)}~devient
+\[
+\frac{d^{2} \theta}{d \alpha^{2}}\, \phi'{}^{2}
+  + \frac{d \theta}{d \alpha} \left( \phi'' + \frac{R'}{R}\, \phi'\right)
+  + \frac{1}{R^{2}}\, \theta = 0.
+\]
+Il faut alors choisir la fonction~$\phi$ de façon que l'on ait
+\[
+\phi'' + \frac{R'}{R}\, \phi' = 0 ,
+\]
+ou
+\[
+\frac{\phi''}{\phi'} = -\frac{R'}{R}.
+\]
+On peut prendre
+\[
+\phi' = \frac{1}{R},
+\]
+et poser
+\[
+ds = R · \alpha.
+\]
+Nous obtenons alors l'équation
+\[
+\frac{d^{2} \theta}{d \alpha^{2}} + \theta = 0,
+\]
+dont l'intégrale générale est
+\[
+\theta = w' = A \cos \alpha + B \sin \alpha = \frac{dw}{ds};
+\]
+d'où
+\[
+w = A \int \cos \alpha · ds + B \int \sin \alpha · ds + c,
+\]
+et enfin
+\[
+u = -\frac{\ds A \int \cos \alpha · ds + B \int \sin \alpha · ds + c}
+          {A \cos \alpha + B \sin \alpha},
+\]
+avec
+\[
+\alpha = \int \frac{ds}{R}.
+\]
+On peut se dispenser d'introduire l'arc~$s$ explicitement. Donc
+les lignes géodésiques d'une surface développable s'obtiennent
+par \Card{3} quadratures au plus. On constate de plus que les deux
+méthodes conduisent aux mêmes calculs.
+%% -----File: 105.png---Folio 97-------
+
+
+\Section{Surfaces réglées gauches. Trajectoires orthogonales des génératrices.}
+{6.}{} Soit la surface
+\[
+x = f(v) + u·l(v),\qquad y = g(v) + u·m(v),\qquad z = h(v) + u·n(v);
+\]
+les génératrices étant des géodésiques, il en résulte que
+\emph{les trajectoires orthogonales des génératrices déterminent
+sur ces génératrices des segments égaux}. Pour obtenir ces
+trajectoires orthogonales, il faut déterminer $u$ en fonction de~$v$
+de façon que l'on ait
+\[
+\sum l\, dx = 0.
+\]
+Pour simplifier nous supposerons que $l, m, n$ soient cosinus directeurs;
+on a alors
+\[
+\sum l^{2} = 1,\qquad \sum l\,dl = 0;
+\]
+et l'équation différentielle devient
+\[
+\sum l · df + du = 0,
+\]
+d'où
+\[
+u = - \int \sum \DPtypo{.l\, df}{l·df}.
+\]
+La détermination des trajectoires orthogonales des génératrices
+d'une surface réglée dépend d'une quadrature.
+
+\Paragraph{Remarque.} On peut rattacher ce fait à la formule qui
+donne la variation d'un segment de droite. Prenons sur la
+droite $MM_{1}$ une direction positive; soit $r$~la distance $MM_{1}$, prise
+\Figure[2.5in]{105a}
+en valeur absolue. Soient $(x, y, z)$ et
+$\DPtypo{}{(}x_{1}, y_{1}, z_{1})$, les coordonnées des deux extrémités,
+qui décrivent deux courbes données. Nous
+avons
+\[
+r^{2} = (x_{1} - x)^{2} + (y_{1} - y)^{2} + (z_{1} - z)^{2}\Add{,}
+\]
+d'où
+\[
+r\, dr = (x_{1} - x)(dx_{1} - dx) + (y_{1} - y)(dy_{1} - dy) % +
++ (z_{1} - z)(dz_{1} - dz)\DPtypo{.}{,}
+\]
+%% -----File: 106.png---Folio 98-------
+d'où
+\begin{multline*}
+dr = \left(\frac{x_{1} - x}{r}\, dx_{1}
+         + \frac{y_{1} - y}{r}\, dy_{1}
+         + \frac{z_{1} - z}{r}\, dz_{1}\right) \\
+   - \left(\frac{x_{1} - x}{r}\, dx
+         + \frac{y_{1} - y}{r}\, dy
+         + \frac{z_{1} - z}{r}\, dz\right).
+\end{multline*}
+Soient $\alpha, \beta, \gamma$, $\alpha_{1}, \beta_{1}, \gamma_{1}$ les tangentes aux courbes en $MM_{1}$ dirigées
+dans le sens des arcs croissants. Soient $\lambda\Add{,} \mu\Add{,} \nu$ les cosinus directeurs
+de la droite~$MM_{1}$. Nous avons
+\[
+dr = ds_{1} (\lambda \alpha_{1} + \mu \beta_{1} + \nu \gamma_{1})
+   - ds (\lambda \alpha + \mu \beta + \nu \gamma);
+\]
+soient $\theta, \theta_{1}$ les angles de $MM_{1}$ avec les \Card{2} tangentes; nous avons
+enfin la \emph{formule importante}
+\[
+dr = ds_{1} · \cos \theta_{1} - ds \Add{·} \cos \theta.
+\]
+
+Supposons la droite $MM_{1}$ tangente à la \Ord{1}{ère} courbe et normale
+à la \Ord{2}{e}, $\theta = 0$, $\theta_{1} = ±\frac{\pi}{2}$\Add{;} nous avons
+\[
+dr = - ds
+\]
+et nous retrouvons ainsi les propriétés des développantes et
+des développées.
+
+Supposons la droite normale aux deux courbes, $\theta = ±\frac{\pi}{2}$,
+$\theta_{1} = ±\frac{\pi}{2}$, alors $dr = 0$, $r = \cte$, et nous retrouvons les \DPtypo{propriétes}{propriétés}
+des trajectoires orthogonales des génératrices.
+
+
+\Section{Cône directeur. Point central. Ligne de striction.}
+{7.}{} On appelle \emph{cône directeur} de la surface le cône
+\[
+x = u · l(v),\qquad y = u · m(v),\qquad z = u · n(v).
+\]
+
+Si ce cône se réduit à un plan, ce plan s'appelle \emph{plan
+directeur}, et les génératrices sont toutes parallèles à ce
+plan.
+
+Le plan tangent en un point quelconque de la surface a
+pour coefficients les déterminants déduits de:
+%% -----File: 107.png---Folio 99-------
+\[
+\Tag{(1)}
+\begin{Vmatrix}
+l           & m           & n \\
+df + u\, dl & dg + u\, dm & dh + u\, dn
+\end{Vmatrix}\Add{.}
+\]
+Le plan tangent au cône directeur le long de la génératrice
+correspondant à celle qui passe par le point considéré a pour
+coefficients les déterminants déduits de
+\[
+\begin{Vmatrix}
+l  & m  & n \\
+dl & dm & dn
+\end{Vmatrix}\Add{.}
+\]
+Ces plans sont parallèles si $u$~est infini. On a alors sur la
+surface le plan tangent au point à l'infini sur la génératrice
+qu'on appelle \emph{plan asymptote}. \emph{Les plans asymptotes sont parallèles
+aux plans tangents au cône directeur le long des génératrices
+correspondantes.}
+
+\emph{Dans une surface à plan directeur, tous les plans asymptotes
+sont parallèles au plan directeur.}
+
+Pour que les deux plans tangents soient rectangulaires,
+il faut que la somme des produits des déterminants précédents
+soit nulle
+\[
+\begin{vmatrix}
+\sum l^{2} & \sum l\, df + u \sum l\, dl \\
+\sum ldl   & \sum dl·df + u \sum dl^{2}
+\end{vmatrix}
+= 0.
+\]
+Nous avons une équation du \Ord{1}{er} degré en~$u$. \emph{Il existe donc en
+général sur toute génératrice un point où le plan tangent est
+perpendiculaire au plan tangent au cône directeur, c'est-à-dire
+au plan asymptote. C'est le point central}, le plan tangent
+en ce point s'appelle \emph{plan central}.
+
+Le lieu des points centraux s'appelle \emph{ligne de striction}.
+
+Nous supposerons pour simplifier $\sum l^{2} = 1$, ce qui écarte
+le cas des surfaces réglées à génératrices isotropes. L'équation
+%% -----File: 108.png---Folio 100-------
+qui donne \DPtypo{le~$u$}{l'$u$} du point central se réduit à
+\[
+u \sum dl^{2} + \sum dl · df = 0;
+\]
+le point central existe donc toujours, sauf si l'on a
+\[
+\sum dl^{2} = 0.
+\]
+Dans ce cas la courbe sphérique base du cône directeur est
+une courbe minima de la sphère, c'est-à-dire, une génératrice
+isotrope. Le cône est alors un plan tangent au cône asymptote
+de la sphère, qui est un cône isotrope, c'est un plan isotrope.
+Les surfaces considérées sont des \emph{surfaces réglées à plan directeur
+isotrope}. Toutes sont imaginaires, sauf le \DPchg{paraboloide}{paraboloïde}
+de révolution.
+
+\Paragraph{Remarque.} Le plan tangent est indéterminé si tous les
+déterminants du tableau~\Eq{(1)} sont nuls. Alors $K$~étant un certain
+facteur, on a
+\[
+df + u\, dl + Kl = 0,\qquad
+dg + u\, dm + Km = 0,\qquad
+dh + u\, dn + Kn = 0\DPtypo{;}{,}
+\]
+ce qui donne
+\[
+\begin{vmatrix}
+df & dl & l \\
+dg & dm & m \\
+dh & dn & n
+\end{vmatrix} = 0\DPtypo{:}{;}
+\]
+la surface est développable. Pour trouver le point où le plan
+tangent est indéterminé multiplions par $dl, dm, dn$ et ajoutons,
+il vient
+\[
+u \sum dl^{2} + \sum dl · df = 0;
+\]
+c'est le point de contact de la génératrice et de l'arête de
+rebroussement. C'est ce qui explique que la formule précédente
+qui donne la ligne de striction pour une surface réglée quelconque,
+donne l'arête de rebroussement pour une surface développable.
+%% -----File: 109.png---Folio 101-------
+
+
+\Section{Variations du plan tangent le long d'une génératrice.}
+{8.}{} Proposons-nous de chercher l'angle des plans tangents
+à une surface réglée en 2 points d'une même génératrice.
+A cet effet, traitons d'abord le problème suivant: on a une
+droite~$\Delta$, de cosinus directeurs $\alpha\Add{,} \beta\Add{,} \gamma$, et \Card{2} droites qui la rencontrent
+$D (p\Add{,} q\Add{,} r)$ et $D' (p'\Add{,} q'\Add{,} r')$. Calculons l'angle~$V$ des \Card{2} plans
+$D \Delta$~et~$D' \Delta$.
+
+\Figure[2.25in]{109a}
+Considérons un trièdre trirectangle
+auxiliaire dont l'un des axes soit~$\Delta$;
+soient $\alpha'\Add{,} \beta'\Add{,} \gamma'$, $\alpha''\Add{,} \beta''\Add{,} \gamma''$ les cosinus directeurs
+des autres axes, et soient dans ce
+système $u\Add{,} v\Add{,} w$ et $u'\Add{,} v'\Add{,} w'$ les coefficients de
+direction de~$\Delta \Delta'$. Nous avons
+\[
+\tg V = \frac{vw' - wv'}{vv' + ww'}\Add{.}
+\]
+Mais on a
+\[
+\begin{alignedat}{3}
+  u  &= \alpha p  &&+ \beta q  &&+ \gamma r,  \\
+  u' &= \alpha p' &&+ \beta q' &&+ \gamma r',
+\end{alignedat}
+\quad
+\begin{alignedat}{3}
+  v  &= \alpha' p  &&+ \beta' q  &&+ \gamma' r, \\
+  v' &= \alpha' p' &&+ \beta' q' &&+ \gamma' r',
+\end{alignedat}
+\quad
+\begin{alignedat}{3}
+  w  &= \alpha'' p  &&+ \beta'' q  &&+ \gamma'' r \\
+  w' &= \alpha'' p' &&+ \beta'' q' &&+ \gamma'' r',
+\end{alignedat}
+\]
+d'où
+\begin{align*}
+vw' - wv' &=
+\begin{vmatrix}
+\begin{alignedat}{3}
+  &\alpha' p  &&+ \beta' q  &&+ \gamma' r \\
+  &\alpha' p' &&+ \beta' q' &&+ \gamma' r'
+\end{alignedat}
+&
+\begin{alignedat}{3}
+  &\alpha'' p  &&+ \beta'' q  &&+ \gamma'' r \\
+  &\alpha'' p' &&+ \beta'' q' &&+ \gamma'' r'
+\end{alignedat}
+\end{vmatrix} \\
+&=
+\begin{Vmatrix}
+\alpha'  & \beta'  & \gamma' \\
+\alpha'' & \beta'' & \gamma''
+\end{Vmatrix}
+\begin{Vmatrix}
+p  & q  & r \\
+p' & q' & r'
+\end{Vmatrix}
+=
+\begin{vmatrix}
+\alpha & \beta & \gamma \\
+p      & q     & r \\
+p'     & q'    & r'
+\end{vmatrix}.
+\end{align*}
+D'ailleurs
+\[
+uu' + vv' + ww' = pp' + qq' + rr',
+\]
+d'où
+\[
+vv' + ww' = pp' + qq' + rr' - \sum \alpha p · \sum \alpha p'.
+\]
+Alors
+\[
+\tg V = \frac{
+\begin{vmatrix}
+\alpha & \beta & \gamma \\
+p      & q     & r \\
+p'     & q'    & r'
+\end{vmatrix}}{\sum pp' - \sum \alpha p · \sum \alpha p'}
+= \frac{D \sum \alpha^{2}}
+       {\sum \alpha^{2} · \sum pp' - \sum \alpha p · \sum \alpha p'}.
+\]
+Sous cette forme, on peut alors introduire les coefficients
+%% -----File: 110.png---Folio 102-------
+directeurs $l, m, n$ de la direction~$\Delta$
+\[
+\Tag{(1)}
+\tg V = \frac{\raisebox{-\baselineskip}{$\sqrt{l^{2} + m^{2} + n^{2}}$}\,
+\begin{vmatrix}
+l  & m  & n \\
+p  & q  & r \\
+p' & q' & r'
+\end{vmatrix}}{\sum l^{2} · \sum pp' - \sum lp · \sum lp'}.
+\]
+
+Appliquons cette formule à l'angle des plans tangents en
+\Card{2} points $M$,~$M'$ d'une même génératrice. On peut prendre pour directions
+$D$,~$D'$ les directions tangentes aux courbes $u = \cte$:
+\begin{align*}
+p  &= df + u\, dl,  & q  &= dg + u\, dm,  & r  &= dh + u\, dn; \\
+p' &= df + u'\, dl, & q' &= dg + u'\, dm, & r' &= dh + u'\, dn;
+\end{align*}
+le déterminant de la formule~\Eq{(1)} devient
+\[
+\begin{vmatrix}
+l & df + u\, dl & df + u'\, dl \\
+m & dg + u\, dm & dg + u'\, dm \\
+n & dh + u\, dn & dh + u'\, dn
+\end{vmatrix}
+=
+\begin{vmatrix}
+l & dl & df \\
+m & dm & dg \\
+n & dn & dh
+\end{vmatrix} (u - u');
+\]
+et
+\[
+\tg V = \frac{\raisebox{-\baselineskip}{$(u' - u) \sqrt{l^{2} + m^{2} + n^{2}}$}\,
+\begin{vmatrix}
+df & dg & dh \\
+dl & dm & dn \\
+l  & m  & n
+\end{vmatrix}}
+{\begin{vmatrix}
+\sum l^{2}           & \sum l (df + u\, dl) \\
+\sum l(df + u'\, dl) & \sum (df + u\, dl) (df + u'\, dl)
+\end{vmatrix}}.
+\]
+Nous poserons
+\[
+D =
+\begin{vmatrix}
+df & dg & dh \\
+dl & dm & dn \\
+l  & m  & n
+\end{vmatrix}.
+\]
+
+Pour simplifier ce résultat, nous prendrons pour $l, m, n$
+les cosinus directeurs de la génératrice ($\sum l^{2} = 1$, $\sum l\, dl = 0$);
+nous supposerons que la courbe $x = f(v)$, $y = g(v)$, $z = h(v)$
+soit trajectoire orthogonale des génératrices, $\sum l\, df = 0$.
+Nous déterminerons~$u$ par la relation
+\[
+u \sum dl^{2} + \sum dl · df = 0
+\]
+ce qui revient à prendre pour l'un des points le point central.
+%% -----File: 111.png---Folio 103-------
+Le dénominateur devient
+\[
+\sum df^{2} + u \sum dl\Add{·} df
+  = \frac{\sum df^{2} · \sum dl^{2} - \left(\sum dl · df\right)^{2}}{\sum dl^{2}};
+\]
+et alors
+\[
+\tg V = \frac{(u' - u) D · \sum dl}{\sum df^{2} · dl^{2} - \left(\sum dl · df\right)^{2}}.
+\]
+Posons
+\[
+K = \frac{\sum dl^{2} · \sum df^{2} - \left(\sum dl · df\right)^{2}}
+         {D · \sum dl^{2}}\DPtypo{:}{;}
+\]
+en remarquant que $u'- u = CM$, on a
+\[
+\Tag{(2)}
+\tg V = \frac{CM}{K},
+\]
+\emph{formule de Chasles}. D'où les conséquences bien connues suivantes,
+et qui ne sont en défaut que pour des génératrices singulières:
+
+\ParItem{\Primo.} \emph{\DPtypo{lorsque}{Lorsque} $M$~décrit la génératrice d'un bout à l'autre,
+le plan tangent~$(P)$ en~$M$ tourne autour de la génératrice toujours
+dans le même sens, et la rotation totale qu'il effectue
+est de~$180°$.} En deux points différents, les plans tangents
+sont différents.
+
+\ParItem{\Secundo.} \emph{La division des points~$M$ et le faisceau des plans~$(P)$
+sont en correspondance homographique.}
+
+\ParItem{\Tertio.} Comme trois couples définissent une homographie,
+\emph{deux surfaces réglées qui ont une génératrice commune, et
+qui sont tangentes en trois points de cette génératrice, sont
+tangentes en tous les autres points de cette génératrice},
+c'est-à-dire se raccordent tout le long de cette génératrice.
+L'expression de~$K$ peut se simplifier; on a:
+%% -----File: 112.png---Folio 104-------
+\[
+D =
+\begin{vmatrix}
+\sum df^{2} & \sum dl·df\DPtypo{.}{} & \sum l·df \\
+\sum dl·df  & \sum dl^{2} & \sum l\Add{·}dl \\
+\sum l·df   & \sum l\Add{·}dl & \sum l^{2}
+\end{vmatrix}
+= \sum dl^{2}·\sum df^{2} - \left(\sum dl\Add{·}df\right)^{2},
+\]
+d'où
+\[
+\Tag{(3)}
+K = \frac{D}{\sum dl^{2}}.
+\]
+Dans le cas général, on trouve de même
+\[
+\Tag{(4)}
+K = \frac{D · \sum l^{2}}{\sum l^{2} · \sum dl^{2} - \left(\sum l\Add{·} dl\right)^{2}}.
+\]
+$K$~est le \emph{paramètre de distribution}; il est rationnel. La formule~\Eq{(2)}
+montre que, si $M$~se déplace dans une direction quelconque
+sur la génératrice, le plan tangent tourne, par rapport
+à cette direction, dans le sens positif de rotation, si
+$K$~est positif; et tourne dans le sens négatif, si $K$~est négatif.
+
+La signe de~$K$ correspond donc à une propriété géométrique
+de la surface. D'après \Eq{(3)}~ou~\Eq{(4)}, \emph{le paramètre de distribution
+est nul pour une surface développable}.
+
+\Paragraph{Remarque.} Soient sur une même génératrice \Card{2} points $M$\Add{,}~$M'$
+où les plans tangents soient rectangulaires. On a
+\[
+\tg V · \tg V' = -1,
+\]
+d'où, en vertu de~\Eq{\DPtypo{(7)}{(2)}}\Add{,}
+\[
+CM · CM' = -K^{2};
+\]
+\emph{les points d'une génératrice où les plans tangents sont rectangulaires
+forment une involution dont $C$ est le point central}.
+
+\Paragraph{Exemple \1.} \emph{Surface engendrée par les binormales d'une
+courbe gauche\Add{.}}
+
+Soit la courbe
+\[
+x = f(s),\qquad
+y = g(s),\qquad
+z = h(s);
+\]
+%% -----File: 113.png---Folio 105-------
+avec les notations habituelles, nous avons
+\begin{alignat*}{3}
+df &= a\, ds,  &dg &= b\, ds, & dh &= c\, ds,\\
+l &= a'',       &m &= b'',    &  n &= c'',\\
+\intertext{et}
+dl &= \frac{a'}{T}\, ds, \qquad &
+dm &= \frac{b'}{T}\, ds, \qquad &
+dn &= \frac{c'}{T}\, ds.
+\end{alignat*}
+Le point central est ici défini par $u = 0$; la courbe est ligne
+de striction. Le paramètre de distribution est
+\[
+K = T^{2}
+\begin{vmatrix}
+a  &   b   &  c \\
+\mfrac{a'}{T}  &   \mfrac{b'}{T} & \mfrac{c'}{T} \\
+a'' &   b''  &  c''
+\end{vmatrix} = T ;
+\]
+le paramètre de distribution est égal au rayon de torsion de
+la courbe au point correspondant. La courbe est ligne de
+striction, trajectoire orthogonale des génératrices et géodésique.
+
+\Paragraph{Exemple \2.} \emph{Surface engendrée par les normales principales
+à une courbe.}
+On a ici
+\begin{alignat*}{3}
+df &= a\, ds,   &  dg &= b\, ds,    & dh &= c\, ds,\\
+ l &= a',   &  m &= b',   &  n &= c',\\
+dl &= - \frac{a}{R} - \frac{a''}{T}\, ds, \qquad &
+dm &= - \frac{b}{R} - \frac{b''}{T}\, ds, \qquad &
+dn &= - \frac{c}{R} - \frac{c''}{T}\, ds;
+\end{alignat*}
+le point central est défini par
+\[
+u = \frac{\displaystyle\sum a\left(\frac{a}{R} + \frac{a''}{T}\right)}
+         {\displaystyle\sum  \left(\frac{a}{R} + \frac{a''}{T}\right)^{2} }
+  = \frac{\dfrac{1}{R}}{\dfrac{1}{R^{2}} + \dfrac{1}{T^{2}}}
+  = \frac{RT^{2}}{R^{2} + T^{2}} = MC,
+\]
+et on a:
+\[
+K = - \frac{R^{2}T^{2}}{R^{2} + T^{2}}
+\begin{vmatrix}
+a   &   b  &    c \\
+\mfrac{a}{R} + \mfrac{a''}{T} &
+\mfrac{b}{R} + \mfrac{b''}{T} &
+\mfrac{c}{R} + \mfrac{c''}{T} \\
+a'  &   b'  &   c'
+\end{vmatrix} = \frac{R^{2}T}{R^{2} + T^{2}}.
+\]
+%% -----File: 114.png---Folio 106-------
+Cherchons le plan tangent au centre de courbure~$O$. Nous avons
+\[
+\tg V = \frac{CO}{K}
+  = \frac{MO - MC}{K}
+  = \frac{1}{K} \left(R - \frac{RT^{2}}{R^{2} + T^{2}}\right)
+  = \frac{1}{K} · \frac{R^{2}}{R^{2} + T^{2}}
+  = \frac{R}{T};
+\]
+pour le point~$M$, qui est sur la courbe on a
+\[
+\tg V = \frac{CM}{K} = - \frac{T}{R},
+\]
+donc
+\[
+\tg V · \tg V'= - 1\Add{.}
+\]
+Les plans tangents en $M$~et~$O$ sont rectangulaires, ce qui est un
+cas particulier d'une proposition que nous verrons plus loin
+(\No12).
+
+\Section{\DPchg{Elément}{Élément} linéaire.}
+{9.}{} Cherchons l'élément linéaire d'une surface réglée:
+\[
+x = f(v) + u·l(v) \qquad
+y = g(v) + u·m(v) \qquad
+z = h(v) + u·n(v)\Add{.}
+\]
+
+En désignant par des accents les dérivées par rapport à~$v$,
+il vient:
+\[
+dx = (f' + ul')\, dv + l\, du, \quad
+dy = (g' + um')\, dv + m\, du, \quad
+dz = (h' + un')\, dv + n\, du
+\]
+et
+\[
+ds^{2} = E\, du^{2} + 2F\, du\, dv + G\, dv^{2},
+\]
+avec:
+\[
+E = \sum l^{2}, \quad
+F = u \sum ll' + \sum lf', \quad
+G = u^{2} \sum l'{}^{2} + 2u \sum l'f' + \sum f'{}^{2}\Add{.}
+\]
+Supposons que $l\Add{,} m\Add{,} n$ soient les cosinus directeurs:
+\begin{gather*}
+\sum l^{2} = 1, \qquad \sum ll' = 0, \\
+E = 1, \qquad F = \sum lf', \qquad
+G = u^{2} \sum l'{}^{2} + 2u \sum l'f' + \sum f'{}^{2}.
+\end{gather*}
+Ces résultats s'obtiennent directement en faisant le changement
+de paramètre
+\[
+\sqrt{E} · u = u';
+\]
+d'où
+\[
+du' = \sqrt{E} · du + u\, \smash[t]{\frac{\dfrac{dE}{dv}}{2 \sqrt{E}}}\,dv.
+\]
+Nous avons alors, en supprimant les accents,
+\[
+ds^{2} = du^{2} + 2F du\,dv + G\,dv^{2}.
+\]
+%% -----File: 115.png---Folio 107-------
+Supposons de plus que la courbe $x = f(v), y = g(v), z = h(v)$
+soit trajectoire orthogonale des génératrices, alors $\sum lf' = 0$,
+$F = 0$, et on a
+\[
+ds^{2} = du^{2} + G\, dv^{2};
+\]
+il est évident que l'élément linéaire doit avoir cette forme,
+car on a un système de coordonnées orthogonales. On arrive
+aussi à cette expression en posant
+\[
+du + F\, dv = du',
+\]
+d'où
+\[
+u' = u + \int F\, dv\Add{,}
+\]
+ce qui exige une quadrature. La variable~$u$ est définie à une
+constante près, c'est une longueur portée sur chaque génératrice
+à partir de la même trajectoire orthogonale. Pour définir
+la variable~$v$, considérons la direction de la génératrice
+$x = l$, $y = m$, $z = n$. Ces équations sont celles de la trace du
+cône directeur sur la sphère de rayon~$1$; nous prendrons pour~$v$
+l'arc de cette courbe; alors $\sum l'{}^{2} = 1$, et
+\[
+G = u^{2} + 2u \sum l'f' + \sum f'{}^{2}.
+\]
+Posons
+\[
+\sum l'f' = G_{0}, \qquad \sum f'{}^{2} = G_{1},
+\]
+nous avons
+\[
+G = u + 2u G_{0} + G_{1};
+\]
+les quantités $G_{0}\Add{,} G_{1}$, ainsi introduites sont liées d'une façon
+simple au point central et au paramètre de distribution. Considérons
+l'involution des points~$M\Add{,} M'$ où les plans tangents sont
+rectangulaires; son point central est le point central de la
+génératrice, et on a, en désignant par~$K$ le paramètre de distribution,
+\[
+CM \cdot CM'= -K^{2}.
+\]
+Le plan tangent en un point~$u$ de la génératrice a pour coefficients
+%% -----File: 116.png---Folio 108-------
+les déterminants déduits du tableau
+\[
+\begin{Vmatrix}
+a         & b        &  c     \\
+f' + ul'  & g' + um' &  h' + un'
+\end{Vmatrix};
+\]
+de même le plan tangent au point~$u'$ aura pour \DPtypo{cofficients}{coefficients} les
+déterminants déduits du tableau
+\[
+\begin{Vmatrix}
+a         & b         & c       \\
+f' + u'l' & g' + u'm' & h' + u'n'
+\end{Vmatrix}.
+\]
+Exprimons que ces plans tangents sont rectangulaires. La somme
+des produits des déterminants précédents, et par suite le
+produit des tableaux, doit être nul, ce qui donne
+\[
+\begin{vmatrix}
+1 & 0                           \\
+0 & G_{1} + (u + u')G_{0} + uu'
+\end{vmatrix} = 0 ;
+\]
+la relation d'involution est donc
+\[
+uu' + (u + u')G_{0} + G_{1} = 0,
+\]
+ou
+\[
+(u + G_{0})(u' + G_{0}) = G_{0}^{2} - G_{1}.
+\]
+$u + G_{0}$ doit représenter~$CM$; si donc $I$~est l'intersection de la
+génératrice avec la trajectoire orthogonale $u = 0$, on a
+\[
+u + G_{0} = CM = IM - IC;
+\]
+mais $IM = u$, donc $G_{0} = -IC$, $-G_{0}$~est l'$u$ du point central; posons
+\[
+P = -G_{0} = -\sum l'f'.
+\]
+De plus
+\[
+G_{0}^{2} - G_{1} = -K^{2},
+\]
+d'où
+\[
+G_{1} = G_{0}^{2} + K^{2} = P^{2} + K^{2} = \sum f'{}^{2};
+\]
+alors
+\[
+G = u^{2} - 2uP + P^{2} + K^{2} = (u - P)^{2} + K^{2}.
+\]
+Finalement, \emph{si $l, m, n$ sont les cosinus directeurs de la génératrice,
+$v$~l'arc de la trace du cône directeur sur la sphère
+de rayon~$1$, $u$~la longueur portée sur la génératrice à partir
+d'une trajectoire orthogonale, on a}
+%% -----File: 117.png---Folio 109-------
+\[
+\Tag{(1)}
+ds^{2} = du^{2} + \bigl[(u - P)^{2} + K^{2}\bigr]\, dv^{2}\Add{,}
+\]
+\emph{$P$ étant l'$u$ du point central et $K$~le paramètre de distribution\Add{.}}
+
+\Paragraph{Remarque.} Ceci peut servir à calculer le paramètre de
+distribution. On~a
+\[
+\begin{vmatrix}
+f' & g' & h' \\
+l' & m' & n' \\
+l  & m  & n
+\end{vmatrix}^{2} =
+\begin{vmatrix}
+G_{1} & G_{0} & 0 \\
+G_{0} & 1     & 0 \\
+0     & 0     & 1
+\end{vmatrix} = G_{1} - G_{0}^{2} = K^{2},
+\]
+et on peut écrire
+\[
+\Tag{(2)}
+K =
+\begin{vmatrix}
+f' & g' & h' \\
+l' & m' & n' \\
+l  & m  & n
+\end{vmatrix}, \qquad P = - \sum l'f'.
+\]
+
+\emph{Réciproquement}, soit une surface dont l'élément linéaire
+soit de la forme
+\[
+ds^{2} = du^{2} + \bigl[(u - P)^{2} + K^{2}\bigr] dv^{2};
+\]
+cherchons s'il y a des surfaces réglées applicables sur cette
+surface; les éléments d'une telle surface réglée seront déterminés
+par les relations
+\[
+\sum l^{2} = 1, \quad
+\sum lf' = 0, \quad
+\sum l'{}^{2}= 1, \quad
+\sum l'f'= -P, \quad
+\sum f'{}^{2} = K^{2} + P^{2};
+\]
+la dernière de ces relations s'écrit, d'après l'expression de~$K$,
+\[
+\sum f' (mn' - nm') = -K.
+\]
+Nous pouvons d'abord nous donner arbitrairement le cône directeur
+de façon à satisfaire $\sum l^{2} = 1$, $\sum l'{}^{2} = 1$. Il reste alors à
+satisfaire à \Card{3} équations linéaires en $f'\Add{,} g'\Add{,} h'$ dont le déterminant
+n'est pas nul; $f', g', h'$ seront alors parfaitement déterminés,
+$f, g, h$ le seront à une constante additive près, ce qui revient
+à ajouter à $x, y, z$ des quantités constantes, c'est-à-dire à
+faire subir à la surface une translation. \emph{Lorsqu'on a une surface
+réglée, il y a donc une infinité de surfaces réglées applicables
+%% -----File: 118.png---Folio 110-------
+sur elle, les génératrices correspondant aux génératrices}
+puisqu'on peut prendre arbitrairement le cône directeur.
+Remarquons que dans l'élément linéaire figure, non pas~$K$,
+mais~$K^{2}$, de sorte qu'en particulier \emph{il existe \Card{2} surfaces réglées
+ayant même cône directeur, des paramètres de distribution
+égaux et de signes contraires et applicables l'une sur
+l'autre}.
+
+Pour avoir explicitement $f, g, h$, résolvons le système des
+équations linéaires
+\[
+\sum lf' = 0, \qquad \sum l'f' = -P, \qquad \sum (mn' - nm')f' = -K;
+\]
+$l\Add{,} m\Add{,} n$, $l'\Add{,} m'\Add{,} n'$ sont ici cosinus directeurs de \Card{2} directions rectangulaires.
+Introduisons une nouvelle direction de cosinus
+$l_{2}\Add{,} m_{2}\Add{,} n_{2}$ formant avec les \Card{2} précédentes un trièdre trirectangle
+\[
+l_{2}= mn' - nm', \qquad m_{2} = nl' - ln', \qquad n_{2} = lm' - ml'.
+\]
+Le système devient
+\[
+\sum lf' = 0, \qquad \sum l'f' = -P, \qquad \sum l_{2} f'= -K;
+\]
+d'où
+\[
+\Tag{(3)}%[** TN: Added brace]
+\left\{
+\begin{alignedat}{2}
+f' &= -Pl' &&- K (mn' - nm'), \\
+g' &= -Pm' &&- K (nl' - ln'), \\
+h' &= -Pn' &&- K (lm' - ml').
+\end{alignedat}
+\right.
+\]
+On a $f, g, h$ par des quadratures.
+
+\Section{La forme \texorpdfstring{$\Psi$}{Psi} et les lignes asymptotiques.}
+{10.}{} Nous avons
+\[
+\Psi(du, dv) = \sum Ad^{2}x =
+\begin{vmatrix}
+d^{2}x  &  d^{2}y  & d^{2}z \\
+\mfrac{\dd x}{\dd u} & \mfrac{\dd y}{\dd u} & \mfrac{\dd z}{\dd u} \\
+\mfrac{\dd x}{\dd v} & \mfrac{\dd y}{\dd v} & \mfrac{\dd z}{\dd v}
+\end{vmatrix} = \begin{vmatrix}
+(f'' + ul'')\,dv^{2} + 2l'\,du\,dv & \dots \\
+1            & \dots \\
+f' + ul'     & \dots
+\end{vmatrix}
+\]
+on a pour~$\Psi$ une expression de la forme
+\[
+\Psi(du, dv) = 2F'\, du\, dv + G'\, dv^{2},
+\]
+%% -----File: 119.png---Folio 111-------
+$F'$~étant fonction de $v$~et~$G'$ un trinôme du \Ord{2}{e} degré en~$u$. Nous
+trouvons naturellement pour lignes asymptotiques les courbes
+$dv = 0$, $v = \cte$ qui sont les génératrices. Les autres lignes
+asymptotiques sont déterminées par l'équation différentielle
+\[
+\frac{du}{dv} = - \frac{G'}{2F'},
+\]
+ou
+\[
+\Tag{(1)}
+\frac{du}{dv} = Ru^{2} + 2Su + T,
+\]
+$R\Add{,} S\Add{,} T$ étant fonctions de~$v$. C'est une \emph{équation de \DPtypo{Ricatti}{Riccati}}. Rappelons
+ses propriétés.
+
+\ParItem[\DPchg{Equation}{Équation} de \DPtypo{Ricatti}{Riccati}.]{\Primo.} \emph{Supposons qu'on ait une solution}~$u_{1}$, de cette équation.
+Posons
+\[
+\Tag{(2)}
+u = u_{1} + \frac{1}{w},
+\]
+d'où
+\[
+du = du_{1} - \frac{dw}{w^{2}}.
+\]
+L'équation~\Eq{(1)} devient
+\[
+\frac{du_{1}}{dv} - \frac{1}{w^{2}} · \frac{dw}{dv}
+  = Ru_{1}^{2} + 2R \frac{u_{1}}{w} + R \frac{1}{w^{2}}
+  + 2Su_{1} + 2S \frac{1}{w} + T;
+\]
+mais $u_{1}$ étant intégrale de~\Eq{(1)}, on a
+\[
+\frac{du_{1}}{dv} = Ru_{1}^{2} + 2Su_{1} + T,
+\]
+de sorte que l'équation devient
+\[
+-\frac{dw}{dv} = 2(Ru_{1} + S) w + R,
+\]
+ou
+\[
+\Tag{(3)}
+\frac{dw}{dv} = Qw - R.
+\]
+C'est une équation linéaire dont \emph{l'intégration s'effectue par
+\Card{2} quadratures.}
+
+\ParItem{\Secundo.} \emph{Supposons qu'on ait \Card{2} intégrales} $u_{2}\Add{,} u_{1}$ de l'équation.
+Posons
+\[
+u_{2} = u_{1} + \frac{1}{w_{0}},
+\]
+d'où
+\[
+w_{0} = \frac{1}{u_{2} - u_{1}};
+\]
+%% -----File: 120.png---Folio 112-------
+$w_{0}$~sera une intégrale de l'équation~\Eq{(3)}. Posons alors
+\[
+\Tag{(4)}
+w = w_{0} + \theta,
+\]
+d'où
+\[
+dw = dw_{0} + d\theta;
+\]
+\Eq{(3)}~devient
+\[
+\frac{dw_{0}}{dv} + \frac{d\theta}{dv} = Qw_{0} + Q\theta - R;
+\]
+ou, comme $w$~est intégrale de~\Eq{(3)}
+\[
+\frac{d\theta}{dv} = Q\theta,
+\]
+équation linéaire sans \Ord{2}{e} membre qui s'intègre immédiatement
+\emph{par une seule quadrature}:
+\begin{gather*}
+\Tag{(5)}
+\frac{d\theta}{\theta} = Q\, dv, \\
+\log \theta = \int Q\, dv, \\
+\theta = e^{Q\, dv}.
+\end{gather*}
+
+\ParItem{\Tertio.} \emph{Supposons qu'on ait \Card{3} intégrales} $u_{1}\Add{,} u_{2}\Add{,} u_{3}$ de l'équation~\Eq{(1)}.
+On a alors \Card{2} intégrales de l'équation~\Eq{(3)}. Soit
+\[
+w_{1} = \frac{1}{u_{0} - u_{1}};
+\]
+$w_{1}$~est intégrale de~\Eq{(3)}, et par suite on a une intégrale~$\theta_{0}$ de~\Eq{(5)}
+\[
+\theta_{0} = w_{1} - w_{0}
+  = \frac{1}{u_{3} - u_{1}} - \frac{1}{u_{2} - u_{1}}
+  = \frac{u_{2} - u_{3}}{(u_{3} - u_{1}) (u_{2} - u_{1})}.
+\]
+Posons
+\begin{gather*}
+\theta = \theta_{0} \psi, \\
+d\theta = \theta_{0} · d\psi + \psi · d\theta_{0};
+\end{gather*}
+\Eq{(5)}~devient
+\[
+\theta_{0}\, \frac{d\psi}{dv} + \psi · \frac{d\theta_{0}}{dv}
+  = Q \psi\theta_{0},
+\]
+ou, comme $\theta_{0}$~est intégrale de~\Eq{(5)}\Add{,}
+\[%[** TN: Set on two lines in original]
+\theta_{0} \frac{d\psi}{dv} = 0, \qquad\qquad
+\frac{d\psi}{dv} = 0.
+\]
+$\psi$~est une constante~$C$, et l'intégrale générale de~\Eq{(5)} est
+%% -----File: 121.png---Folio 113-------
+\[
+\Tag{(6)}
+\theta = C\, \theta_{0}.
+\]
+\emph{L'équation s'intègre complètement par des opérations algébriques.}
+Si nous cherchons l'expression de l'intégrale générale~$u$
+en fonction des intégrales particulières $u_{1}\Add{,} u_{2}\Add{,} u_{3}$, nous avons,
+en vertu de \Eq{(2)}\Add{,}~\Eq{(4)}\Add{,}~\Eq{(6)},
+\[
+u = u_{1} + \frac{1}{w}
+  = u_{1} + \frac{1}{\dfrac{1}{u_{2} - u_{1}} + \theta}
+  = u_{1} + \frac{1}{\dfrac{1}{u_{2} - u_{1}}
+            + C \dfrac{u_{2} - u_{3}}{(u_{3} - u_{1}) (u_{2} - u_{1})}}\Add{,}
+\]
+d'où
+\[
+\frac{1}{u - u_{1}}
+  = \frac{1}{u_{2} - u_{1}}
+  + \frac{C(u_{2} - u_{3})}{(u_{3} - u_{1})(u_{2} - u_{1})}
+  = \frac{u_{3} - u_{1} + C (u_{2} - u_{3})}{(u_{3} - u_{1}) (u_{2} - u_{1})},
+\]
+d'où
+\begin{gather*}
+C(u_{2} - u_{3})
+  = \frac{(u_{3} - u_{1}) (u_{2} - u_{1})}{u - u_{1}} - (u_{3} - u_{1})
+  = \frac{(u_{3} - u_{1}) (u_{2} - u)}{(u - u_{1})}, \\
+\Tag{(7)}
+C = \frac{u - u_{2}}{u - u_{1}} : \frac{u_{3} - u_{2}}{u_{3} - u_{1}},
+\end{gather*}
+ou
+\[
+\Ratio{u}{u_{1}}{u_{2}}{u_{3}} = C.
+\]
+\emph{Ainsi le rapport anharmonique de \Card{4} intégrales quelconques d'une
+équation de Riccati est constant.} En remarquant que, dans
+le cas présent, ces intégrales sont précisément les~$u$ des
+points d'intersection d'une génératrice quelconque avec les
+asymptotiques, on voit que \emph{\Card{4} lignes asymptotiques coupent les
+génératrices suivant un rapport anharmonique constant.}
+
+\Paragraph{Remarque.} L'équation~\Eq{(7)} résolue par rapport à~$u$ donne
+\[
+\Tag{(8)}
+u = \frac{VC + V_{0}}{V_{1}C + V_{2}}\Add{,}
+\]
+$V, V_{0}, V_{1}\Add{,} V_{2}$ étant fonctions de~$v$. La constante arbitraire figure
+donc dans l'intégrale générale par une fraction du \Ord{1}{e} degré.
+Inversement toute fonction de la forme~\Eq{(8)} satisfait à une
+équation de Riccati, car si on élimine la constante~$C$ au
+moyen d'une différentiation, on retrouve une équation
+%% -----File: 122.png---Folio 114-------
+différentielle de la forme~\Eq{(1)}.
+
+\MarginNote{Cas
+particuliers.}
+Si la surface réglée a une directrice rectiligne, cette
+directrice est une asymptotique, on a une solution particulière
+de l'équation de \DPtypo{Ricatti}{Riccati}. La détermination des lignes asymptotiques
+se fait au moyen de \Card{2} quadratures. C'est le cas
+des \emph{surfaces réglées à plan directeur}. Si la surface admet
+\Card{2} directrices rectilignes, ces \Card{2} droites sont des asymptotiques,
+et on a \Card{2} solutions particulières de l'équation de Riccati.
+C'est le cas des \emph{surfaces conoïdes à plan directeur}. Il ne
+faut plus alors qu'une quadrature pour déterminer les lignes
+asymptotiques. En réalité, on peut les obtenir sans quadrature\Add{.}
+Considérons en effet une surface réglée admettant \Card{2} directrices
+rectilignes. On peut effectuer une transformation homographique
+de façon que l'une des directrices s'en aille à l'infini,
+la surface se transforme en un conoïde à plan directeur.
+Soit
+\[
+z = \phi \left(\frac{y}{x}\right)
+\]
+l'équation d'un tel conoïde. Posons
+\[
+x = u, \qquad
+y = uv, \qquad
+z = \Err{(v) \phi}{\phi(v)};
+\]
+les lignes asymptotiques sont telles que le plan osculateur
+\DPtypo{coincide}{coïncide} avec le plan tangent; ses coefficients doivent donc
+satisfaire aux relations
+\[
+A \frac{\dd x}{\dd u} + B \frac{\dd y}{\dd u} + C \frac{\dd z}{\dd u} = 0,\qquad
+A \frac{\dd x}{\dd v} + B \frac{\dd y}{\dd v} + C \frac{\dd z}{\dd v} = 0,
+\]
+ou
+\[
+A + B v = 0, \qquad
+B u + C \phi'(v) = 0;
+\]
+équations satisfaites si l'on prend $C = - u$, $B = \phi'(v)$,
+$A = -v\phi'(v)$. On a alors
+%% -----File: 123.png---Folio 115-------
+\[
+\Psi(du, dv) = A\,d^{2} x + B\, d^{2} y + C\,d^{2} z = 0;
+\]
+mais $A\Add{,} B\Add{,} C$~étant les coefficients du plan tangent, on a
+\[
+A\, dx + B\, dy + C\, dz = 0;
+\]
+en différentiant cette relation, on voit qu'on peut mettre
+l'équation différentielle des lignes asymptotiques sous la
+forme
+\[
+dA·dx + dB·dy + dC·dz = 0,
+\]
+ou
+\begin{gather*}
+-du\bigl[\phi'(v)\,dv + v\phi''(v)\, dv\bigr]
+  + \phi''(v)\, dv (v\, du + u\, dv) - du · \phi'(v) · dv = 0, \\
+u\phi''(v) · dv^{2} - 2\phi'(v)\, du\, dv = 0;
+\end{gather*}
+nous trouvons la solution $v = \cte$ qui nous donne les génératrices,
+et il reste
+\[
+\frac{\phi''(v)\, dv}{\phi '(v)} = \frac{2\, du}{u},
+\]
+d'où
+\[
+L\phi'(v) = Lu^{2} - LC,
+\]
+d'où
+\[
+u^{2} = \DPtypo{C\phi'(v)}{C + \phi'(v)};
+\]
+\emph{on a ainsi les lignes asymptotiques d'un conoïde sans quadrature}.
+
+\Paragraph{Remarque.} S'il y a trois directrices rectilignes, la
+surface est une surface du second degré, et est doublement réglée.
+
+\MarginNote{Calcul de $\Psi$.}
+Cherchons l'expression générale de la forme~$\Psi$. Introduisons
+pour cela les variables \DPtypo{Canoniques}{canoniques}~$u\Add{,} v$ qui nous ont permis
+d'arriver à la forme de l'élément linéaire. Considérons le
+trièdre de Serret de la courbe~$(\Sigma)$ \DPtypo{trace}{tracé} du cône directeur sur
+la sphère de rayon~$1$. La génératrice~$(l\Add{,} m\Add{,} n)$ est dans le plan
+normal à cette courbe: soit $\theta$~son angle avec la normale principale;
+avec les notations habituelles, nous avons:
+%% -----File: 124.png---Folio 116-------
+\[%[** TN: Set on separate lines in original]
+l = a' \cos\theta + a'' \sin\theta, \qquad
+m = b' \cos\theta + b'' \sin\theta, \qquad
+n = c' \cos\theta + c'' \sin\theta;
+\]
+d'où
+\[
+l' = a
+  = \theta'(-a' \sin\theta + a'' \cos\theta)
+  - \cos\theta \left(\frac{a}{R} + \frac{a''}{T}\right)
+  + \sin\theta · \frac{a'}{T},
+\]
+et les analogues; %[** TN: Omitted newline]
+d'où
+\[
+\frac{\Cos\theta}{R} = - 1, \qquad
+\theta' = \frac{1}{T}.
+\]
+Nous avons alors
+\begin{alignat*}{4}
+mn' - nm' = &mc &&- nb &&= a' \sin \theta &&- a'' \cos \theta, \\
+&nl' &&- ln' &&= b' \sin \theta &&- b'' \cos \theta, \\
+&lm' &&- ml' &&= c' \sin \theta &&- c'' \cos \theta;
+\end{alignat*}
+et nous obtenons, au moyen des formules~\Eq{(3)} du \No9,
+\begin{alignat*}{2}%[** TN: Completed last two equations]
+f' + ul' &= (u - P)l' - K(mn'- nm')
+  &&= (u - P)a - K \sin \theta · a' + K \cos \theta · a'', \\
+g' + um' &= (u - P)m' - K(nl'- ln')
+  &&= (u - P)b - K \sin \theta · b' + K \cos \theta · b'', \\
+h' + un' &= (u - P)n' - K(lm'- ml')
+  &&= (u - P)c - K \sin \theta · c' + K \cos \theta · c'',
+\end{alignat*}
+puis
+\begin{multline*}
+f'' + ul''\DPtypo{)}{}
+  = - P'a + \frac{u - P}{R} a' - K' \sin\theta · a'
+  - \frac{K \cos \theta}{T} a'
+  + K \sin \theta \left(\frac{a}{R} + \frac{a''}{T}\right) \\
+  + K'\cos \theta · a'' - \frac{K \sin \theta}{T} a''
+  + K · \cos \theta · \frac{a'}{T},
+\end{multline*}
+ou
+\begin{alignat*}{4}%[** TN: Completed last two equations, fixed typos in original]
+f'' &+ ul'' &&= a \left(\frac{K \sin \theta}{R} - P'\right)
+  &&+ a' \left(\frac{u - P}{R} - K' \sin \theta\right)
+  &&+ a'' · K'\cos \theta, \\
+g'' &+ um'' &&= b \left(\frac{K \sin \theta}{R} - P'\right)
+  &&+ b' \left(\frac{u - P}{R} - K' \sin \theta\right)
+  &&+ b'' · K'\cos \theta, \\
+h'' &+ un'' &&= c \left(\frac{K \sin \theta}{R} - P'\right)
+  &&+ c' \left(\frac{u - P}{R} - K' \sin \theta\right)
+  &&+ c'' · K'\cos \theta,
+\end{alignat*}
+Alors
+\[
+\Psi = \begin{vmatrix}
+  \left[
+    \begin{aligned}
+      2a · du\, dv
+      &+ dv^{2} \biggl[a \left(\mfrac{K \sin \theta}{R} - P'\right) \\
+      &\quad+ a' \left(\mfrac{u - P}{R} - K' \sin \theta\right)
+       + a'' · K'\cos \theta\biggr]
+    \end{aligned}
+  \right] & \dots & \dots \\
+  \vphantom{\bigg|}
+  (u - P) a - K \sin \theta · a' + K \cos \theta · a'' & \dots & \dots \\
+  a' \cos \theta + a'' \sin \theta & \dots & \dots
+\end{vmatrix},
+\]
+%% -----File: 125.png---Folio 117-------
+ou
+\[
+\Psi = \begin{vmatrix}
+2\, du\, dv + \left(\mfrac{K \sin \theta}{R} - P'\right)\, dv^{2}
+  &  \left(\mfrac{u - P}{R} - K' \sin \theta\right)\, dv^{2} &  K' \cos \theta · dv^{2} \\
+u - P & - K \sin\theta & K \cos\theta \\
+0     &     \Cos\theta &   \sin\theta
+\end{vmatrix}\Add{,}
+\]
+c'est-à-dire
+\[
+\Psi = -\left[2\, du\, dv
+  + \left(\frac{K \sin \theta}{R} - P'\right) dv^{2}\right] K
+  - (u - P) \left[- K' + (u-P) \frac{\sin \theta}{R}\right] dv^{2},
+\]
+ou enfin
+\[
+\Psi = - 2K · du\, dv + \left\{(u - P) K' + KP'
+  - \frac{\sin \theta}{R} \left[(u - P)^{2} + K^{2}\right]\right\} dv^{2}.
+\]
+Le seul élément nouveau qui intervient est la courbure géodésique
+$\dfrac{\sin \theta}{R}$ de la courbe~$(\Sigma)$ sur la sphère. Cet élément suffit
+à déterminer~$(\Sigma)$; supposons en effet
+\[
+\frac{\sin \theta}{R} = \Phi (v);
+\]
+nous avons
+\[
+\frac{\Cos \theta}{R} = - 1, \qquad
+\frac{1}{T} = \theta';
+\]
+nous en déduisons
+\[
+\Tag{(1)}
+\tg \theta = - \Phi (v), \qquad
+R = - \Cos \theta, \qquad
+T = \frac{dv}{d \theta};
+\]
+nous avons ainsi tous les éléments de la courbe~$(\Sigma)$.
+
+\Paragraph{Remarque.} Les formules~(\DPtypo{'}{1}) nous permettent de trouver
+la condition pour qu'une courbe soit tracée sur la sphère de
+rayon~$1$. Nous avons en effet
+\[
+\frac{dR}{dv} = +\sin\theta · \frac{d\theta}{dv} = \frac{\sin\theta}{R},
+\]
+d'où
+\[
+R^{2} + T^{2} \left(\frac{dR}{dv}\right)^{2} = 1;
+\]
+%% -----File: 126.png---Folio 118-------
+Ce qui exprime que le rayon de la sphère osculatrice est égal
+à~$1$.
+
+\Section{Lignes de courbure.}
+{11.}{} L'équation différentielle des lignes de courbure est
+\[
+\begin{vmatrix}
+\mfrac{\dd ds^{2}}{\dd\, du} & \mfrac{\dd ds^{2}}{\dd\, dv} \\
+\mfrac{\dd \Psi}{\dd\, du}   & \mfrac{\dd \Psi}{\dd\, dv}
+\end{vmatrix} = 0,
+\]
+ou
+\[
+\begin{vmatrix}
+du     & \bigl[(u - P)^{2} + K^{2}\bigr] dv \\
+- K\, dv &
+- K\, du + \left[(u - P) K' + KP'
+    - \mfrac{\sin\theta}{R} \bigl[(u - P)^{2} + K^{2}\bigr]\DPtypo{}{\right]}dv
+\end{vmatrix} = 0,
+\]
+c'est-à-dire
+\begin{multline*}
+- K du^{2} + \Bigl[(u - P)K' + KP'
+  - \Phi \bigl[(u - P)^{2} + K^{2}\bigr]\Bigr]du\,dv \\ % +
+  + K \left[(u - P)^{2} + K^{2}\right]dv^{2} = 0.
+\end{multline*}
+Telle est l'équation différentielle des lignes de courbure,
+où $\Phi$~représente la courbure géodésique de la courbe~$(\Sigma)$.
+
+\Section{Centre de courbure géodésique.}
+{12.}{} Considérons une trajectoire orthogonale des génératrices,
+par exemple $u = 0$,
+\[
+x = f(v), \qquad
+y = g(v), \qquad
+z = h(v);
+\]
+cherchons son centre de courbure géodésique. C'est le point où
+la droite polaire rencontre le plan tangent. Or\Add{,} la génératrice
+étant normale à sa trajectoire orthogonale\DPtypo{,}{} est l'intersection
+du plan normal et du plan tangent; \emph{le centre de courbure
+géodésique est donc à l'intersection de la droite polaire avec
+la génératrice}. Le plan normal est
+\[
+\sum (x - f) f' = 0;
+\]
+la caractéristique est définie par l'équation précédente et
+%% -----File: 127.png---Folio 119-------
+par
+\[
+\sum (x - f) f'' - \sum f'{}^{2} = 0.
+\]
+Pour déterminer le centre de courbure géodésique, il suffit de
+déterminer l'$u$~du point d'intersection de la droite précédente
+avec la génératrice
+\[
+x = f(v) + ul(v), \qquad
+y = g(v) + um(v), \qquad
+z = h(v) + un(v).
+\]
+La \Ord{1}{ère} équation se réduit à une identité, la \Ord{2}{e} donne
+\[
+u \sum lf'' - \sum f'{}^{2} = 0;
+\]
+mais on a
+\[
+\sum lf' = 0,
+\]
+d'où
+\[
+\sum l'f' + \sum lf'' = 0;
+\]
+et l'équation qui donne l'$u$~du point cherché devient
+\[
+u \sum l'f' + \sum f'{}^{2} = 0,
+\]
+ou
+\[
+- u P + P^{2} + K^{2} = 0;
+\]
+ou enfin:
+\[
+P (u - P) = K^{2}.
+\]
+\DPtypo{si}{Si} $C$~est le point central, $M$~le point considéré sur la trajectoire
+orthogonale, $M'$~le centre de courbure géodésique, l'équation
+précédente donne
+\[
+CM \cdot CM' = - K^{2}.
+\]
+Donc les plans tangents en $M\Add{,} M'$ sont rectangulaires. Ainsi le
+\emph{centre de courbure géodésique en un point~$M$ d'une trajectoire
+orthogonale des génératrices d'une surface réglée est le point
+de la génératrice où le plan tangent est perpendiculaire au
+plan tangent en~$M$}.
+
+Si nous considérons maintenant une courbe~$(\DPtypo{C}{c})$ tracée sur
+une surface quelconque~$(S)$ les normales à~$(c)$ tangentes à~$(\DPtypo{s}{S})$
+engendrent une surface réglée~$(\Sigma)$; les surfaces $(\DPtypo{s}{S})\Add{,} (\Sigma)$ étant
+tangentes tout le long de~$(c)$, la courbe~$(c)$ a même centre de
+%% -----File: 128.png---Folio 120-------
+courbure géodésique sur~$(\DPtypo{s}{S})$ et sur~$(\Sigma)$; ce qui permet de construire
+le centre de courbure géodésique d'une courbe tracée
+sur une surface quelconque.
+
+
+\ExSection{V}
+
+\begin{Exercises}
+\item[21.] Trouver les points de contact des plans isotropes \DPtypo{menes}{menés} par
+une génératrice quelconque d'une surface réglée. Quelles relations
+ont-ils avec le point central et le paramètre de
+distribution?
+
+\item[22.] Trouver les surfaces réglées dont les lignes asymptotiques
+interceptent sur les génératrices des segments égaux.
+
+\item[23.] Trouver les surfaces réglées dont les lignes de courbure
+interceptent sur les génératrices des segments égaux.
+
+\item[24.] Trouver les surfaces réglées dont les rayons de courbure
+principaux sont fonctions l'un de l'autre.
+
+\item[25.] Trouver les lignes de courbure et les lignes géodésiques de
+l'\DPchg{hélicoide}{hélicoïde} développable.
+
+\item[26.] Montrer que les lignes d'une surface~$(S)$ quelconque, pour
+lesquelles: $ds - R_{g}\, d\phi = 0$, sont caractérisées par cette
+propriété que, si l'on mène par chacun des points de l'une
+% [** TN: Regularized, "constant" in original]
+d'elles une tangente à la courbe $v = \const.$, la surface réglée
+ainsi obtenue a pour ligne de striction la courbe considérée
+(voir \hyperref[exercice11]{exercice~11}).
+
+\item[27.] \DPchg{Etant}{Étant} donnée une surface~$S$ et une courbe~$C$ de cette surface,
+on considère la surface \DPtypo{reglée}{réglée}~$G$ engendrée par les normales~$MN$
+menées à~$S$ aux divers points $M$~de~$C$. Le point central de~$MN$
+s'appelle le \emph{métacentre} de~$S$, correspondant au point~$M$ et à
+la tangente~$MT$ de~$C$.
+
+%[** TN: Regularized formatting of parts]
+\Primo. Déterminer ce métacentre, le plan
+asymptote, le paramètre de distribution. Discuter la variation
+du métacentre quand la courbe~$(C)$ varie, en passant toujours
+en~$M$.
+
+\Secundo. Montrer que le métacentre est le centre de
+courbure de la section droite du cylindre circonscrit à~$S$, et
+dont les génératrices sont perpendiculaires au plan asymptote
+de~$G$.
+
+\Tertio. On suppose qu'on ait plusieurs surfaces~$S$, et que
+l'on affecte chacune d'elles d'un coefficient \DPtypo{numerique}{numérique}~$a$.
+On considère comme homologues sur ces diverses surfaces les
+points~$M$ (pris un sur chaque surface) pour lesquels les plans
+tangents à ces diverses surfaces sont parallèles; soit~$M_{0}$ le
+centre des moyennes distances d'un tel système de points~$M$
+homologues, et relatif au système des coefficients~$a$. Soit~$S_{0}$
+la surface lieu des points~$M_{0}$. Montrer qu'elle correspond
+à chacune des surfaces~$S$ par plans tangents parallèles; et que
+si~$I_{0}$ est le métacentre de~$S_{0}$ correspondant aux divers métacentres~$I$
+des surfaces~$S$ qui se trouvent associés dans la
+correspondance \DPtypo{considerée}{considérée}, on a $(\sum a) · M_{0} I_{0} = \sum (a · MI)$.
+
+\item[28.] On donne une courbe gauche~$R$, arête de rebroussement d'une
+développable~$\Delta$. Déterminer toutes les surfaces réglées satisfaisant
+aux conditions suivantes: chacune des génératrices~$G$
+d'une telle surface est perpendiculaire à un plan tangent~$P$ de~$\Delta$,
+et le point de rencontre de~$G$ et de~$P$ est le point central
+de~$G$. Soit alors~$\Sigma$ l'une de ces surfaces réglées, chacun des
+plans isotropes passant par une de ses génératrices enveloppe
+une développable. Montrer que le lieu des milieux des segments
+dont les extrémités décrivent, indépendamment l'un de
+l'autre, les arêtes de rebroussement de ces deux développables
+est une surface minima inscrite dans~$\Delta$.
+\end{Exercises}
+%% -----File: 129.png---Folio 121-------
+
+
+\Chapitre{VI}{Congruences de Droites.}
+
+\Section{Points et plans focaux.}
+{1.}{} On appelle \emph{congruence} un ensemble de droites dépendant
+de \Card{2} paramètres; toutes les droites rencontrant \Card{2} droites
+fixes constituent une congruence; de même les droites passant
+par un point fixe, les normales à une surface; si sur une
+surface on considère une famille de courbes dépendant d'un paramètre,
+l'ensemble de leurs tangentes constitue une congruence.
+
+Une droite d'une congruence pourra se représenter par les
+équations
+\[%[** TN: Set on one line in original; added brace]
+\Tag{(1)}
+\left\{
+\begin{alignedat}{2}
+x &= f(v, w) &&+ u · a(v, w), \\
+y &= g(v, w) &&+ u · b(v, w), \\
+z &= h(v, w) &&+ u · c(v, w).
+\end{alignedat}
+\right.
+\]
+Les équations
+\[
+\Tag{(2)}
+x = f(v, w), \qquad
+y = g(v, w), \qquad
+z = h(v, w)
+\]
+définissent le \emph{support} de la congruence, $a, b, c$ définissent les
+directions des droites de la congruence passant par chaque
+point du support. Ce support sera en général une surface, et
+la congruence sera constituée par les droites de directions
+données passant par tous les points d'une surface. Il peut
+arriver que $f, g, h$ ne dépendent que d'un seul paramètre, le
+support est alors une courbe, et par tout point de la courbe
+passent une infinité de droites de la congruence, qui constituent
+un cône. Enfin $f, g, h$ peuvent se réduire à des constantes,
+et la congruence est constituée par toutes les droites
+%% -----File: 130.png---Folio 122-------
+passant par le point fixe de coordonnées~$f, g, h$.
+
+Supposons qu'on établisse une relation entre~$v, w$; cela
+revient à choisir $\infty^{1}$~droites de la congruence, qui constituent
+une surface réglée de la congruence. On retrouverait ainsi les
+équations générales d'une surface réglée. Considérons toutes
+les surfaces réglées passant par une droite~$D$ de la congruence\Add{.}
+Deux de ces surfaces se raccordent en \Card{2} points de la droite~$D$.
+Nous allons montrer que ces \Card{2} points sont indépendants des
+surfaces réglées que l'on considère. En d'autres termes \emph{sur
+chaque droite~$D$ de la congruence il existe \Card{2} points $F, F'$ auxquels
+correspondent \Card{2} plans $P, P'$ passant par la droite~$D$ et
+tels que toutes les surfaces réglées de la congruence passant
+par la droite~$D$ ont pour plans tangents en~$F, F'$ respectivement
+les plans~$P, P'$}. Ces points~$F, F'$ s'appellent \emph{foyers} ou \emph{points focaux}
+de la droite~$D$, les plans~$P, P'$ sont les \emph{plans focaux} associés
+à~$F, F'$. Pour démontrer la proposition, cherchons le plan
+tangent en un point quelconque de la génératrice~\Eq{(1)}. Les paramètres
+$A\Add{,} B\Add{,} C$ du plan tangent satisfont aux équations
+\begin{gather*}
+\Tag{(3)}
+Aa + Bb + Cc = 0, \\
+\Tag{(3')}
+A(df + u\,da) + B(dg + u\,db) + C(dh + u\,dc) = 0.
+\end{gather*}
+On peut choisir~$u$ de façon que le plan tangent soit indépendant
+des différentielles $dv, dw$, et par suite indépendant de la relation
+existant entre~$v\Add{,} w$, c'est-à-dire indépendant de la surface
+réglée. Développons la \Ord{2}{e} équation~\Eq{(3)}
+\begin{alignat*}{5}
+0 = &\biggl[
+  &&  A\left(\frac{\dd f}{\dd v} + u \frac{\dd a}{\dd v}\right)
+  &&+ B\left(\frac{\dd g}{\dd v} + u \frac{\dd b}{\dd v}\right)
+  &&+ C\left(\frac{\dd h}{\dd v} + u \frac{\dd c}{\dd v}\right)
+  &&\biggr] dv \\
+  + &\biggl[
+  && A\left(\frac{\dd f}{\dd w} + u \frac{\dd a}{\dd w}\right)
+  &&+ B\left(\frac{\dd g}{\dd w} + u \frac{\dd b}{\dd w}\right)
+  &&+ C\left(\frac{\dd h}{\dd w} + u \frac{\dd c}{\dd w}\right)
+  &&\biggr] dw.
+\end{alignat*}
+%% -----File: 131.png---Folio 123-------
+Pour que le plan tangent soit indépendant de $dv, dw$, il suffit
+que l'on ait
+\[
+\Tag{(4)}
+\left\{
+\begin{alignedat}{4}
+  &   A\left(\frac{\dd f}{\dd v} + u\frac{\dd a}{\dd v}\right)
+  &&+ B\left(\frac{\dd g}{\dd v} + u\frac{\dd b}{\dd v}\right)
+  &&+ C\left(\frac{\dd h}{\dd v} + u\frac{\dd c}{\dd v}\right)
+  &&= 0 \\
+  &   A\left(\frac{\dd f}{\dd w} + u\frac{\dd a}{\dd w}\right)
+  &&+ B\left(\frac{\dd g}{\dd w} + u\frac{\dd b}{\dd w}\right)
+  &&+ C\left(\frac{\dd h}{\dd w} + u\frac{\dd c}{\dd w}\right)
+  &&= 0
+\end{alignedat}
+\right.
+\]
+Les relations~\Eq{(4)} et la \Ord[f]{1}{e} des relations~\Eq{(3)} doivent être
+satisfaites pour des valeurs non toutes nulles de~$A\Add{,} B\Add{,} C$, donc
+on doit avoir
+\[
+\Tag{(5)}
+\begin{vmatrix}
+a    &   b   &    c  \\
+\mfrac{\dd f}{\dd v} + u\mfrac{\dd a}{\dd v} &
+\mfrac{\dd g}{\dd v} + u\mfrac{\dd b}{\dd v} &
+\mfrac{\dd h}{\dd v} + u\mfrac{\dd c}{\dd v} \\
+\mfrac{\dd f}{\dd w} + u\mfrac{\dd a}{\dd w} &
+\mfrac{\dd g}{\dd w} + u\mfrac{\dd b}{\dd w} &
+\mfrac{\dd h}{\dd w} + u\mfrac{\dd c}{\dd w}
+\end{vmatrix} = 0.
+\]
+Telle est l'équation qui donne les~$u$ des points focaux; elle
+est du \Ord{2}{e} degré, donc il y a \Card{2} points focaux; le plan focal
+correspondant à chacun d'eux aura pour coefficients les valeurs
+de~$A, B, C$ satisfaisant aux équations \Eq{(3)}~et~\Eq{(4)}.
+
+\MarginNote{Surfaces focales.
+Courbes focales.}
+Le lieu des foyers s'obtiendra sans difficulté. Il
+suffit de tirer~$u$ de~\Eq{(5)} et de porter sa valeur dans~\Eq{(1)}.
+L'équation~\Eq{(5)} étant du \Ord{2}{e} degré donne pour~$u$ \Card{2} valeurs, de
+sorte que le lieu se compose de \Card{2} parties distinctes dans le
+voisinage de la droite~$D$. Considérons l'une de ces parties;
+elle peut être une surface, que l'on appellera \emph{surface focale},
+ou une courbe, que l'on appellera \emph{courbe focale}, ou bien elle
+peut se réduire à un point, et la congruence comprend alors
+toutes les droites passant par ce point. En écartant ce cas,
+on voit que le lieu des foyers peut se composer de \Card{2} surfaces,
+d'une courbe et d'une surface, ou de deux courbes.
+
+\ParItem{\Primo.} Supposons qu'une portion du lieu des foyers soit une
+%% -----File: 132.png---Folio 124-------
+surface~$(\Phi)$. Prenons cette surface comme support de la congruence;
+l'équation~\Eq{(5)} a pour racine $u = 0$, on a donc
+\[
+\begin{vmatrix}
+a                    & b                    & c  \\
+\mfrac{\dd f}{\dd v} & \mfrac{\dd g}{\dd v} & \mfrac{\dd h}{\dd v}  \\
+\mfrac{\dd f}{\dd w} & \mfrac{\dd g}{\dd w} & \mfrac{\dd h}{\dd w}
+\end{vmatrix} = 0.
+\]
+Ceci exprime que la droite~$D$ est dans le plan tangent à la
+surface $(\Phi)$ au point~$M$ ($u = 0$), qui est l'un des foyers, soit~$F$.
+\emph{Ainsi les droites de la congruence sont tangentes à la surface
+focale au foyer correspondant.} Cherchons le plan focal
+correspondant à~$F$. Ses coefficients $A\Add{,} B\Add{,} C$ sont déterminés par
+les équations
+\[
+\left\{
+\begin{aligned}
+Aa + Bb + Cc = 0\Add{,}& \\
+\begin{alignedat}{4}
+&A \mfrac{\dd f}{\dd v} &&+ B \mfrac{\dd g}{\dd v}
+  &&+ C \mfrac{\dd h}{\dd v} &&= 0\Add{,} \\
+&A \mfrac{\dd f}{\dd w} &&+ B \mfrac{\dd g}{\dd w}
+  &&+ C \mfrac{\dd h}{\dd w} &&= 0\Add{,}
+\end{alignedat}&
+\end{aligned}
+\right.
+\]
+d'après la condition précédemment écrite, ces équations se réduisent
+à~$2$, et expriment que \emph{le plan focal correspondant au
+foyer~$F$ est le plan tangent en~$F$ à la surface~$(\Phi)$. Toutes les
+surfaces réglées de la congruence sont circonscrites à la surface
+focale}.
+
+\emph{Si le lieu des foyers $F, F'$ comprend deux surfaces focales~$(\Phi)\Add{,} (\Phi')$,
+les droites de la congruence sont tangentes aux
+\Card{2} surfaces focales, les foyers~$F, F'$ sont les points de contact,
+les plans focaux sont les plans tangents aux surfaces focales
+aux foyers correspondants. Le lieu des foyers coïncide avec
+l'enveloppe des plans focaux.}
+%% -----File: 133.png---Folio 125-------
+
+\emph{Réciproquement}, étant données \Card{2} surfaces quelconques $(\Phi)\Add{,}
+(\Phi')$, leurs tangentes communes dépendent de \Card{2} paramètres. Soit~$F$
+un point de~$(\Phi)$. Considérons le plan tangent en~$F$ à~$(\Phi)$; il
+coupe~$(\Phi')$ suivant une certaine courbe; si nous menons de~$F$ des
+tangentes à cette courbe, ces droites, qui sont tangentes aux
+\Card{2} surfaces $(\Phi)\Add{,} (\Phi')$ sont déterminées quand le point~$F$ est déterminé;
+elles dépendent d'autant de paramètres que le point~$F$,
+donc de \Card{2} paramètres; elles constituent une congruence,
+dont les surfaces réglées sont circonscrites aux surfaces $(\Phi)\Add{,}
+(\Phi')$ qui sont les surfaces focales.
+
+Si les surfaces $(\Phi)\Add{,} (\Phi')$ constituent \Card{2} nappes d'une même
+surface~$(S)$, comme cela arrive en général, la congruence sera
+constituée par les tangentes doubles de la surface~$(S)$.
+
+\ParItem{\Secundo.} Supposons qu'une portion du lieu des foyers soit une
+courbe~$(\phi)$, que nous prendrons pour support de la congruence.
+Nous pouvons alors supposer que $f\Add{,} g\Add{,} h$ ne dépendent que d'un paramètre,
+$v$~par exemple; alors $\dfrac{\dd f}{\dd w}, \dfrac{\dd g}{\dd w}, \dfrac{\dd h}{\dd w}$ sont nuls, et $u = 0$ est
+racine de l'équation~\Eq{(5)}. \emph{Si les droites d'une congruence rencontrent
+une courbe fixe, les points de cette courbe sont des
+foyers pour les droites de la congruence qui y passent.} Cherchons
+le plan focal correspondant. Nous avons
+\[
+\left\{
+\begin{aligned}
+Aa + Bb + Cc &= 0\Add{,} \\
+A \frac{\dd f}{\dd v} + B \frac{\dd g}{\dd v} + C \frac{\dd h}{\dd v} &= 0\Add{,}
+\end{aligned}
+\right.
+\]
+\emph{le plan focal passe par la droite~$D$ et est tangent à la courbe
+focale. Toutes les surfaces réglées de la congruence passent
+par la courbe focale, et en un point~$M$ de cette courbe sont
+tangentes au plan tangent à cette courbe passant par la droite~$D$.}
+%% -----File: 134.png---Folio 126-------
+
+Supposons qu'il y ait une surface focale~$(\Phi)$ et une
+courbe focale~$(\phi')$; \emph{la congruence est constituée par les droites
+rencontrant~$(\phi')$ et tangentes à~$(\Phi)$}. On a immédiatement les
+foyers et les plans focaux, d'après ce qui précède. \emph{Réciproquement,
+les droites rencontrant une courbe~$(\phi')$ et tangentes
+à une surface~$(\Phi)$ constituent une congruence qui admet $(\phi')$~et~$(\Phi)$
+pour lieu de ses foyers.}
+
+Supposons qu'il y ait \Card{2} courbes focales $(\phi)\Add{,} (\phi')$. \emph{La
+congruence est constituée par les droites rencontrant $(\phi)\Add{,} (\phi')$,
+et ses surfaces réglées contiennent les \Card{2} courbes focales. Réciproquement
+les droites rencontrant \Card{2} courbes données constituent
+une congruence qui admet ces \Card{2} courbes comme courbes focales.}
+Si $(\phi)\Add{,} (\phi')$ constituent \Card{2} parties d'une même courbe~$(c)$,
+la congruence est constituée par les droites rencontrant~$(c)$
+en \Card{2} points, c'est-à-dire les cordes de~$(c)$.
+
+\MarginNote{Cas singuliers.}
+Voyons dans quels cas les \Card{2} foyers sont confondus sur
+toutes les droites de la congruence.
+
+Examinons d'abord le cas de \Card{2} surfaces focales confondues\Add{.}
+Prenons cette surface~$(\Phi)$ comme support; en chaque point~$F$ de
+cette surface est tangente une droite~$D$ de la congruence. Si
+on considère ces points focaux et les droites correspondantes,
+on peut trouver sur la surface une famille de courbes tangentes
+en chacun de leurs points à la droite correspondante de la
+congruence. Soit la droite~$D$, elle est tangente à la surface,
+donc ses coefficients directeurs sont:
+\[
+a = \lambda \frac{\dd f}{\dd v} + \mu \frac{\dd f}{\dd w}, \qquad
+b = \lambda \frac{\dd g}{\dd v} + \mu \frac{\dd g}{\dd w}, \qquad
+c = \lambda \frac{\dd h}{\dd v} + \mu \frac{\dd h}{\dd w}\Add{.}
+\]
+%% -----File: 135.png---Folio 127-------
+Soit une courbe de la surface~$(\Phi)$ définie en exprimant~$v\Add{,} w$ en
+fonction d'un paramètre; les coefficients directeurs de la
+tangente sont
+\[
+dx = \frac{\dd f}{\dd v} · dv + \frac{\dd f}{\dd w} · dw, \quad
+dy = \frac{\dd g}{\dd v} · dv + \frac{\dd g}{\dd w} · dw, \quad
+dz = \frac{\dd h}{\dd v} · dv + \frac{\dd h}{\dd w} · dw;
+\]
+pour que cette tangente soit la droite~$D$, il faut que l'on
+ait
+\[
+\frac{dv}{\lambda} = \frac{dw}{\mu}.
+\]
+Pour déterminer l'un des paramètres~$v\Add{,} w$ en fonction de l'autre,
+on a à intégrer une équation différentielle du \Ord{1}{er} ordre. La
+famille de courbes dépend d'un paramètre, soit $w = \cte[]$. On
+aura alors
+\[
+a = \frac{\dd f}{\dd v}, \qquad
+b = \frac{\dd g}{\dd v}, \qquad
+c = \frac{\dd h}{\dd v};
+\]
+et \Eq{(5)}~devient
+\[
+\begin{vmatrix}
+\mfrac{\dd f}{\dd v} &
+\mfrac{\dd g}{\dd v} &
+\mfrac{\dd h}{\dd v} \\
+\mfrac{\dd f}{\dd v} + u \mfrac{\dd^{2} f}{\dd v^{2}} &
+\mfrac{\dd g}{\dd v} + u \mfrac{\dd^{2} g}{\dd v^{2}} &
+\mfrac{\dd h}{\dd v} + u \mfrac{\dd^{2} h}{\dd v^{2}} \\
+\mfrac{\dd f}{\dd w} + u \mfrac{\dd^{2} f}{\dd v \dd w} &
+\mfrac{\dd g}{\dd w} + u \mfrac{\dd^{2} g}{\dd v \dd w} &
+\mfrac{\dd h}{\dd w} + u \mfrac{\dd^{2} h}{\dd v \dd w}
+\end{vmatrix} = 0.
+\]
+\DPtypo{en}{En} retranchant la \Ord{1}{ère} ligne de la \Ord{2}{e}, $u$~vient en facteur:
+pour que les points focaux soient confondus, il faut que le
+déterminant s'annule encore pour $u = 0$, ce qui donne
+\[
+\begin{vmatrix}
+\mfrac{\dd f}{\dd v} &
+\mfrac{\dd g}{\dd v} &
+\mfrac{\dd h}{\dd v} \\
+\mfrac{\dd^{2} f}{\dd v^{2}} &
+\mfrac{\dd^{2} g}{\dd v^{2}} &
+\mfrac{\dd^{2} h}{\dd v^{2}} \\
+\mfrac{\dd f}{\dd w} &
+\mfrac{\dd g}{\dd w} &
+\mfrac{\dd h}{\dd w}
+\end{vmatrix} = 0\Add{,}
+\]
+ou $E' = 0$. Alors l'équation des lignes asymptotiques de la surface~$(\Phi)$,
+qui est
+\[
+E'\,dv^{2} + 2F'\,dv · dw + G'\,dw^{2} = 0\Add{,}
+\]
+%% -----File: 136.png---Folio 128-------
+est satisfaite pour $dw = 0$; les courbes $w = \cte$ sont des asymptotiques
+de la surface~$(\Phi)$. Ainsi \emph{les congruences à surface
+focale double peuvent s'obtenir en prenant les tangentes
+aux lignes asymptotiques d'une surface quelconque}.
+
+Considérons maintenant le cas de \Card{2} courbes focales confondues.
+Prenons cette courbe pour support. Nous pouvons supposer
+que $f\Add{,} g\Add{,} h$ ne dépendent plus de~$w$. Exprimons alors que l'équation~\Eq{(5)}
+admet pour racine double $u = 0$, nous avons
+\[
+\begin{vmatrix}
+a                    & b                    & c \\
+\mfrac{\dd f}{\dd v} & \mfrac{\dd g}{\dd v} & \mfrac{\dd h}{\dd v} \\
+\mfrac{\dd a}{\dd w} & \mfrac{\dd b}{\dd w} & \mfrac{\dd c}{\dd w}
+\end{vmatrix} = 0\Add{.}
+\]
+Les droites~$D$ de la congruence passant par un point~$F$ de la
+courbe~$(\phi)$ engendrent un cône. Le plan tangent à ce cône a
+pour coefficients les déterminants déduits du tableau
+\[
+\begin{Vmatrix}
+a                    & b                    & c \\
+\mfrac{\dd a}{\dd w} & \mfrac{\dd b}{\dd w} & \mfrac{\dd c}{\dd w}
+\end{Vmatrix}\Add{,}
+\]
+et la condition précédente exprime que la tangente~$FT$ à la
+courbe focale est dans le plan tangent au cône; ceci devant
+avoir lieu quelle que soit la génératrice du cône que l'on
+considère, tous les plans tangents au cône passent par~$FT$,
+et le cône se réduit à un plan. \emph{Une congruence à courbe focale
+double est engendrée par les droites qui en chaque point~$F$
+d'une courbe sont situées dans un plan passant par la tangente.}
+Ici l'enveloppe des plans focaux ne coïncide plus avec le
+lieu des points focaux.
+%% -----File: 137.png---Folio 129-------
+
+\Section{Développables de la congruence.}
+{2.}{} Cherchons si l'on peut associer les droites de la
+congruence de façon à obtenir des surfaces développables.
+Reprenons les équations de la droite
+\begin{alignat*}{2}%[** TN: Set on one line in original]
+x &= f(v,w) &&+ u · a(v,w), \\
+y &= g(v,w) &&+ u · b(v,w), \\
+z &= h(v,w) &&+ u · c(v,w);
+\end{alignat*}
+la condition pour que cette droite engendre une surface développable
+est
+\[
+\begin{vmatrix}
+a   &  b   &  c   \\
+da  &  db  &  dc  \\
+df  &  dg  &  dh
+\end{vmatrix} = 0,
+\]
+ou
+\[
+\Tag{(1)}
+\begin{vmatrix}
+a    &    b     &       c  \\
+\mfrac{\dd a}{\dd v} dv + \mfrac{\dd a}{\dd w} dw &
+\mfrac{\dd b}{\dd v} dv + \mfrac{\dd b}{\dd w} dw &
+\mfrac{\dd c}{\dd v} dv + \mfrac{\dd c}{\dd w} dw  \\
+%
+\mfrac{\dd f}{\dd v} dv + \mfrac{\dd f}{\dd w} dw &
+\mfrac{\dd g}{\dd v} dv + \mfrac{\dd g}{\dd w} dw &
+\mfrac{\dd h}{\dd v} dv + \mfrac{\dd h}{\dd w} dw
+\end{vmatrix} = 0.
+\]
+Telle est l'équation différentielle qui exprime que la droite
+de la congruence engendre une surface développable. Elle est
+de la forme
+\[
+A\, dv^{2} + 2 B\, dv · dw + C\, dw^{2} = 0;
+\]
+elle donne \Card{2} valeurs de~$\dfrac{dv}{dw}$, il y a donc \Card{2} familles de développables,
+qu'on appelle \emph{développables de la congruence}. \emph{Par chaque
+droite de la congruence passent \Card{2} développables de la congruence.}
+Cherchons les points de contact de cette droite avec
+les arêtes de rebroussement. Les coordonnées de l'un de ces
+points vérifient les équations
+\[
+\left\{
+\begin{alignedat}{3}
+df &+ u\, da &&+ a\, d\sigma &&= 0, \\
+dg &+ u\, db &&+ b\, d\sigma &&= 0, \\
+dh &+ u\, dc &&+ c\, d\sigma &&= 0;
+\end{alignedat}
+\right.
+\]
+%% -----File: 138.png---Folio 130-------
+ou
+\[
+\left\{
+\begin{aligned}
+  \left(\frac{\dd f}{\dd v} + u \frac{\dd a}{\dd v}\right) dv
++ \left(\frac{\dd f}{\dd w} + u \frac{\dd a}{\dd w}\right) dw
++ a\, d\sigma &= 0, \\
+  \left(\frac{\dd g}{\dd v} + u \frac{\dd b}{\dd v}\right) dv
++ \left(\frac{\dd g}{\dd w} + u \frac{\dd b}{\dd w}\right) dw
++ b\, d\sigma &= 0, \\
+  \left(\frac{\dd h}{\dd v} + u \frac{\dd c}{\dd v}\right) dv
++ \left(\frac{\dd h}{\dd w} + u \frac{\dd c}{\dd w}\right) dw
++ c\, d \sigma &= 0.
+\end{aligned}
+\right.
+\]
+\DPchg{Eliminons}{Éliminons} entre ces équations $dv, dw, d\sigma$, nous avons pour déterminer
+l'$u$~du point de contact de la droite avec l'arête de rebroussement,
+l'équation qui donne les points focaux. Donc \emph{les
+points où une droite~$D$ de la congruence touche les arêtes de
+rebroussement des deux développables de la congruence qui passent
+par cette droite sont les foyers de la droite~$D$}.
+
+\MarginNote{Développables
+et surface
+focale.}
+Supposons que le lieu des points focaux comprenne une surface~$(\Phi)$
+que nous prendrons pour support
+\[
+x = f(v,w), \qquad
+y = g(v,w), \qquad
+z = h(v,w).
+\]
+En chaque point~$F$ de la surface~$(\Phi)$ passe une droite~$D$ de la
+congruence tangente en~$F$ à~$(\Phi)$ et admettant~$F$ pour foyer. Nous
+avons trouvé sur la surface~$(\Phi)$ une famille de courbes tangentes
+aux droites~$D$. La développable qui a pour arête de rebroussement
+une de ces courbes~$(A)$ est une développable de la congruence.
+
+\Illustration[2.25in]{138a}
+Nous avons ainsi une des familles de développables.
+Considérons alors les courbes~$(c)$
+formant avec~$(A)$ un réseau conjugué.
+Considérons la développable enveloppe des
+plans tangents à~$(\Phi)$ tout le long d'une
+courbe~$(c)$; la génératrice de cette développable
+en un point~$F$ de~$(c)$ est la caractéristique
+du plan tangent, c'est la tangente
+%% -----File: 139.png---Folio 131-------
+conjuguée de la tangente à~$(c)$, c'est la droite~$D$. Nous
+avons la \Ord{2}{e} famille de développables en prenant l'enveloppe
+des plans tangents à~$(\Phi)$ en tous les points des courbes~$(c)$
+conjuguées des courbes~$(A)$.
+
+Supposons que les courbes $w = \cte$ soient précisément les
+courbes~$(A)$. On a
+\[
+a = \frac{\dd f}{\dd v}, \qquad
+b = \frac{\dd g}{\dd v}, \qquad
+c = \frac{\dd h}{\dd v};
+\]
+l'équation~\Eq{(1)} devient
+\[
+\begin{vmatrix}%[** TN: Added elided columns]
+\mfrac{\dd f}{\dd v} &
+\mfrac{\dd g}{\dd v} &
+\mfrac{\dd h}{\dd v} \\
+%
+\mfrac{\dd^{2} f}{\dd v^{2}}·dv + \mfrac{\dd^{2} f}{\dd v · \dd w}\, dw &
+\mfrac{\dd^{2} g}{\dd v^{2}}·dv + \mfrac{\dd^{2} g}{\dd v · \dd w}\, dw &
+\mfrac{\dd^{2} h}{\dd v^{2}}·dv + \mfrac{\dd^{2} h}{\dd v · \dd w}\, dw \\
+%
+\mfrac{\dd f}{\dd v} · dv       + \mfrac{\dd f}{\dd w}\, dw &
+\mfrac{\dd g}{\dd v} · dv       + \mfrac{\dd g}{\dd w}\, dw &
+\mfrac{\dd h}{\dd v} · dv       + \mfrac{\dd h}{\dd w}\, dw
+\end{vmatrix} = 0.
+\]
+Retranchons la \Ord{1}{ère} ligne de la \Ord{3}{e}: $dw$~vient en facteur, et
+l'équation prend la forme
+\[
+dw (E'\, dv + F'\, dw) = 0;
+\]
+nous trouvons d'abord $dw = 0$, (courbes~$A$); la relation
+\[
+E'\, dv + F'\, dw = 0
+\]
+définit précisément les courbes~$(c)$ conjuguées des courbes
+$w = \cte[]$. Nous retrouvons les résultats précédents.
+
+\MarginNote{Développables
+et courbe
+focale.}
+Examinons maintenant le cas d'une courbe focale~$(\phi)$ que
+nous prendrons pour support:
+\[
+x = f(v), \qquad
+y = g(v), \qquad
+z = h(v),
+\]
+alors $\dfrac{\dd f}{\dd w}, \dfrac{\dd g}{\dd w}, \dfrac{\dd h}{\dd w}$ sont nuls, et l'équation~\Eq{(1)} devient
+\[%[** TN: Filled in last two columns]
+\begin{vmatrix}
+a & b & c \\
+\mfrac{\dd a}{\dd v}\, dv + \mfrac{\dd a}{\dd w}\, dw &
+\mfrac{\dd b}{\dd v}\, dv + \mfrac{\dd b}{\dd w}\, dw &
+\mfrac{\dd c}{\dd v}\, dv + \mfrac{\dd c}{\dd w}\, dw \\
+\mfrac{\dd f}{\dd v}\, dv &
+\mfrac{\dd g}{\dd v}\, dv &
+\mfrac{\dd h}{\dd v}\, dv
+\end{vmatrix} = 0;
+\]
+%% -----File: 140.png---Folio 132-------
+$dv$~est en facteur. L'une des familles de développables est
+formée par les droites $v = \cte$, c'est-à-dire par toutes les
+droites de la congruence passant par un même point~$F$ de~$(\phi)$.
+Ce sont des cônes.
+
+\MarginNote{Examen des diverse
+cas possibles.}
+Examinons alors tous les cas possibles relativement à la
+nature du lieu des foyers.
+
+\ParItem{\Primo.} Supposons qu'il y ait \Card{2} surfaces focales $(\Phi), (\Phi')$.
+Toute droite~$D$ de la congruence est tangente à $(\Phi), (\Phi')$ aux
+\Card{2} points $F\Add{,} F'$ foyers de~$D$. Considérons une des développables
+ayant pour arête de rebroussement l'une des
+\Figure[3.25in]{140a}
+courbes~$(A)$. Toutes ses génératrices sont
+tangentes à $(\Phi')$, cette développable est
+circonscrite à~$(\Phi')$ le long d'une courbe~$(c')$
+que nous appellerons \emph{courbe de contact}.
+Le plan focal correspondant à~$F$ est
+le plan tangent en~$F$ à la surface~$(\Phi)$. Le
+\Ord{2}{e} plan focal est le plan tangent en~$F'$ à~$(\Phi')$,
+et comme la développable est circonscrite à~$(\Phi')$, ce plan
+tangent est le plan tangent à la développable au point~$F'$,
+c'est-à-dire le long de la génératrice~$D$; c'est le plan osculateur
+à l'arête de rebroussement~$(A)$ au point~$F$. Il y a
+évidemment réciprocité entre $(\Phi), (\Phi')$. L'autre série de développables
+aura pour arêtes de rebroussement les enveloppes
+des droites~$D$ sur la surface~$(\Phi')$. Soient $(A')$ ces arêtes de
+rebroussement, et ces développables seront circonscrites à~$(\Phi)$
+le long des courbes de contact~$(C)$. Nous avons ainsi déterminé
+sur~$(\Phi)\Add{,} (\Phi')$ \Card{2} réseaux conjugués qui se correspondent
+de manière qu'aux courbes~$(A)$ correspondent les courbes~$(c')$
+%% -----File: 141.png---Folio 133-------
+et aux courbes~$(c)$ les courbes~$(A')$, l'une des familles de courbes
+correspondantes étant constituée par des arêtes de rebroussement,
+et l'autre par des courbes de contact. Le \Ord{2}{e} foyer~$F'$
+est le point de contact de la droite~$D$ avec son enveloppe
+quand on se \DPtypo{deplace}{déplace} sur la courbe~$(c)$.
+
+\ParItem{\Secundo.} Supposons une surface focale~$(\Phi)$ et une courbe focale~$(\phi')$.
+Une des séries de développables est constituée par
+\Figure[3.25in]{141a}
+des cônes ayant leurs sommets
+sur~$(\phi')$. Les courbes~$(c)$ sur~$(\Phi)$
+sont les courbes de contact des
+cônes circonscrits à~$(\Phi)$ par les
+divers points de~$(\phi')$. Les plans
+focaux sont le plan osculateur à~$(A)$
+au point~$F$ et le plan tangent
+à~$(\Phi)$ au point~$F$, c'est-à-dire
+le plan tangent à~$(\phi')$ passant
+par~$(D)$, et le plan tangent au cône de la congruence de
+sommet~$F'$, le long de~$(D)$. Les courbes~$(c), (A)$ forment un réseau
+conjugué sur~$(\Phi)$.
+
+\ParItem{\Tertio.} Supposons enfin \Card{2} courbes focales $(\phi)\Add{,} (\phi')$; les deux
+familles de développables sont des cônes passant par l'une des
+courbes et ayant leurs sommets sur l'autre.
+
+\MarginNote{Cas singuliers.}
+Voyons maintenant le cas des foyers confondus.
+
+\ParItem{\Primo.} Il y a une \emph{surface focale double}. Dans ce cas la congruence
+est constituée par les tangentes à une famille d'asymptotiques
+de cette surface. Il n'y a plus qu'une famille de
+développables ayant pour arêtes de rebroussement ces asymptotiques.
+%% -----File: 142.png---Folio 134-------
+Prenons cette surface pour support, et pour courbes $w = \cte$
+ces asymptotiques. L'équation différentielle qui détermine
+les développables est
+\[
+dw (E'\, dv + F'\, dw) = 0.
+\]
+L'équation des lignes asymptotiques est
+\[
+E'\, dv^{2} + 2F'\, dv · dw + G'\, dw^{2} = 0;
+\]
+elle doit être vérifiée pour $dw = 0$; donc $E' = 0$, et l'équation
+qui détermine les développables devient $dw^{2} = 0$, ce qui démontre
+le \DPtypo{resultat}{résultat} précédemment énoncé.
+
+\ParItem{\Secundo.} Il y a une \emph{courbe focale double}~$(\phi)$. Les droites de
+la congruence sont dans des plans tangents aux divers points
+de~$(\phi)$. Une famille de ces développables est donc constituée
+par ces plans. On aperçoit immédiatement deux autres développables,
+l'enveloppe des plans tangents précédents, et la développable
+qui a pour arête de rebroussement la courbe~$(\phi)$. Il
+est facile de voir qu'il n'y en a pas d'autre. Soit la courbe~$(\phi)$:
+\[
+x = f(v), \qquad
+y = g(v), \qquad
+z = h(v);
+\]
+la tangente a pour coefficients directeurs $\dfrac{\dd f}{\dd v}, \dfrac{\dd g}{\dd v}, \dfrac{\dd h}{\dd v}$; donnons-nous
+en chaque point les coefficients directeurs d'une droite
+de la congruence $\alpha(v), \beta(v), \gamma(v)$. Une droite quelconque de la
+congruence aura pour coefficients directeurs
+\[
+a = \frac{\dd f}{\dd v} + w \alpha (v), \qquad
+b = \frac{\dd g}{\dd v} + w \beta (v), \qquad
+c = \frac{\dd h}{\dd v} + w \gamma (v).
+\]
+L'équation des développables est
+\[%[** TN: Filled in last two columns]
+\begin{vmatrix}
+f' + w · \alpha & g' + w · \beta & h' + w · \gamma \\
+(f'' + w \alpha')\, dv + \alpha · dw &
+(g'' + w \beta')\, dv + \beta · dw &
+(h'' + w \gamma')\, dv + \gamma · dw \\
+f'\, dv & g'\, dv & h'\, dv
+\end{vmatrix} = 0;
+\]
+%% -----File: 143.png---Folio 135-------
+$dv$~est en facteur; en retranchant la \Ord{3}{e} ligne de la \Ord{1}{ère}, $w$~est
+en facteur, et l'équation se réduit à
+\[%[** TN: Filled in last two columns]
+w · dv^{2}
+\begin{vmatrix}
+\alpha          & \beta & \gamma \\
+f'' + w \alpha' & g'' + w \beta' & h'' + w \gamma' \\
+f'              & g' & h'
+\end{vmatrix} = 0.
+\]
+Nous trouvons $dv = 0$ correspondant aux plans tangents; $w = 0$
+correspondant à la développable d'arête de rebroussement~$\phi$,
+et enfin
+\[
+\begin{vmatrix}
+\alpha & \beta & \gamma \\
+f''    & g''   & h''    \\
+f'     & g'    & h'
+\end{vmatrix} + w \begin{vmatrix}
+\alpha  & \beta  & \gamma  \\
+\alpha' & \beta' & \gamma' \\
+f'      & g'     & h'
+\end{vmatrix} = 0\Add{.}
+\]
+Le plan tangent considéré en un point de la courbe~$(\phi)$ a pour
+équation
+\[
+\begin{vmatrix}
+x - f  & y - g & z - h \\
+f'     & g'    & h'    \\
+\alpha & \beta & \gamma
+\end{vmatrix} = 0;
+\]
+\DPtypo{Cherchons}{cherchons} son enveloppe. La caractéristique est dans le plan
+\[
+\begin{vmatrix}
+x - f  & y - g & z - h \\
+f''    & g''   & h''   \\
+\alpha & \beta & \gamma
+\end{vmatrix} + \begin{vmatrix}
+x - f   & y - g  & z - h \\
+f'      & g'     & h'    \\
+\alpha' & \beta' & \gamma'
+\end{vmatrix} = 0.
+\]
+La droite~$D$ est
+\[%[** TN: Filled in last two equations]
+x = f + u \left[\frac{\dd f}{\dd v} + w \alpha (v)\right]\!\!, \quad
+y = g + u \left[\frac{\dd g}{\dd v} + w \beta (v)\right]\!\!, \quad
+z = h + u \left[\frac{\dd h}{\dd v} + w \gamma (v)\right]\!\!.
+\]
+Exprimons que cette droite est dans le \Ord{2}{e} plan qui contient la
+caractéristique, nous avons
+\[%[** TN: Filled in last two columns]
+\begin{vmatrix}
+f' + w \alpha & g' + w \beta & h' + w \gamma \\
+f''           & g'' & h'' \\
+\alpha        & \beta & \gamma
+\end{vmatrix} + \begin{vmatrix}
+f' + w \alpha & g' + w \beta & h' + w \gamma \\
+f'            & g' & h' \\
+\alpha'       & \beta' & \gamma'
+\end{vmatrix} = 0,
+\]
+%% -----File: 144.png---Folio 136-------
+condition qui se réduit à
+\[
+\begin{vmatrix}
+f'      & g'     & h'     \\
+f''     & g''    & h''    \\
+\alpha  & \beta  & \gamma
+\end{vmatrix} + w \begin{vmatrix}
+\alpha  & \beta  & \gamma \\
+f'      & g'     & h'     \\
+\alpha' & \beta' & \gamma'
+\end{vmatrix} = 0.
+\]
+C'est précisément l'équation qui définit la \Ord{3}{e} développable,
+qui est donc l'enveloppe des plans qui contiennent les droites
+de la congruence.
+
+\Section{Sur le point de vue corrélatif.}
+{3.}{} Nous avons trouvé comme cas particulier du lieu des
+foyers une courbe. En examinant la question au point de vue
+corrélatif, nous sommes conduits à examiner \emph{le cas où l'enveloppe
+des plans focaux est une surface développable}, soit~$\Phi$.
+Soit~$\Phi'$ l'autre nappe le la surface focale. Les droites de la
+congruence sont tangentes à $\Phi, \Phi'$; or\Add{,} une tangente à la développable~$\Phi$
+doit être dans l'un des plans tangents qui enveloppent
+cette développable; les droites de la congruence sont donc les
+tangentes à~$\Phi'$ qui sont dans les plans tangents à~$\Phi$, ce sont
+les tangentes aux sections de~$\Phi'$ par les plans qui enveloppent~$\Phi$.
+Dans ce cas les arêtes de rebroussement~$(A')$ sur la surface~$\Phi'$
+sont des courbes planes, les développables correspondantes
+étant les plans de ces courbes. Les foyers d'une droite~$D$ sont
+le point de contact avec~$\Phi'$, et le point d'intersection avec
+la caractéristique du plan tangent à la développable~$\Phi$. L'autre
+famille de développables aura ses arêtes de rebroussement
+sur la surface~$\Phi$, et correspondant aux courbes~$(c')$ \DPtypo{conjuguees}{conjuguées}
+des courbes~$(A')$.
+
+\emph{Réciproquement, si les arêtes de rebroussement des développables
+%% -----File: 145.png---Folio 137-------
+situées sur une des nappes de la surface focale
+sont des courbes planes, les développables correspondantes
+seront des plans, et leur enveloppe sera la \Ord{2}{e} nappe de la
+surface focale.}
+
+Pour avoir une congruence de cette espèce on peut prendre
+arbitrairement la développable~$\Phi$, et sur cette développable,
+une famille de courbes quelconque. Les tangentes à ces courbes
+engendrent une congruence de l'espèce considérée, car l'une
+des familles de développables est évidemment constituée par
+les plans tangents à la développable~$\Phi$; les courbes de contact
+sur la développable sont les génératrices, qui peuvent
+être considérées comme conjuguées à toute famille de courbes.
+
+\emph{Supposons les \Card{2} nappes de la surface focale développables\Add{.}}
+Il suffit de partir d'une développable~$\Phi$, de la couper par
+une famille de plans quelconques. Les sections seront les
+courbes~$A$, et les plans de ces sections envelopperont l'autre
+développable focale. On peut dire dans ce cas que l'on a \Card{2} familles
+de plans à un paramètre, les droites de la congruence
+étant les intersections de chaque plan d'une famille avec chaque
+plan de l'autre.
+
+Les \emph{\Card{2} cas singuliers} se correspondent à eux-mêmes au
+point de vue corrélatif. Les asymptotiques d'une surface se
+correspondent à elles-mêmes; car une asymptotique est telle
+que le plan osculateur en l'un de ses points est tangent à la
+surface, et au point de vue corrélatif, un point d'une courbe
+se transforme en plan osculateur et inversement.
+%% -----File: 146.png---Folio 138-------
+
+\MarginNote{Congruences de
+Koenigs.}
+Il y a un \emph{autre cas particulier corrélatif de lui-même},
+c'est le \emph{cas de Koenigs}. On appelle \emph{élément de contact} le système
+constitué par un point~$M$ et un plan passant par ce point.
+Les surfaces et les courbes sont alors engendrées de la même
+façon au moyen des \DPtypo{élément}{éléments} de contact: en chaque point d'une
+surface, il y a un plan tangent et un seul, ce qui donne
+$\infty^{2}$~éléments de contact; sur une courbe, il y a $\infty^{1}$~points, et
+en chaque point $\infty^{1}$~plans tangents, ce qui donne encore $\infty^{2}$~éléments
+de contact; pour les développables, nous avons $\infty^{1}$~plans
+et $\infty^{2}$~points, donnant $\infty^{2}$~éléments de contact. Une droite est
+de même constituée par $\infty^{2}$~éléments de contact, $\infty^{1}$~points sur
+la droite et $\infty^{1}$~plans passant par la droite. Dans la \DPtypo{Théorie}{théorie}
+des congruences, \emph{un foyer et le plan focal correspondant constituent
+un élément de contact}, et les surfaces focales, courbes
+focales, développables focales, ou comme l'on dit plus généralement,
+les \emph{multiplicités focales, sont engendrées par les
+éléments de contact focaux}. Nous voyons alors que nous avons
+considéré tous les cas possibles, sauf celui où l'une des multiplicités
+focales est une droite.
+
+\Illustration{146a}
+La droite peut être considérée comme le lieu de $\infty^{1}$~points
+ou comme l'enveloppe de $\infty^{1}$~plans; c'est donc à la fois une
+courbe et une développable; il en résulte qu'une des familles
+de développables de la congruence est constituée par des cônes
+ayant leurs sommets sur la droite, et l'autre par des plans
+passant par la droite. Si en particulier la congruence a pour
+multiplicités focales une droite~$D$ et une
+surface~$\Phi$, les séries de développables
+seront d'une part les cônes circonscrits à~$\Phi$
+%% -----File: 147.png---Folio 139-------
+par les différents points de~$D$, ce qui donne les courbes de
+contact~$(c)$; et les plans passant par~$D$, qui coupent suivant
+les arêtes de rebroussement~$(A)$, et $(A)\Add{,} (c)$ forment un système
+de courbes conjuguées. On obtient ainsi le \emph{Théorème de Koenigs:
+Les courbes de contact des cônes circonscrits à une surface
+par les divers points d'une droite~$D$, et les sections de
+cette surface par les plans passant par~$D$ constituent un réseau
+conjugué}.
+
+\MarginNote{Congruences
+linéaires.}
+Si les multiplicités focales sont \Card{2} droites, la congruence
+est constituée par les droites rencontrant ces \Card{2} droites.
+C'est une \emph{congruence linéaire}.
+
+Il peut encore arriver qu'il y ait une droite focale double;
+il suffira alors d'associer à chaque point~$A$ de la droite
+un plan~$P$ passant par cette droite, et la congruence sera
+constituée par les droites~$D$ situées dans les plans~$P$ et passant
+par les points~$A$.
+
+\Section{Application. Surfaces de Joachimsthal.}
+{}{Rechercher les surfaces dont les lignes de courbure d'un
+système sont dans des plans passant par une droite fixe~$\Delta$.}
+
+Soit $S$ une surface répondant à la question; considérons
+les tangentes aux lignes de courbure; ces tangentes~$D$ constituent
+une congruence, et comme les lignes de courbure sont
+dans des plans passant par~$\Delta$, ces droites~$D$ rencontrent la
+droite~$\Delta$; $S$~est une des nappes de la surface focale; les développables
+comprennent, d'une part les plans des lignes de
+courbure, et d'autre part les cônes circonscrits à~$S$ par les
+différents points de $\Delta$, dont les courbes de contact constituent
+%% -----File: 148.png---Folio 140-------
+un système conjugué du \Ord{1}{er} système de lignes de courbure,
+et par suite forment le \Ord{2}{e} système de lignes de courbure.
+Si nous considérons ce \Ord{2}{e} système de lignes de courbure, le
+cône circonscrit coupe la surface~$S$ suivant un angle constamment
+nul; la courbe de contact, qui est une ligne de courbure
+de~$S$, est donc aussi une ligne de courbure du cône circonscrit,
+d'après le Théorème de Joachimsthal; c'est donc une trajectoire
+orthogonale des génératrices, donc l'intersection du
+cône avec une sphère ayant son centre au sommet; le \Ord{2}{e} système
+de lignes de courbure est donc constitué par des courbes
+sphériques, et les sphères correspondantes coupent orthogonalement
+la surface~$S$ le long des lignes de courbure. \emph{La surface~$S$
+est donc trajectoire orthogonale d'une famille de sphères
+ayant leurs centres sur $\Delta$.} Cette propriété est caractéristique
+de la surface~$S$; supposons en effet une famille de sphères
+ayant leurs centres sur~$\Delta$, et une surface~$S$ orthogonale
+à chacune de ces sphères tout le long de la courbe d'intersection;
+l'intersection est une ligne de courbure de la sphère,
+et comme l'angle d'intersection de~$S$ et de la sphère est constamment
+droit, c'est une ligne de courbure de~$S$. Si on joint
+le centre~$A$ de la sphère à un point~$M$ de la ligne de courbure,
+cette droite est normale à la sphère, donc tangente à la surface~$S$,
+de sorte que la ligne de courbure est la courbe de
+contact du cône circonscrit à~$S$ par le point~$A$.
+
+Nous sommes ainsi conduits à rechercher les surfaces coupant
+à angle droit une famille de sphères. Considérons les lignes
+de courbure du \Ord{1}{er} système; chacune d'elles est tangente
+à la droite~$D$ correspondante, donc normale à la sphère, et
+%% -----File: 149.png---Folio 141-------
+comme elle est dans un plan passant par~$\Delta$, elle est trajectoire
+orthogonale pour le grand cercle section de la sphère
+par ce plan. Si donc on considère les sections de toutes les
+sphères de la famille par un même plan passant par~$\Delta$, la ligne
+de courbure située dans ce plan sera trajectoire orthogonale
+de la famille de cercles obtenue. Si on considère un autre
+plan, la ligne de courbure dans ce plan sera aussi trajectoire
+orthogonale de la famille de cercles. En rabattant le \Ord{2}{e} plan
+sur le \Ord{1}{er} on aura une autre trajectoire orthogonale de
+la même famille de cercles. \emph{On considère donc une famille de
+cercles ayant leurs centres sur~$\Delta$, on en détermine les trajectoires
+orthogonales, et on fait tourner chacune de ces trajectoires
+orthogonales autour de~$\Delta$ d'un angle qui lui corresponde
+et qui varie d'une manière continue quand on passe d'une
+trajectoire à la trajectoire infiniment voisine.} Le lieu des
+courbes ainsi obtenues est une surface qui sera la surface~$S$
+si la loi de rotation est convenablement choisie. Quelle que
+soit d'ailleurs cette loi on obtient toujours une surface répondant
+à la question; cette surface sera en effet engendrée
+par des courbes qui couperont orthogonalement la famille de
+sphères ayant pour grands cercles les cercles considérés, et
+par conséquent la surface coupera à angle droit toutes ces
+sphères tout le long des courbes d'intersection.
+
+Nous allons donc chercher les trajectoires orthogonales
+d'une famille de cercles ayant leurs centres sur une droite~$\Delta$.
+Cherchons plus généralement les trajectoires orthogonales d'une
+famille de cercles quelconque, que nous définirons en donnant
+%% -----File: 150.png---Folio 142-------
+les coordonnées $a\Add{,}b$ du centre et le rayon~$R$ en fonction
+d'un paramètre~$u$\DPtypo{;}{.} Considérons une trajectoire orthogonale rencontrant
+un des cercles en un point~$M$. Les coordonnées du
+point~$M$ sont, en fonction du paramètre~$u$
+\[
+\Tag{(1)}
+x = a + R \cos\phi, \qquad
+y = b + R \sin \phi,
+\]
+$\phi$~étant une fonction de~$u$ convenablement choisie. Tout revient
+à déterminer cette fonction de~$u$ de façon que la courbe représentée
+par les équations~\Eq{(1)} soit normale à tous les cercles.
+La normale au cercle a pour paramètres directeurs $x - a$, $y - b$;
+elle doit être tangente à la courbe, donc
+\[
+\Tag{(2)}
+\frac{dx}{x-a} = \frac{dy}{y-b}.
+\]
+Or:
+\begin{gather*}
+dx = da + \cos \phi · dR - R \sin \phi · d \phi, \quad
+dy = db + \sin \phi · dR + R \cos \phi · d \phi, \\
+x - a = R \cos \phi, \quad
+y - b = R \sin \phi.
+\end{gather*}
+L'équation~\Eq{(2)} devient
+\[
+\begin{vmatrix}
+da + \cos\phi · dR - R \sin\phi · d\phi &
+db + \sin\phi · dR + R \cos\phi · d\phi \\
+R \cos\phi   &    R \sin\phi
+\end{vmatrix} = 0,
+\]
+ou:
+\[
+\sin\phi · da - \cos\phi · db - R\, d\phi = 0,
+\]
+ou:
+\[
+\Tag{(3)}
+\frac{d\phi}{du} = \frac{a'}{R} \sin\phi - \frac{b'}{R} \cos\phi.
+\]
+Si nous posons
+\[
+\tg \frac{\phi}{2} = w,
+\]
+d'où
+\[
+d\phi = \frac{2\, dw}{1 + w^{2}},
+\]
+l'équation différentielle devient
+\[
+\frac{1}{du}\, \frac{2\, \Err{du}{dw}}{1 + w^{2}}
+  = A \frac{2w}{1 + w^{2}} + B \frac{1 - w^{2}}{1 + w^{2}},
+\]
+$A\Add{,} B$ étant fonctions de~$u$; de sorte que l'équation est de la
+forme
+\[
+\frac{dw}{du} = Aw + \frac{B}{2} (1 - w^{2}).
+\]
+%% -----File: 151.png---Folio 143-------
+C'est une équation de Riccati. Le rapport anharmonique de \Card{4} solutions~$w$
+est constant. Or\Add{,} si $M$~est un point d'une trajectoire
+orthogonale, $\tg \dfrac{\phi}{2}$ est le coefficient angulaire de la
+droite~$AM$. Si on considère \Card{4} trajectoires
+orthogonales $M, M', M'', M'''$, les \Card{4} valeurs de~$u$
+correspondantes sont les coefficients angulaires
+des \Card{4} droites $AM, AM', AM'', AM'''$, et
+le rapport anharmonique des \Card{4} solutions~$u$
+est le rapport anharmonique du faisceau
+$(A, M, M', M'' M''')$. Ce rapport est indépendant de la position du
+point~$A$ sur le cercle. Il en résulte que \emph{\Card{4} trajectoires orthogonales
+d'une famille de cercles coupent tous les cercles de
+la famille suivant le même rapport anharmonique}.
+
+%[** TN: Setting inset illos side-by-side and floating]
+\begin{figure}[hbt]
+\centering
+\Input{151a}\hfil\hfil
+\Input[3in]{151b}
+\end{figure}
+Dans le cas particulier où les cercles ont leurs centres
+sur une droite~$\Delta$, les points $M'\Add{,} M''$ d'intersection du cercle avec~$\Delta$
+correspondent à \Card{2} trajectoires orthogonales; on a donc \Card{2} solutions
+de l'équation de Riccati, et la détermination des trajectoires
+orthogonales se ramène à une quadrature. Pour définir
+la famille, au lieu de se donner $a, b, R$ en fonction d'un paramètre,
+on peut se donner une trajectoire orthogonale~$\Gamma$, on
+aura alors \Card{3} solutions de l'équation de \DPtypo{Ricatti}{Riccati}, et la solution
+la plus générale s'obtiendra en écrivant que son rapport
+anharmonique avec les \Card{3} solutions connues est constant.
+
+Supposons que $(\Delta)$ soit
+l'axe~$\DPtypo{OX}{Ox}$, et donnons
+nous $(\Gamma)$ par ses tangentes~$(T)$. L'une d'elles
+a pour équations:
+%% -----File: 152.png---Folio 144-------
+\begin{align*}
+x &= a + \rho\cos u, \\
+y &= \rho\sin u,
+\end{align*}
+$a$~étant une fonction de~$u$. Pour déterminer le point de contact
+avec~$(\Gamma)$, on a, en différentiant:
+\[
+da - \rho\sin u\, du + \cos u\, d\rho = 0, \quad
+\rho\cos u\, du + \sin u\, d\rho = 0,
+\]
+d'où, pour la valeur de~$\rho$, c'est-à-dire du rayon~$R$ du cercle,
+\[
+R = \rho = \frac{da}{du} \sin u.
+\]
+Une trajectoire orthogonale quelconque est donc représentée
+par
+\[
+\Tag{(4)}
+x = a + \frac{da}{du} \sin u · \cos\phi, \qquad
+y = \frac{da}{du} \sin u · \sin\phi,
+\]
+$\phi$~étant lié à~$u$ par la constance du rapport anharmonique
+$(M, M', M'', M''')$, ce qui donne simplement
+\[
+\Tag{(5)}
+\tg \frac{\phi}{2} = m · \tg \frac{u}{2}.
+\]
+Si maintenant on fait tourner la courbe~\Eq{(4)} d'un angle~$v$ autour
+de~$\DPtypo{ox}{Ox}$, en supposant~$m$ fonction de~$v$, et posant
+\[
+a = f(u), \qquad
+m = g(v),
+\]
+on obtiendra une trajectoire orthogonale quelconque de la famille
+de sphères ayant pour grands cercles les cercles considérés:
+\[
+\Tag{(6)}
+\left\{
+  \begin{aligned}
+x &= f(u) + f'(u)\sin u \cos\phi, \\
+y &= f'(u) \sin u \sin\phi \cos v,
+     \text{ (avec $\tg \tfrac{\phi}{2} = g(v)\tg \tfrac{u}{2}$)}  \\
+z &= f'(u) \sin u \sin\phi \sin v.
+\end{aligned}
+\right.
+\]
+Et en considérant dans ces équations $u$~et~$v$ comme des paramètres
+arbitraires, elles représentent la surface de Joachimsthal
+la plus générale.
+%% -----File: 153.png---Folio 145-------
+
+\Section{Détermination des développables d'une congruence.}
+{4.}{} Nous avons vu que la détermination des développables
+d'une congruence dépend de l'intégration d'une équation \DPtypo{difféférentielle}{différentielle}
+du \Ord{1}{er} ordre et du \Ord{2}{e} degré. Cette intégration
+peut se simplifier dans certains cas.
+
+On obtient les développables sans quadrature si la congruence
+admet \Card{2} courbes focales, ou corrélativement deux développables
+focales. Dans le \Ord{1}{e} cas, on obtient des cônes, et
+dans le \Ord{2}{e}, des plans tangents, comme on l'a vu précédemment.
+
+Si on a une courbe focale, ou corrélativement une développable
+focale, on a immédiatement une des familles de développables
+de la congruence; pour avoir l'autre, on a à intégrer
+une équation différentielle du \Ord{1}{e} ordre et du \Ord{1}{er}~degré.
+
+\Illustration[2.25in]{153a}
+Cette équation a des propriétés particulières dans un cas
+corrélatif de lui-même, \emph{cas où l'on a une courbe focale et une
+développable focale}. Soit $(\alpha)$ l'arête de rebroussement de la
+développable focale~$(\Phi)$; considérons
+une génératrice quelconque~$C$
+de cette développable; les droites
+de la congruence rencontrent
+la courbe focale~$(\phi')$ et sont dans
+les plans tangents à~$(\Phi)$. Considérons
+un plan tangent à~$(\Phi)$ qui
+rencontre $(\phi')$ en~$F'$; toutes les droites du plan tangent qui
+passent par~$F'$ sont des droites de la congruence. Considérons
+les développables de la congruence passant par une de ces
+droites~$D$; il y a d'abord le plan qui enveloppe la développable,
+et qui admet pour courbe de contact la génératrice~$C$. Les
+foyers de la droite~$D$ sont $F'$~sur~$(\phi')$ et $F$~sur~$C$. La \Ord{2}{e} développable
+%% -----File: 154.png---Folio 146-------
+a pour arête de rebroussement une courbe~$(A)$ de~$(\Phi)$
+dont les tangentes vont rencontrer~$(\phi')$. Le problème revient
+donc à \emph{trouver les courbes d'une développable~$(\Phi)$ dont les
+tangentes vont rencontrer une courbe~$(\phi')$}. Nous allons chercher
+directement les développables de la congruence, que nous définirons
+en partant de la courbe~$(\phi')$ et en associant à chacun de
+ses points un certain plan dans lequel seront toutes les droites
+de la congruence passant par ce point; la développable~$(\Phi)$
+sera l'enveloppe de ce plan. Soit la courbe~$(\phi')$
+\[
+x = f(v), \qquad
+y = g(v), \qquad
+z = h(v);
+\]
+pour définir un plan passant par un de ses points, il suffit
+de se donner \Card{2} directions $\alpha(v), \beta(v), \gamma(v)$ et $\alpha_{1}(v), \beta_{1}(v), \gamma_{1}(v)$.
+
+On a ainsi le plan contenant toutes les droites de la
+congruence; les coefficients directeurs d'une telle droite
+sont alors:
+\[
+\bar{a} = \alpha + w \alpha_{1}, \quad
+\bar{b} = \beta  + w \beta_{1}, \quad
+\bar{c} = \gamma + w \gamma_{1}.
+\]
+L'équation aux développables
+\[
+\begin{vmatrix}
+\bar{a}   & \bar{b}   & \bar{c}  \\
+d\bar{a}  & d\bar{b}  & d\bar{c} \\
+df        & dg        & dh
+\end{vmatrix} = 0
+\]
+devient ici
+\[%[** TN: Filled in last two columns]
+dv \begin{vmatrix}
+\alpha + w \alpha_{1}  & \beta + w \beta_{1}  & \gamma + w \gamma_{1}  \\
+f'(v)                  & g'(v)                  & h'(v)               \\
+(\alpha' + w \alpha_{1}')\, dv + \alpha_{1} dw  &
+(\beta'  + w \beta_{1}')\, dv + \beta_{1} dw  &
+(\gamma' + w \gamma_{1}')\, dv + \gamma_{1} dw
+\end{vmatrix} = 0.
+\]
+Nous trouvons $dv = 0, v = \cte$ ce qui nous donne les plans des
+droites de la congruence. L'autre solution s'obtiendra par
+l'intégration de l'équation:
+%% -----File: 155.png---Folio 147-------
+\[%[** TN: Filled in last two columns]
+dw \begin{vmatrix}
+\alpha + w \alpha_{1} & \beta + w \beta_{1} & \gamma + w \gamma_{1} \\
+f'                    & g'                    & h'                   \\
+\alpha_{1}            & \beta_{1}            & \gamma_{1}
+\end{vmatrix} + dv \begin{vmatrix}
+\alpha + w \alpha_{1}   & \beta + w \beta_{1}   & \gamma + w \gamma_{1} \\
+f'                      & g'                      & h'                   \\
+\alpha' + w \alpha_{1}' & \beta' + w \beta_{1}' & \gamma' + w \gamma_{1}'
+\end{vmatrix} = 0,
+\]
+équation de la forme
+\[
+\frac{dw}{dv} = Pw^{2} + Qw + R,
+\]
+$P\Add{,} Q\Add{,} R$ étant fonctions de $v$~seulement. C'est une équation de Riccati.
+
+Cherchons dans quels cas on peut avoir des solutions particulières
+de cette équation. Si la courbe~$(\phi')$ est plane, si
+on coupe~$(\Phi)$ par son plan, la section est une courbe dont les
+tangentes rencontrent~$(\phi')$, c'est une courbe~$(A)$; on a une solution
+particulière, le problème s'achève au moyen de \Card{2} quadratures.
+En particulier si $(\phi')$ est le cercle imaginaire à
+l'infini, on a à déterminer sur~$(\Phi)$ des courbes dont les tangentes
+rencontrent le cercle imaginaire à l'infini, ce sont
+les courbes minima. \emph{La détermination des courbes minima d'une
+développable se ramène à \Card{2} quadratures.}
+
+Corrélativement, si $(\Phi)$ est un cône, considérons le cône
+de même sommet et qui a pour base~$(\phi')$; c'est une développable
+de le \Ord{2}{e} famille; on a une solution particulière, et le problème
+s'achève par \Card{2} quadratures.
+
+Si $(\Phi)$ est un cône et $(\phi')$~une courbe plane, on a \Card{2} solutions
+particulières, donc une seule quadrature.
+
+Supposons encore que les plans~$P$ précédemment définis
+soient normaux à la courbe~$(\phi')$. Nous avons la \emph{congruence des
+normales} à la courbe~$(\phi')$, et la recherche des développables
+conduira à celle des \emph{développées} de~$(\phi')$. Le plan normal à~$(\phi')$
+%% -----File: 156.png---Folio 148-------
+en l'un de ses points~$F'$ est perpendiculaire à la tangente~$F'T$.
+Si on \DPtypo{onsidère}{considère} le cône isotrope~$J$ de sommet~$F'$, le plan normal
+est le plan polaire de la tangente par rapport à ce cône isotrope;
+parmi les normales il y a donc les \Card{2} génératrices de
+contact des plans tangents menés par la tangente au cône isotrope.
+Soit $G$ l'une d'elles, on l'obtient algébriquement; considérons
+la surface réglée qu'elle engendre lorsque $F'$~décrit
+la courbe~$(\phi')$. Le plan asymptote, plan tangent à l'infini sur~$G$,
+est le plan tangent au cône isotrope~$J$ le long de~$G$; la
+surface réglée contient la courbe~$(\phi')$, et le plan tangent au
+point~$F'$ est le plan~$G·F'·T$, qui est encore le plan tangent au
+cône isotrope le long de~$G$. Ce plan tangent est donc le même
+tout le long de la génératrice~$G$, et cette droite engendre une
+surface développable. Ainsi \emph{les droites isotropes des plans
+normaux à une courbe gauche décrivent \Card{2} développables et enveloppent
+\Card{2} développées de la courbe gauche}. Nous avons ainsi
+\Card{2} solutions particulières, et la détermination des développées
+doit s'achever par une seule quadrature.
+
+Effectivement, en supposant que $\DPtypo{v}{w}$~est l'arc~$s$ de~$(\phi')$, que
+$\alpha, \beta, \gamma$; $\alpha_{1}, \beta_{1}, \gamma_{1}$, sont les cosinus directeurs $a'\Add{,} b'\Add{,} c'$ de la normale
+principale et $a''\Add{,} b''\Add{,} c''$ de la binormale, l'équation générale
+se réduit, en désignant par $a, b, c$ les cosinus directeurs de la
+tangente,
+\[%[** TN: Filled in last two columns]
+dw \begin{vmatrix}
+a'  & b'  & c' \\
+a   & b   & c  \\
+a'' & b'' & c''
+\end{vmatrix} + ds \begin{vmatrix}
+a' + wa'' & b' + wb'' & c' + wc'' \\
+a & b & c \\
+-\mfrac{a}{R} - \mfrac{a''}{T} + w\mfrac{a'}{T} &
+-\mfrac{b}{R} - \mfrac{b''}{T} + w\mfrac{b'}{T} &
+-\mfrac{c}{R} - \mfrac{c''}{T} + w\mfrac{c'}{T}
+\end{vmatrix} = 0,
+\]
+%% -----File: 157.png---Folio 149-------
+c'est-à-dire
+\[
+- dw + \frac{ds}{T} (1 + w^{2}) = 0,
+\]
+ou enfin
+\[
+%[** TN: "tang." in original]
+w = \tg \int \frac{ds}{T}.
+\]
+On vérifie bien que l'équation différentielle en~$w$ admet les
+deux solutions: $w = ±i$, qui correspondent aux développables
+isotropes.
+
+Si on remarque de plus que la surface focale de la congruence
+des normales est la surface polaire de~$(\phi')$, c'est-à-dire
+que les points de contact des normales avec les développées
+sont sur la droite polaire, on retrouve tous les résultats
+essentiels sur la détermination des développées.
+
+
+\ExSection{VI}
+
+\begin{Exercises}
+\item[29.] On considère la congruence des tangentes communes aux deux
+surfaces $x^{2} + y^{2} = 2az$, $x^{2} + y^{2} = -2az$. Déterminer les développables
+de cette congruence: étudier leurs arêtes de rebroussement,
+leurs courbes de contact, leurs traces sur le plan
+$z = 0$.
+
+\item[30.] Si les deux multiplicités focales d'une congruence sont des
+développables isotropes (congruence isotrope), toutes les
+surfaces réglées qui passent par une même droite de la congruence
+ont même point central et même paramètre de distribution.
+Le plan perpendiculaire à chaque droite de la congruence
+\DPtypo{mene}{mène} à \DPtypo{egale}{égale} distance des deux points focaux enveloppe
+une surface minima. On peut obtenir ainsi la surface minima la
+plus générale.
+
+\item[31.] On suppose que les droites $D$~et~$D'$ de deux congruences se correspondent
+de manière que deux droites correspondantes soient
+parallèles. Si alors les développables des deux congruences se
+correspondent, les plans focaux de~$D$ sont parallèles à ceux
+de~$D'$; les droites $\Delta, \Delta_{1}$, qui joignent les points focaux
+correspondants se coupent en un point~$M$; le lieu de ce point
+admet $\Delta$~et~$\Delta_{1}$, pour tangentes conjuguées, et les courbes conjuguées
+enveloppées par ces droites correspondent aux développables
+des deux congruences.
+\end{Exercises}
+%% -----File: 158.png---Folio 150-------
+
+
+\Chapitre{VII}{Congruences de Normales.}
+
+\Section{Propriété caractéristique des congruences de normales.}
+{1.}{} Considérons une surface, les coordonnées d'un de ses
+points dépendent de deux paramètres; l'ensemble des normales
+à cette surface dépend de deux paramètres, et constitue
+une congruence. Pour obtenir les développables, il suffit
+de considérer sur la surface les \Card{2} \DPtypo{series}{séries} de lignes de
+courbure, puisque les normales à une surface en tous les
+points d'une ligne de courbure engendrent une surface développable.
+Le plan tangent à une développable passe par la normale~$D$
+et par la tangente à la ligne de courbure correspondante.
+C'est l'un des plans focaux de la droite~$D$. Ainsi \emph{les plans
+focaux sont les plans des sections principales de la surface.
+Les plans focaux d'une congruence de normales sont rectangulaires}.
+Il en résulte qu'une congruence quelconque ne peut pas
+en général être considérée comme formée des normales à une
+%[** TN: Added parentheses between \gamma, \gamma']
+surface. Considérons les \Card{2} lignes de courbure $(\gamma)\Add{,}(\gamma')$ passant
+par un point~$M$ de la surface; à la développable
+de~$(\gamma)$ correspond une arête de
+rebroussement~$(A)$ dont le plan osculateur
+est le plan focal, le point de contact~$F$
+de~$A$ et de la droite~$D$ est un des points
+focaux. On peut considérer l'arête de rebroussement~$(A)$
+comme étant l'enveloppe de
+la droite~$D$ quand le point~$M$ se déplace
+%% -----File: 159.png---Folio 151-------
+sur la courbe~$(\gamma)$; le point~$F$ est alors l'un des centres de
+courbure principaux de la surface au point~$M$. Le plan focal
+associé est le \Ord{2}{e} plan de section principale~$FMT'$. On aura de
+même une \Ord{2}{e} arête de rebroussement~$(A')$ en \DPtypo{considerant}{considérant} la courbe~$(\gamma')$.
+
+%[** TN: Exchanged diagram labels for (T') and (\gamma')]
+\Illustration[1.5in]{158a}
+On verra facilement que ces propriétés des centres de
+courbure principaux et des plans de sections principales subsistent,
+quelle que soit la nature des multiplicités focales
+de la congruence considérée.
+
+\emph{Réciproque}. Prenons une congruence constituée par les
+droites~$D$
+\begin{alignat*}{2}%[** TN: Stacked to accommodate figure]
+x &= f(v,w) &&+ u · a(v,w), \\
+y &= g(v,w) &&+ u · b(v,w), \\
+z &= h(v,w) &&+ u · c(v,w).
+\end{alignat*}
+Cherchons à quelles conditions on peut déterminer sur la droite~$D$
+un point~$M$ dont le lieu soit une surface constamment normale
+à~$D$. Il suffit que l'on puisse déterminer~$u$ en fonction
+de~$v\Add{,}w$ de façon que l'on ait
+\[
+\sum a\,dx = 0,
+\]
+ou
+\[
+\sum a(df + u\, da + a\, du) = 0.
+\]
+On peut supposer que $a\Add{,} b\Add{,}c$ soient les cosinus directeurs; alors
+$\sum a^{2} = 1$, et $u$~représentera la distance du point~$P$ où la droite
+rencontre le support, au point~$M$. On a en même temps $\sum a\, da = 0$
+et la condition précédente devient
+\[
+du + \sum a\, df = 0;
+\]
+\DPtypo{Cette}{cette} équation peut encore s'écrire
+\[
+\Tag{(1)}
+-du = \sum a\, df.
+\]
+Elle exprime que $\sum a\, df$ est une différentielle totale exacte;
+or\Add{,} on a
+\[
+\sum a\, df
+  = \sum a \frac{\dd f}{\dd v}\, dv
+  + \sum a \frac{\dd f}{\dd w}\, dw;
+\]
+%% -----File: 160.png---Folio 152-------
+la condition est donc:
+\[
+\frac{\dd}{\dd w} \sum a\Add{·} \frac{\dd f}{\dd v}
+  = \frac{\dd}{\dd v} \sum a\Add{·} \frac{\dd f}{\dd w},
+\]
+ou:
+\[
+\sum \frac{\dd a}{\dd w}\Add{·} \frac{\dd f}{\dd v}
+  = \sum \frac{\dd a }{\dd v} · \frac{\dd f}{\dd w},
+\]
+ou enfin:
+\[
+\Tag{(2)}
+\sum \left(\frac{\dd a}{\dd w}\Add{·} \frac{\dd f}{\dd v}
+         - \frac{\dd a}{\dd v} · \frac{\dd f}{\dd w}\right) = 0.
+\]
+Nous trouvons une condition unique. Or\Add{,} nous avons trouvé précédemment
+comme condition nécessaire l'orthogonalité des plans
+focaux. Nous sommes donc conduits à comparer les deux conditions\Add{.}
+Les coefficients $A\Add{,} B\Add{,} C$ d'un plan focal vérifient les relations
+\begin{gather*}
+%[** TN: Moved equation number per errata list]
+\Tag{(3)}
+Aa + Bb + Cc = 0, \\
+\left\{
+\begin{alignedat}{4}
+  &   A\left(\frac{\dd f}{\dd v} + u \frac{\dd a}{\dd v}\right)
+  &&+ B\left(\frac{\dd g}{\dd v} + u \frac{\dd b}{\dd v}\right)
+  &&+ C\left(\frac{\dd h}{\dd v} + u \frac{\dd c}{\dd v}\right) &&= 0, \\
+%
+  &   A\left(\frac{\dd f}{\dd w} + u \frac{\dd a}{\dd w}\right)
+  &&+ B\left(\frac{\dd g}{\dd w} + u \frac{\dd b}{\dd w}\right)
+  &&+ C\left(\frac{\dd h}{\dd w} + u \frac{\dd c}{\dd w}\right) &&= 0.
+\end{alignedat}
+\right.
+\end{gather*}
+\DPchg{Eliminant}{Éliminant} $u$ entre les \Card{2} dernières équations, nous avons
+\[
+\Tag{(4)}
+\begin{vmatrix}
+\sum A \mfrac{\dd f}{\dd v} & \sum A \mfrac{\dd a}{\dd v} \\
+\sum A \mfrac{\dd f}{\dd w} & \sum A \mfrac{\dd a}{\dd w}
+\end{vmatrix} = 0.
+\]
+Les coefficients de direction des normales aux plans focaux
+sont définis par \Eq{(3)}\Add{,}~\Eq{(4)}. Si nous considérons $A\Add{,} B\Add{,} C$ comme coordonnées
+courantes, \Eq{(3)}~représente un plan passant par l'origine,
+\Eq{(4)}~un cône ayant pour sommet l'origine; et les génératrices
+d'intersection sont précisément les normales cherchées.
+Exprimons que ces deux droites sont rectangulaires; le plan~\Eq{(3)}
+est perpendiculaire à la droite~$(a\Add{,} b\Add{,} c)$, qui est sur le cône~\Eq{(4)},
+car on a, puisque $\sum a^{2} = 1$ et $\sum a\, da = 0$
+\[
+\sum a \frac{\dd a}{\dd v} = 0, \qquad
+\sum a \frac{\dd a}{\dd w} = 0;
+\]
+donc les \Card{2} normales sont perpendiculaires à la droite~$(a\Add{,} b\Add{,} c)$;
+si elles sont rectangulaires, c'est que le cône~\Eq{(4)} est capable
+d'un trièdre trirectangle inscrit, ce qui donne la condition
+%% -----File: 161.png---Folio 153-------
+\[
+\sum \left(\frac{\dd f}{\dd v} · \frac{\dd a}{\dd w}
+         - \frac{\dd f}{\dd w} · \frac{\dd a}{\dd v}\right) = 0;
+\]
+ce qui est précisément la condition~\Eq{(2)}. De sorte que \emph{la condition
+nécessaire et suffisante pour que la congruence soit une
+congruence de normales, c'est que les plans focaux soient rectangulaires}.
+
+Supposons satisfaite la condition~\Eq{(2)}. Pour obtenir la
+surface normale à toutes les droites de la congruence, il suffit
+de calculer~$u$ en fonction de~$v\Add{,}w$, ce qui se fait par l'équation~\Eq{(1)}
+\[
+du = d\Phi(v, w),
+\]
+d'où
+\[
+u = \Phi(v, w) + \cte[.]
+\]
+Il y a donc une infinité de surfaces répondant à la question;
+si un point~$M$ décrit une surface~$(S)$ et un point~$M'$ une surface~$(S')$
+répondant à la question, on a $u = PM$, $u' = PM'$, la distance~$MM'$
+sera une quantité constante. Les surfaces~$(S)\Add{,} (S')$ sont appelées
+\emph{surfaces parallèles} et \emph{une famille de surfaces parallèles
+admet pour chaque normale mêmes centres de courbure principaux
+et mêmes multiplicités focales}; ces multiplicités focales
+constituent la \emph{développée} de l'une quelconque de ces surfaces.
+
+\Section{Relations entre une surface et sa développée.}
+{2.}{} Considérons une nappe de la développée d'une
+surface~$(S)$. Supposons d'abord que ce soit une surface~$(\Phi)$.
+Considérons une droite~$D$ de la congruence des normales à~$(S)$;
+cette droite est tangente en~$F$ à l'arête de rebroussement~$(A)$
+qui appartient à~$(\Phi)$; les plans focaux associés à~$D$ sont
+le plan osculateur à~$(A)$ et le plan tangent à~$(\Phi)$. Pour que la
+%% -----File: 162.png---Folio 154-------
+congruence soit une congruence
+de normales, il faut et il suffit
+que le plan osculateur à~$(A)$
+soit normal à~$(\Phi)$, donc que $(A)$
+soit une géodésique de~$(\Phi)$. \emph{La
+congruence des normales à la
+surface~$(S)$ est constituée par
+les tangentes à une famille de
+géodésiques de sa développée~$(\Phi)$\Add{.}
+Et réciproquement les tangentes
+à une famille de $\infty^{1}$~géodésiques d'une surface quelconque~$(\Phi)$
+constituent une congruence de normales.} Soit $M$~le point où la
+droite~$D$ coupe la surface~$(S)$; lorsque la droite~$D$ roule sur
+l'arête de rebroussement~$(A)$, le point~$M$ décrit une ligne de
+courbure~$(\gamma)$ de~$(S)$. A chaque point~$M$ de~$(S)$ correspond un
+point~$F$ de~$(\Phi)$; il y a correspondance point par point entre
+les \Card{2} surfaces; à la famille de lignes de courbure~$(\gamma)$ de~$(S)$
+correspond une famille de géodésiques de~$(\Phi)$. Voyons maintenant
+les courbes de contact~$(c)$ de~$(\Phi)$; considérons la tangente~$F\theta$
+à~$(c)$, c'est la caractéristique du plan tangent à~$(\Phi)$
+lorsque le point~$M$ décrit~$(\gamma)$; or\Add{,} ce plan tangent à~$(\Phi)$ est le
+\Ord{2}{e} plan focal, c'est le plan perpendiculaire au plan~$FMT$ passant
+par~$FM$, c'est donc le plan normal à~$(\gamma)$ au point~$M$. Donc~$F\theta$
+est la caractéristique du plan normal à~$(\gamma)$, c'est la droite
+polaire de~$(\gamma)$. \emph{Les courbes de contact de~$(\Phi)$ sont les
+courbes tangentes aux droites polaires des différents points
+des courbes~$(\gamma)$.} $F\theta$~étant dans le plan normal à~$(\gamma)$ rencontre
+la tangente à la \Ord{2}{e} section principale; elle passe au centre
+%% -----File: 163.png---Folio 155-------
+de courbure géodésique de~$(\gamma)$ sur~$(S)$.
+
+\begin{figure}[hbt]
+\centering
+\Input[1.75in]{162a}\hfil\hfil
+\Input[2.5in]{163a}
+\end{figure}
+\MarginNote{Surface Canal\Add{.}}
+Supposons que l'une des nappes de la développée se réduise
+à une courbe~$(\phi)$. La droite~$D$ rencontre~$(\phi)$ en l'un des
+points focaux~$F$. L'une des développables passant par~$D$ est un
+cône de soumet~$F$; l'une des lignes
+de courbure~$(\gamma)$ de~$(S)$ passant
+par~$M$ est située sur un
+cône de sommet~$F$. Or\Add{,} $(\gamma)$~est
+constamment normale à~$D$, c'est
+donc une trajectoire orthogonale
+des génératrices du cône; c'est
+l'intersection de ce cône avec une sphère de centre~$F$. Cette
+sphère en chaque point~$M$ a pour normale la droite~$D$, elle est
+donc tangente à la surface~$(S)$ tout le long de la courbe~$(\gamma)$.
+A chaque point~$F$ de~$(\phi)$ correspond une sphère ayant ce point
+pour centre et tangente à~$(S)$ tout le long de la ligne de
+courbure correspondante. \emph{La surface~$(S)$ est l'enveloppe d'une
+famille de sphères dépendant d'un paramètre.} Nous l'appellerons
+une \emph{surface canal}. La réciproque est vraie, comme on le
+verra plus loin. La courbe~$(\gamma)$ est alors l'intersection d'une
+sphère avec une sphère infiniment voisine; c'est un cercle.
+Le cône~$F$ est de révolution, l'axe de ce cône est la position
+limite de la ligne des centres, c'est la tangente~$Fu$ à~$(\phi)$.
+Considérons la tangente~$MT$ à~$(\gamma)$: $MT$~tangente en un point du
+cercle est orthogonale à~$Fu$, $Fu$ est donc dans le \Ord{2}{e} plan de
+section principale. \emph{Les congruences considérées sont donc formées
+des génératrices de $\infty^{1}$~cônes de révolution, dont les axes
+sont tangents à la courbe lieu des sommets} de ces cônes. Et
+%% -----File: 164.png---Folio 156-------
+\emph{réciproquement toute congruence ainsi constituée est une congruence
+de normales}, car les plans focaux sont les plans tangents
+et les plans méridiens de ces cônes, et sont par conséquent
+rectangulaires.
+
+\MarginNote{Cyclide de Dupin.}
+Voyons si les \Card{2} nappes de la développée peuvent se réduire
+à \Card{2} courbes $(\phi)\Add{,} (\phi')$. Les développables de la congruence sont
+les cônes ayant leur sommet sur l'une des courbes et passant
+par l'autre. Tous les cônes~$F$ de révolution doivent passer par la
+courbe~$(\phi')$. Cette courbe~$(\phi')$ est telle qu'il passe par
+cette courbe une infinité de cônes de révolution. De même~$(\phi)$;
+$(\phi), (\phi')$ ne peuvent donc être que des biquadratiques gauches ou
+leurs éléments de décomposition. Aucune de ces courbes ne peut
+être une biquadratique gauche, sans quoi par chacune d'elles
+il passerait \Card{4} cônes du \Ord{2}{e} degré seulement. Voyons si l'une
+d'elles peut être une cubique gauche; les cônes du \Ord{2}{e} degré
+passant par~$(\phi')$ ont leurs sommets sur~$\Err{(\phi)}{(\phi')}$: les \Card{2} courbes $(\phi)\Add{,}
+(\phi')$ seraient confondues. Voyons donc s'il peut exister des cubiques
+gauches telles que les cônes du \Ord{2}{e} degré qui les contiennent
+soient de révolution. Un tel cône aurait pour axe
+la tangente~$Fu$; or\Add{,} il contient cette
+tangente, donc il se décompose. Donc ni~$(\phi)$
+ni~$(\phi')$ ne peuvent être des cubiques
+gauches. Supposons donc que~$(\phi')$ soit une
+conique; le lieu des sommets des cônes de
+révolution passant par cette conique est
+une autre conique, qui est dite focale de la \Ord{1}{ère}. Il y a réciprocité
+entre ces coniques, et les cônes de révolution ont
+%% -----File: 165.png---Folio 157-------
+pour axes les tangentes aux focales. Donc \emph{les droites rencontrant
+\Card{2} coniques focales l'une de l'autre constituent une congruence
+de normales}. Les surfaces normales à ces droites s'appellent
+\emph{Cyclides de Dupin. Leurs \Card{2} systèmes de lignes de courbure
+sont des cercles}.
+
+\Illustration[1.25in]{164a}
+Supposons en particulier que $(\phi')$ soit un cercle; alors
+le lieu les sommets des cônes de révolution passant par $(\phi')$~est
+l'axe~$(\phi)$ de ce cercle, et nous voyons que toutes les
+droites qui s'appuient sur $(\phi)\Add{,} (\phi')$ sont normales à une famille
+de surfaces. Ces surfaces sont des \emph{tores} de révolution autour
+de l'axe~$(\phi)$, le lieu du centre du cercle méridien étant le
+cercle~$(\phi')$.
+
+Supposons que $(\phi')$ soit une droite, la surface est l'enveloppe
+d'une famille de sphères ayant leurs centres sur cette
+droite. C'est une surface de révolution autour de~$(\phi')$; la \Ord{1}{ère} nappe
+de la développée est la droite~$(\phi')$, la \Ord{2}{e} est engendrée
+par la rotation de la développée de la méridienne principale;
+pour que ce soit une courbe, il faut que la développée soit un
+point, donc que la méridienne soit un cercle, et nous retombons
+sur le cas du tore.
+
+\MarginNote{Cas singulier.}
+Cherchons enfin si les \Card{2} nappes de la développée peuvent
+être confondues. Alors les \Card{2} familles de lignes de courbure de
+la surface~$(S)$ sont confondues. C'est le cas des \emph{surfaces réglées
+à génératrices isotropes}. Les \Card{2} nappes se réduisent à
+une seule courbe, comme on le verra au paragraphe suivant.
+%% -----File: 166.png---Folio 158-------
+
+\Section{\DPchg{Etude}{Étude} des surfaces enveloppes de sphères.}
+{3.}{} Nous avons été amenés dans ce qui précède à considérer
+les surfaces enveloppes de sphères. Nous allons
+maintenant étudier les réciproques des propriétés précédentes.
+
+Considérons une surface~$(S)$ enveloppe de $\infty^{1}$ sphères~$(\Sigma)$.
+Chaque sphère coupe la sphère infiniment voisine suivant un
+cercle, et les normales à~$(S)$ en tous les points de ce cercle
+passent par le centre de la sphère. Le lieu des centres des
+sphères est une courbe rencontrée par les normales à~$(S)$,
+c'est une des nappes de la développée. D'autre part, la sphère~$(\Sigma)$
+étant tangente à la surface~$(S)$ tout le long du cercle caractéristique,
+ce cercle est une ligne de courbure de la
+surface~$(S)$, d'après le Théorème de Joachimsthal. \emph{Les surfaces
+enveloppes de sphères ont une famille de lignes de courbure
+circulaires. Réciproquement toute surface ayant une famille
+de lignes de courbure circulaires est une enveloppe de sphères\Add{.}}
+Considérons une ligne de courbure circulaire~$(K)$; toute sphère
+passant par~$(K)$ coupe la surface~$(S)$ sous un angle constant,
+d'après le Théorème de Joachimsthal. Or\Add{,} il est possible de
+trouver une sphère passant par~$(K)$ et tangente à~$(S)$ en l'un
+des points de ce cercle; cette sphère sera alors tangente à~$(S)$
+en tous les points du cercle~$(K)$, et toute ligne de courbure
+circulaire est courbe de contact d'une sphère avec la
+surface. La surface est l'enveloppe des sphères ainsi déterminées.
+
+Soit une sphère de centre $(a,b,c)$ et de rayon~$r$, $a,b,c,r$
+étant fonctions d'un même paramètre.
+%% -----File: 167.png---Folio 159-------
+La sphère a pour équation
+\[%[** TN: Omitted large brace grouping next two equations]
+(x - a)^{2} + (y - b)^{2} + (z - c)^{2} - r^{2} = 0;
+\]
+la caractéristique est en outre définie par l'équation
+\[
+(x - a)\, da + (y - b)\, db + (z - c)\, dc + r\, dr = 0;
+\]
+On vérifie bien que c'est un cercle dont le plan est perpendiculaire
+à la direction~$da, db, dc$, de la tangente au lieu des
+centres des sphères.
+
+Nous venons de considérer les surfaces dont une famille
+de lignes de courbure est constituée par des cercles. Voyons
+si les \Card{2} familles de lignes de courbure peuvent être circulaires.
+La surface correspondante pourra être considérée de
+\Card{2} façons différentes comme l'enveloppe de $\infty^{1}$~sphères. Les \Card{2} nappes
+de la développée sont des courbes. Nous obtenons la
+Cyclide de Dupin, que nous allons étudier à un point de vue
+nouveau.
+
+\MarginNote{Correspondance
+entre les
+droites et les
+sphères.}
+Les droites et les sphères sont des éléments géométriques
+dépendant de \Card{4} paramètres. Ce fait seul permet de prévoir
+qu'il y aura une espèce de correspondance entre l'étude des
+systèmes de droites et celle des systèmes de sphères. Cette
+correspondance trouve son expression analytique dans une
+transformation, due à Sophus Lie, et que nous exposerons plus
+tard. Mais nous la verrons se manifester auparavant dans diverses
+questions. C'est ainsi que l'on peut considérer dans la
+géométrie des sphères les enveloppes de $\infty^{1}$~sphères comme correspondant
+aux surfaces réglées, lieux de $\infty^{1}$~droites; la cyclide
+de Dupin correspond alors aux surfaces doublement réglées,
+%% -----File: 168.png---Folio 160-------
+donc aux surfaces réglées du \Ord{2}{e} degré. Nous allons voir
+l'analogie se développer dans l'étude qui suit.
+
+%[** TN: Several {1}-like superscripts rendered as prime accents]
+Soit $(\Sigma)$ une sphère de la \Ord{1}{ère} famille, $(\Sigma')$~une sphère
+de la \Ord{2}{e} famille, $(\Sigma)$~touche~$(S)$ suivant un cercle~$(K)$, $\Sigma'$~touche~$(S)$
+suivant un cercle~$(K')$. La surface~$(S)$ étant engendrée
+par le cercle~$(c)$ ou par le cercle~$(c')$, il en résulte que ces
+\Card{2} cercles ont au moins un point commun~$M$\DPtypo{.}{}; soient $O\Add{,} O'$ les centres
+des sphères $(\Sigma)\Add{,} (\Sigma')$, $OM$~et~$O'M$ sont normales aux sphères
+$(\Sigma)\Add{,} (\Sigma')$ et par suite normales en~$M$ à la surface. Donc elles
+coïncident, $O\Add{,} M\Add{,} O'$~sont sur une même droite; les sphères $(\Sigma)\Add{,} (\Sigma')$
+sont tangentes en~$M$. \emph{Une sphère de l'une des familles est
+tangente à une sphère quelconque de l'autre famille.} (Deux génératrices
+de systèmes différents d'une quadrique se rencontrent).
+
+Considérons \Card{3} sphères fixes $(\Sigma), (\Sigma_{1}), (\Sigma_{2})$ d'une des familles.
+Elles sont tangentes à toutes les sphères de l'autre famille,
+et par suite \emph{la surface est l'enveloppe des sphères
+tangentes à \Card{3} sphères fixes}. (Une quadrique est le lieu d'une
+droite rencontrant \Card{3} droites fixes). Les \Card{3} sphères $(\Sigma), (\Sigma_{1}), (\Sigma_{2})$
+se coupent en \Card{2} points qui peuvent être considérés comme des
+sphères de rayon nul tangentes à $(\Sigma), (\Sigma_{1}), (\Sigma_{2})$; donc il y a \Card{2} sphères
+de rayon nul dans chaque famille de sphères enveloppées
+par la cyclide. Les sphères de l'autre famille devant
+être tangentes à ces \Card{2} sphères de rayon nul passent par leurs
+centres. Ces deux points sont sur le lieu des centres des
+sphères, donc sur les coniques focales; \emph{si donc nous considérons
+les \Card{2} coniques focales, les sphères d'une des familles
+ont leurs centres sur l'une des coniques et passent par \Card{2} points
+%% -----File: 169.png---Folio 161-------
+fixes de l'autre, symétriques par rapport au plan de
+la \Ord{1}{ère}.} Il est alors facile, avec cette génération, de trouver
+l'équation de la cyclide.
+
+\Section{\DPchg{Equation}{Équation} de la
+\DPtypo{Cyclide}{cyclide} de
+Dupin.}
+{}{\normalfont\Primo.} Supposons d'abord que l'une des coniques soit une
+ellipse, par exemple: l'autre est une hyperbole. Prenons pour
+axes $\DPtypo{ox, oy}{Ox, Oy}$ les axes de l'ellipse, dont l'équation dans son
+plan est:
+\[
+\Tag{(E)}
+\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - 1 = 0.
+\]
+
+\Illustration{169a}
+\noindent L'hyperbole focale est dans le plan
+$y = 0$. Elle a pour équations
+\[
+\Tag{(H)}
+y = 0, \qquad
+\frac{x^{2}}{a^{2} - b^{2}} - \frac{z^{2}}{b^{2}} - 1 = 0\Add{.}
+\]
+Un point~$\omega$ de l'ellipse~\Eq{(E)} a pour
+coordonnées
+\[
+x = a \cos\phi, \qquad
+y = b \sin\phi, \qquad
+z = 0.
+\]
+Soit sur l'hyperbole~\Eq{(H)} le point fixe~$A$ de coordonnées
+\[
+x_{0}, \qquad
+y_{0} = 0, \qquad
+z_{0}^{2} = b^{2} \left(\frac{x_{0}^{2}}{a^{2} - b^{2}} - 1\right).
+\]
+L'équation d'une sphère~$\Sigma$ ayant pour centre~$\omega$ et passant par
+le point~$A$ sera
+\[
+\DPtypo{(x - a \cos\phi^{2})}{(x - a \cos\phi)^{2}}
+  + (y - b \sin\phi)^{2} + z^{2}
+  = (x_{0} - a \cos\phi)^{2} + b^{2} \sin^{2} \phi
+  + b^{2} \left(\frac{x_{0}^{2}}{a^{2} - b^{2}} - 1\right),
+\]
+ou
+\[
+x^{2} + y^{2} + z^{2} - 2ax \cos\phi - 2by \sin\phi
+  = x_{0}^{2} + b^{2} \frac{x_{0}^{2}}{a^{2} - b^{2}}
+  - b^{2} - 2ax_{0} \cos\phi,
+\]
+ou
+\[
+2a (x - x_{0}) \cos\phi + 2by \sin\phi
+  = x^{2} + y^{2} + z^{2} + b^{2} - \frac{a^{2} x_{0}^{2}}{c^{2}},
+\]
+en posant comme d'habitude
+\[
+c^{2} = a^{2} - b^{2}.
+\]
+L'équation de la sphère est de la forme
+\[
+A \cos\phi + B \sin\phi = C\Add{,}
+\]
+%% -----File: 170.png---Folio 162-------
+la condition pour qu'il y ait une racine double, c'est-à-dire,
+l'équation de l'enveloppe, est
+\[
+A^{2} + B^{2} = C^{2}\Add{.}
+\]
+Donc la cyclide a pour équation:
+\[
+4a^{2} (x - x_{0})^{2} + 4b^{2} y^{2}
+  = \left(x^{2} + y^{2} + z^{2} + b^{2} - \frac{a^{2} x_{0}^{2}}{c^{2}}\right)^{2}.
+\]
+
+\ParItem{\Secundo.} Supposons maintenant qu'une des coniques soit une
+parabole. L'autre est aussi une parabole. Prenons les axes ordinaires,
+nous avons pour équations
+des \Card{2} coniques
+\begin{align*}
+\Tag{(P)}
+z &= 0, \qquad y^{2} = 2px, \\
+\Tag{(P')}
+y &= 0, \qquad x^{2} + z^{2} = (x - p)^{2}.
+\end{align*}
+
+%\Illustration{170a}
+\begin{wrapfigure}[11]{O}{2.125in}
+\smash[t]{\raisebox{-1.75in}{\Input{170a}}}
+\end{wrapfigure}
+\noindent Le centre~$C$ de la sphère sur la parabole~$P$
+a pour coordonnées
+\[
+x = 2p \lambda^{2}, \qquad
+y = 2p \lambda, \qquad
+z = 0.
+\]
+Le point fixe~$A$ sur la parabole~$P'$ a pour coordonnées
+\[
+x_{0}, \qquad
+y_{0} = 0, \qquad
+z_{0}^{2} = (x_{0} - p)^{2} - x_{0}^{2}.
+\]
+L'équation de la sphère est
+\[
+(x - 2p \lambda^{2})^{2} + (y - 2p \lambda)^{2} + z^{2}
+  = (x_{0} - 2p \lambda^{2})^{2} + 4p^{2} \lambda^{2}
+  + (x_{0} - p)^{2} - x_{0}^{2},
+\]
+ou
+\[
+x^{2} + y^{2} + z^{2}
+  - (x_{0} - p)^{2} - 4p \lambda y - 4p (x - x_{0}) \lambda^{2} = 0,
+\]
+et l'équation de l'enveloppe, c'est-à-dire de la cyclide, est
+\[
+\bigl[x^{2} + y^{2} + z^{2} - (x_{0} - p)^{2}\bigr] (x - x_{0}) + py^{2} = 0.
+\]
+L'ordre de la surface, qui est on général~$4$, s'abaisse ici
+à~$3$.
+
+\MarginNote{Surface canal
+isotrope.}
+Parmi les surfaces réglées, nous avons considéré les surfaces
+développables, où chaque génératrice rencontre la génératrice
+infiniment voisine. Le cas correspondant pour les enveloppes
+de sphères sera celui où chaque sphère est tangente
+à la sphère infiniment voisine. Pour qu'il en soit ainsi, il
+faut que le plan radical des \Card{2} sphères leur soit tangent.
+%% -----File: 171.png---Folio 163-------
+Prenons la sphère
+\[
+\Tag{(1)}
+(x - a)^{2} + (y - b)^{2} + (z - c)^{2} - r^{2} = 0.
+\]
+Le plan radical de cette sphère et de la sphère infiniment
+voisine est
+\[
+\Tag{(2)}
+(x - a)\, da + (y - b)\, db + (z - c)\, dc + r\, dr = 0;
+\]
+pour qu'il soit tangent à la sphère~\Eq{(1)} il faut et il suffit
+que sa distance au centre~$(a\Add{,} b\Add{,} c)$ soit égale à~$±r$, donc que
+l'on ait
+\[
+\frac{r\, dr}{\sqrt{da^{2} + db^{2} + dc^{2}}} = ±r,
+\]
+ou
+\[
+\Tag{(3)}
+da^{2} + db^{2} + dc^{2} = dr^{2}.
+\]
+
+\Illustration[1.75in]{171a}
+\noindent Donc $r$~n'est autre que l'arc~$S$ de la
+courbe~$(c)$ lieu des centres des sphères,
+cet arc étant compté à partir
+d'une origine arbitraire. Cherchons
+le point de contact de la sphère
+avec la sphère infiniment voisine.
+Les coordonnées satisfont aux équations
+\[
+\frac{x - a}{da} = \frac{y - b}{db} = \frac{z - c}{dc}
+  = \frac{-r\, dr}{\Err{dr}{dr^2}} = -\frac{r}{dr} = - \frac{s}{ds};
+\]
+d'où
+\[
+x = a - s \frac{da}{ds} = a - s \alpha, \quad
+y = b - s \beta, \quad
+z = c - s \gamma,
+\]
+$\alpha\Add{,} \beta\Add{,} \gamma$ étant les cosinus directeurs de la tangente. On obtient
+ainsi le point~$I$, qui décrit une développante~$(r)$ de la courbe~$(c)$.
+L'intersection d'une sphère avec la sphère infiniment
+voisine n'est autre que l'intersection de cette sphère avec
+un de ses plans tangents: c'est un couple de droites isotropes
+se coupant au point~$I$. \emph{L'enveloppe se compose de deux surfaces
+réglées à génératrices isotropes. Nous l'appellerons une
+surface canal isotrope. Réciproquement une surface réglée à
+%% -----File: 172.png---Folio 164-------
+génératrices isotropes est une nappe d'une enveloppe de sphères.}
+Considérons en effet une génératrice~$D$ et le cercle de
+l'infini. Par la génératrice isotrope~$D$ passent une infinité
+de sphères; ces sphères contiennent la droite~$D$ et le cercle
+imaginaire à l'infini, ce qui donné \Card{7} conditions; elles dépendent
+de \Card{2} paramètres arbitraires. Nous pouvons faire en sorte
+que la sphère soit tangente à la surface considérée $(S)$ en \Card{2} points
+à distance finie de la droite~$D$; la sphère est alors
+déterminée; mais de plus elle est tangente à la surface~$(S)$ au
+point à l'infini sur~$D$, donc elle se raccorde avec~$(S)$ tout le
+long de la génératrice~$D$. La surface~$(S)$ fera partie de l'enveloppe
+de ces sphères.
+
+Sur une telle surface, les \Card{2} systèmes de lignes de courbure
+sont confondus avec les génératrices isotropes, les \Card{2} nappes
+de la développée sont confondues avec la courbe~$(C)$. La
+courbe~$(r)$ joue ici un rôle analogue à l'arête de rebroussement
+des surfaces développables. En effet, pour une développable,
+il y a un élément de contact (point de l'arête de rebroussement
+et plan osculateur en ce point) commun à une génératrice
+et à la génératrice infiniment voisine. Ici, c'est
+l'élément de contact constitué par le point~$I$ et le plan tangent
+à la sphère en ce point, plan normal à~$I\omega$, qui est commun
+à la sphère~$(\Sigma)$ et à la sphère infiniment voisine. Le point~$I$
+est un ombilic de la surface~$(S)$. La ligne~$(r)$ en est une ligne
+double, c'est un lieu d'ombilics. Nous l'appellerons la
+\emph{ligne ombilicale} de la surface canal isotrope.
+
+%[** TN: In the manuscript, S_{0} is a tiny 0 set directly below the S]
+\Section{Lignes de courbure et lignes asymptotiques.}
+{4.}{} Considérons une surface~$(S_{0})$ et une ligne asymptotique.
+Les tangentes en chacun des points de cette ligne
+%% -----File: 173.png---Folio 165-------
+engendrent une développable, et l'élément de contact commun à
+une génératrice et à la génératrice infiniment voisine, comprenant
+un point de la ligne et le plan osculateur, qui est
+tangent à~$(S_{0})$, est un élément de contact de~$(S_{0})$. Considérons
+maintenant une ligne de courbure~$(\DPtypo{r}{\Gamma})$: la normale en chaque
+point engendre une développable. Soit $(c)$
+l'arête de rebroussement; $OI$~est égal à
+l'arc de~$(c)$; si donc nous considérons les
+sphères de centres~$\DPtypo{o}{O}$ et de rayons~$OI$, chacune
+de ces sphères touche la sphère infiniment
+voisine, et l'élément de contact
+$(I,P)$ commun à ces \Card{2} sphères est un élément de contact de la
+surface~$(S_{0})$.
+
+%[** TN: Fixed diagram label (T) -> (\Gamma) as per errata list]
+\Illustration[1.5in]{173a}
+Appelons \emph{sphère de courbure} de~$(S_{0})$ toute sphère ayant
+pour centre un centre de courbure principale et pour rayon le
+rayon de courbure principal correspondant. Et nous pourrons
+dire:
+
+%[** TN: Not marked for italicization in the original]
+\emph{Les sphères de courbure de~$(S_{0})$ qui correspondent à une
+même ligne de courbure~$(\Gamma)$, enveloppent une surface canal isotrope,
+ayant~$(\Gamma)$ pour ligne ombilicale.}
+
+\emph{Réciproquement}, si une surface canal isotrope~$(S)$ est
+circonscrite à la surface~$(S_{0})$ le long de sa ligne ombilicale,
+celle-ci étant ligne de courbure pour~$(S)$ sera ligne de courbure
+pour~$(S_{0})$, d'après le théorème de Joachimsthal.
+
+Les choses s'énoncent d'une manière plus nette en substituant
+à la notion de courbe la notion de \emph{bande}. Une bande
+sera, par définition, formée de $\infty^{1}$~éléments de contact appartenant
+à une même multiplicité: le lieu des points (de ces
+%% -----File: 174.png---Folio 166-------
+éléments de contact) sera une courbe, et les plans (de ces
+éléments de contact) seront tangents à la courbe aux points
+correspondants. Une bande appartenant à une surface sera formée
+des points d'une courbe tracée sur la surface, associés
+aux plans tangents à la surface en ces points. On appellera
+\emph{bande de rebroussement} d'une surface développable le lieu des
+éléments de contact communs à chaque génératrice et à la génératrice
+infiniment voisine. Et on appellera \emph{bande ombilicale}
+d'une surface canal isotrope le lieu des éléments de contact
+communs à chacune des sphères inscrites à la surface et à la
+sphère infiniment voisine.
+
+Appelant de même \emph{bandes asymptotiques}, \emph{bandes de courbure}
+les lieux des éléments de contact d'une surface appartenant
+aux lignes asymptotiques ou aux lignes de courbure de cette
+surface, on concluera:
+
+\emph{Une bande asymptotique d'une surface est la bande de rebroussement
+d'une développable; une bande de courbure d'une
+surface est la bande ombilicale d'une surface canal isotrope.
+Et réciproquement}: toute bande de rebroussement (d'une développable),
+qui appartient à une surface~$(S_{0})$, est bande asymptotique
+de~$(S_{0})$; toute bande ombilicale (d'une surface canal
+isotrope), qui appartient à une surface~$(S_{0})$, est bande de
+courbure pour~$(S_{0})$.
+
+On voit ainsi, qu'au point de vue de la correspondance
+entre droites et sphères, les lignes asymptotiques correspondent
+aux lignes de courbure.
+%% -----File: 175.png---Folio 167-------
+
+\Section{Bandes asymptotiques et \DPtypo{Bandes}{bandes} de \DPtypo{Courbure}{courbure}.}
+{}{Remarque \1.} Sur chaque élément de contact~$(M, P)$
+d'une bande, il y a \emph{deux éléments
+linéaires} à considérer. (Un élément
+linéaire étant formé d'un point et
+d'une droite passant par ce point).
+\Figure{175a}
+C'est \emph{l'élément linéaire tangent} formé
+du point~$M$ de l'élément et de la
+tangente~$(T)$ à la courbe qui sert de \emph{support} à la bande,
+qu'on peut appeler simplement la \emph{courbe de la bande}; et \emph{l'élément
+linéaire caractéristique} formé du point~$M$ et de la caractéristique~$(K)$
+du plan~$(P)$, c'est-à-dire de la génératrice
+rectiligne de la développable enveloppée par les plans~$(P)$,
+ou \emph{développable de la bande}. Ces deux éléments linéaires sont
+corrélatifs, au point de vue de la dualité; une bande est corrélative
+d'une bande.
+
+Dans une \emph{bande asymptotique}, $(T)$~et~$(K)$ sont confondues
+dans une \emph{bande de courbure}, $(T)$~et~$(K)$ sont rectangulaires;
+ces termes ont donc un sens par eux-mêmes, sans supposer une
+surface~$(S_{0})$ à laquelle appartienne la bande considérée. Si la
+bande de rebroussement est donnée, la développable correspondante
+est la développable de la bande. Si la bande de courbure
+est donnée, sa courbe~$(\gamma)$ est ligne de courbure de la développable
+de la bande; et la surface canal isotrope dont la
+bande ombilicale se confond avec cette bande de courbure est
+l'enveloppe des sphères de courbure de la développable, construites
+aux divers points~$M$. Les mots bande ombilicale, bande
+de courbure sont donc équivalents; de même que ceux de bande
+%% -----File: 176.png---Folio 168-------
+asymptotique et bande de rebroussement.
+
+Remarquons encore que, si l'on se donne une bande de
+courbure, la sphère de courbure qui correspond à un élément
+de contact~$(M,P)$ de la bande est définie par la condition
+d'admettre~$(M,P)$ pour un de ses éléments de contact et d'avoir
+son centre sur la droite polaire de la courbe~$(\gamma)$ lieu
+des points~$M$ (Voir \No2~et~\No3). Cette seconde condition exprime
+que la sphère a avec~$(\gamma)$ un contact du second ordre; de
+même que dans une bande asymptotique chaque plan~$(P)$ est osculateur
+à~$(\gamma)$. C'est donc une nouvelle analogie entre les bandes
+de courbure et les bandes asymptotiques.
+
+\Section{Lignes de courbure des enveloppes de sphères.}
+{5.}{} Nous connaissons déjà une des familles de lignes
+de courbure, celle constituée par les caractéristiques
+des sphères. Déterminons la \Ord{2}{e} famille. Soit $(c)$ le lieu
+des centres des sphères. Exprimons
+ses coordonnées en fonction
+de l'arc~$(S)$; l'une des sphères
+de centre~$\omega$ rencontre la sphère
+infiniment voisine suivant un
+cercle~$(K)$ dont le plan est normal
+à la tangente~$\omega T$. Introduisons
+le trièdre de Serret au
+point~$\omega$ de la courbe~$(c)$, et
+définissons par rapport à ce trièdre les coordonnées d'un
+point~$M$ de la surface, c'est-à-dire du cercle~$(K)$. Appelons $\theta$
+l'angle de~$\omega M$ avec~$\omega T$, cet angle est le même pour tous les
+points du cercle~$(K)$. Projetons $M$~en~$P$ sur le plan normal, et
+soit $\phi$~l'angle de~$\omega P$ avec~$\omega N$. Les coordonnées de~$M$ par
+%% -----File: 177.png---Folio 169-------
+rapport au trièdre de Serret sont, en appelant~$r$ le rayon de
+la sphère,
+\[
+\Tag{(1)}
+\xi   = r \cos\theta, \qquad
+\eta  = r \sin\theta \cos\phi, \qquad
+\zeta = r \sin\theta \sin\phi.
+\]
+Par rapport à un système d'axes quelconques, ces coordonnées
+sont, en appelant $x\Add{,}y\Add{,}z$ les coordonnées de~$\omega$
+\[
+\Tag{(2)}
+X = x + a \xi + a' \eta + a'' \zeta, \quad
+Y = y + b \xi + b' \eta + b'' \zeta, \quad
+Z = z + c \xi + c' \eta + c'' \zeta;
+\]
+
+\Illustration[2.5in]{176a}
+\noindent $r, \theta$ sont fonctions de~$S$; les paramètres variables sont $s$~et~$\phi$.
+\DPchg{Ecrivons}{Écrivons} que $(K)$~est le cercle caractéristique, nous avons
+\[
+\left\{
+\begin{gathered}
+\sum (X - x)^{2}- r^{2}= 0, \\
+\sum a(X - x) + r \frac{dr}{ds} = 0.
+\end{gathered}\right.
+\]
+En supposant que le trièdre de coordonnées coïncide avec le
+trièdre de Serret, cette équation devient:
+\[
+\xi + r \frac{dr}{ds} = 0,
+\]
+c'est-à-dire:
+\[
+r \cos\theta + r \frac{dr}{ds} = 0,
+\]
+ou
+\[
+\Tag{(3)}
+\Cos\theta = -\frac{dr}{ds}.
+\]
+$\theta$~est ainsi défini en fonction de~$S$.
+
+Une surface enveloppe de sphères est engendrée par des
+cercles; c'est une surface cerclée. Inversement, on peut chercher
+si une surface cerclée est une enveloppe de sphères. Le
+calcul précédent montre que, pour qu'il en soit ainsi, il
+faut que les axes des cercles engendrent une surface développable,
+et on outre, que l'on ait la condition~\Eq{(3)}.
+
+Cherchons les lignes de courbure. Ce sont les trajectoires
+orthogonales des cercles~$(K)$ définis par $S = \cte[]$. La tangente
+à une ligne quelconque passant par $M$ a pour coefficients
+directeurs
+\begin{alignat*}{7}%[** TN: Filled in last two equations]
+dX &= a\, ds &&+ \xi \frac{a'}{R}\, ds
+  &&+ \Err{\zeta}{\eta}
+          \left(-\frac{a}{R} - \frac{a''}{T} - \dots\right)ds
+  &&+ \zeta \frac{a'}{T}\, ds &&+ a\, d\xi &&+ a'\, d\eta  &&+ a''\, d\zeta, \\
+dY &= b\, ds &&+ \xi \frac{b'}{R}\, ds
+  &&+ \eta  \left(-\frac{b}{R} - \frac{b''}{T} - \dots\right)ds
+  &&+ \zeta \frac{b'}{T}\, ds &&+ b\, d\xi &&+ b'\, d\eta  &&+ b''\, d\zeta, \\
+dZ &= c\, ds &&+ \xi \frac{c'}{R}\, ds
+  &&+ \eta  \left(-\frac{c}{R} - \frac{c''}{T} - \dots\right)ds
+  &&+ \zeta \frac{c'}{T}\, ds &&+ c\, d\xi &&+ c'\, d\eta  &&+ c''\, d\zeta.
+\end{alignat*}
+%% -----File: 178.png---Folio 170-------
+En prenant de nouveau le trièdre de Serret pour trièdre de
+coordonnées, ces coefficients directeurs deviennent:
+\[
+\left(1 - \frac{\eta}{R}\right) ds + d\xi, \qquad
+\left(\frac{\xi}{R} + \frac{\zeta}{T}\right) ds + d\eta, \qquad
+- \frac{\eta}{T} ds + d\zeta.
+\]
+Pour la tangente au cercle~$(K)$, on a $ds = 0$, et les coefficients
+directeurs sont:
+\[
+\dd \xi= 0, \qquad
+\dd \eta = -r \sin\theta \sin\phi\, d\phi, \qquad
+\dd \zeta = r \sin\theta \cos\phi\, d\phi.
+\]
+La condition qui définit les trajectoires orthogonales des
+cercles~$(K)$ est donc
+\[
+-\left[\left(\frac{\xi}{R} + \frac{\zeta}{T}\right) ds + d \eta \right] \sin\phi
+  + \left[-\frac{\eta}{T}\, ds + d\zeta\right] \cos\phi = 0.
+\]
+Telle est l'équation différentielle des lignes de courbure.
+On peut l'écrire
+\begin{align*}%[** TN: Re-breaking]
+-\frac{r \cos\theta}{T}\, ds\Add{·} \sin\phi
+  &- \frac{ds}{T} \left(\zeta \sin\phi + \eta \cos\phi\right)
+     - d\eta \sin\phi + d\zeta \cos\phi = 0\Add{,}
+\\
+- \frac{r \cos\theta}{R}\, ds\Add{·} \sin\phi
+  &\begin{aligned}[t]- \frac{r \sin\theta}{T}\, ds
+    &- \sin\phi \bigl[d(r\sin\theta) · \cos\phi - r\sin\theta\Add{·} \sin\phi · d\phi\bigr] \\
+    &+ \cos\phi \bigl[d(r\sin\theta) \Add{·}\sin\phi + r\sin\theta · \cos\phi · d\phi\bigr] = 0,
+\end{aligned} \\
+-\frac{r \cos\theta}{R}\, ds · \sin\phi
+&-\frac{r \sin\theta}{T}\, ds + r \sin\theta · d\phi = 0,
+\end{align*}
+ou
+\[
+\frac{d\phi}{ds} = \frac{1}{T} + \frac{\cotg\theta · \sin\phi}{R};
+\]
+équation de la forme
+\[
+\frac{d\phi}{ds} = A \sin\phi + \beta.
+\]
+\DPtypo{si}{Si} on prend comme fonction inconnue $\tg \frac{\phi}{2}$, on est ramené à
+une équation de Riccati. Mais l'angle~$\phi$ est l'angle du rayon~$IM$
+avec un rayon fixe. Donc \emph{\Card{4} lignes de courbure du \Ord{2}{e} système
+coupent les cercles caractéristiques en \Card{4} points dont le rapport
+anharmonique est constant}. Nouvelle analogie avec les lignes
+asymptotiques d'une surface réglée.
+
+On a les simplifications connues si on a \textit{à~priori} une
+solution de l'équation. Ainsi si on considère l'enveloppe de
+%% -----File: 179.png---Folio 171-------
+sphères~$(\Sigma)$ ayant leurs centres dans un plan, tous les cercles
+caractéristiques sont orthogonaux à la section de la surface
+par ce plan, qui est alors une ligne de courbure. La détermination
+des lignes de courbure se ramène dans ce cas à \Card{2} quadratures.
+
+\Paragraph{Remarque.} Plus généralement, la détermination des trajectoires
+orthogonales d'une famille de $\infty^{1}$~cercles dépend de
+l'intégration d'une équation de Riccati. D'où des conclusions
+analogues aux précédentes.
+
+\Section{Cas où une des nappes de la développée est une développable.}
+{6.}{} Nous venons de considérer le cas où une des nappes
+de la développée d'une surface est une courbe. Corrélativement,
+considérons maintenant le cas où une des
+nappes de la développée est une surface développable.
+Alors les plans tangents à cette développable constituent une
+des familles de développables de la congruence; un tel plan~$P$
+coupe la surface suivant une courbe normale à toutes les droites
+de la congruence situées dans ce plan et qui sera une
+ligne de courbure. En tout point de cette ligne, la normale à
+la surface est dans le plan~$P$. Donc le plan~$P$ coupe orthogonalement
+la surface~$(S)$ tout le long de la ligne de courbure.
+
+\Illustration[1.875in]{180a}
+Réciproquement, si une surface coupe orthogonalement une
+famille de plans, ses sections par ces plans sont des lignes
+de courbure, d'après le Théorème de Joachimsthal, et ces
+plans, constituant une des familles de développables de la
+congruence des normales, enveloppent une développable, qui
+est une des nappes de la développée de la surface.
+
+Considérons la \Ord{2}{e} ligne de courbure passant par un point~$M$;
+%% -----File: 180.png---Folio 172-------
+sa tangente~$MU$ est perpendiculaire à
+la tangente~$MT$ à la \Ord{1}{ère}~ligne de courbure
+et à la normale~$MN$ à la surface; ces \Card{2} droites
+étant dans le plan~$P$, $MU$~est perpendiculaire
+au plan~$P$. \emph{Les lignes de
+courbure de la \Ord{2}{e} famille sont trajectoires
+orthogonales des plans~$P$.}
+
+Considérons une de ces trajectoires orthogonales~$(K)$; les
+plans~$P$ sont normaux à la courbe~$(K)$: l'une des nappes de la
+développée, celle qui est une développable, est ainsi l'enveloppe
+des plans normaux, ou la surface polaire de la courbe~$(K)$.
+\emph{Toutes les lignes de courbure~$(K)$ non planes ont donc
+même surface polaire, qui est l'enveloppe des plans des lignes
+de courbure planes. L'arête de rebroussement de cette surface
+est le lieu des centres des sphères osculatrices à la courbe~$(K)$.}
+La ligne~$(K)$ étant une ligne de courbure, les normales à
+la surface en tous les points de~$(K)$ forment une développable,
+et par suite enveloppent une développée de la courbe~$(K)$, qui
+est une géodésique de sa surface polaire. Si donc on part des
+plans~$P$, pour avoir les courbes~$(K)$ on est ramené à la recherche
+des géodésiques d'une surface développable, ce qui se réduit
+à des quadratures; et comme la surface cherchée peut
+être considérée comme engendrée par les courbes~$(K)$ on voit
+qu'on obtiendra cette surface par des quadratures.
+
+Partons des plans~$P$, et cherchons leurs trajectoires orthogonales.
+Considérons l'arête de rebroussement~$(A)$ de l'enveloppe
+des plans~$P$, et introduisons le trièdre de Serret en
+%% -----File: 181.png---Folio 173-------
+chaque point~$\omega$ de cette courbe, soit~$\DPchg{(\omega\xi\eta\zeta)}{\Tri{\omega}{\xi}{\eta}{\zeta}}$. Le plan~$P$ est
+le plan osculateur~$\xi\omega\eta$, et nous voulons chercher dans ce
+plan un point~$M(\xi, \eta)$ dont le lieu soit normal à~$P$. Les coordonnées
+de~$M$ sont
+\[
+X = x + a \xi + a' \eta, \qquad
+Y = y + b \xi + b' \eta, \qquad
+Z = z + c \xi + c' \eta,
+\]
+la direction de la tangente au lieu du point~$M$ est définie par
+\begin{alignat*}{5}%[** TN: Filled in last two equations]
+dX &= a · ds &&+ \xi \frac{\DPtypo{\alpha'}{a'}}{R}\, ds
+  &&- \eta \left(\frac{a}{R} + \frac{a''}{T}\right) ds
+  &&+ a\, d\xi &&+ a'\, d\eta, \\
+dY &= b · ds &&+ \xi \frac{b'}{R}\, ds
+  &&- \eta \left(\frac{b}{R} + \frac{b''}{T}\right) ds
+  &&+ b\, d\xi &&+ b'\, d\eta, \\
+dZ &= c · ds &&+ \xi \frac{c'}{R}\, ds
+  &&- \eta \left(\frac{c}{R} + \frac{c''}{T}\right) ds
+  &&+ c\, d\xi &&+ c'\, d\eta,
+\end{alignat*}
+expressions de la forme
+\[
+dX = Aa + Ba' + Ca'', \quad
+dY = Ab + Bb' + Cb'', \quad
+dZ = Ac + Bc' + Cc''.
+\]
+
+\DPchg{Ecrivons}{Écrivons} que cette direction est normale au plan~$\xi\omega\eta$,
+c'est-à-dire parallèle à la binormale $(a'', b'', c'')$. Nous avons
+$A = B = 0$, ou
+\[
+ds - \frac{\eta}{R} · ds + d\xi = 0, \qquad
+\frac{\xi}{R}\, ds + d\eta = 0;
+\]
+ou
+\[
+\frac{d\xi}{ds} = \frac{\eta}{R} - 1, \qquad
+\frac{d\eta}{ds} = - \frac{\xi}{R};
+\]
+$\xi, \eta$ sont donnés par \Card{2} équations différentielles du \Ord{1}{er} ordre.
+Il en résulte que par chaque point du plan~$P$ passe une trajectoire
+orthogonale et une seule. Il existe ainsi une correspondance
+point par point entre les divers plans~$P$, les points
+correspondants étant sur une même trajectoire orthogonale.
+
+\Illustration{181a}
+Considérons dans un plan~$P$ \Card{2} points
+$M, N$; et soit $D$ la droite~$MN$; lorsque
+le plan~$P$ varie, la droite~$D$ engendre
+une surface réglée sur laquelle les
+lieux des points $M$~et~$N$ sont trajectoires
+orthogonales des génératrices;
+or\Add{,} les trajectoires orthogonales interceptent
+sur les génératrices des
+%% -----File: 182.png---Folio 174-------
+segments égaux; il en résulte que si l'on considère \Card{2} positions
+$P, P'$, et les positions $MN, M'N'$ correspondantes, on a $MN =
+M'N'$. La correspondance entre les plans~$P$ transforme une courbe
+du plan~$P$ en une courbe égale. En particulier, les plans~$P$
+contenant les lignes de courbure planes, \emph{toutes ces lignes de
+courbure planes sont égales. La surface~$(S)$ est donc engendrée
+par le mouvement d'une courbe plane de forme invariable}. Pour
+la définir, il suffit de \DPchg{connaitre}{connaître} le mouvement de son plan~$P$.
+
+Pour cela, reprenons les équations
+\[
+\Tag{(1)}
+\frac{d\xi}{ds} - \frac{\eta}{R} + 1 = 0, \qquad
+\frac{d\eta}{ds} + \frac{\xi}{R} = 0,
+\]
+et intégrons-les. Considérons d'abord les équations sans \Ord{2}{e} membre
+\[
+R \frac{d\xi}{ds} - \eta = 0, \qquad
+R \frac{d\eta}{ds} + \xi = 0.
+\]
+Posons
+\[
+\frac{R}{ds} = \frac{1}{d\phi},
+\]
+d'où
+\[
+\Tag{(2)}
+d\phi = \frac{ds}{R};
+\]
+les équations deviennent
+\[
+\frac{d\xi}{d\phi} - \eta = 0, \qquad
+\frac{d\eta}{d\phi} + \xi = 0,
+\]
+équations linéaires sans \Ord{2}{e} membre à coefficients constants,
+dont la solution générale est
+\[
+\Tag{(3)}
+\xi = A \cos\phi + B \sin\phi, \qquad
+\eta = -A \sin\phi + B \cos\phi.
+\]
+Passons alors au système avec \Ord{2}{e} membre
+\[
+\Tag{(4)}
+\frac{d\xi}{d\phi} = \eta - R, \qquad
+\frac{d\eta}{d\phi} = - \xi.
+\]
+Considérons dans~\Eq{(3)} $AB$~comme des fonctions de~$\phi$, et cherchons
+à satisfaire au système~\Eq{(4)}. Nous avons
+\begin{alignat*}{4}
+\frac{d\xi}{d\phi}
+  &= \phantom{-}\eta &&+ \frac{dA}{d\phi} \cos\phi &&+ \frac{dB}{d\phi} \sin\phi
+  &&= \phantom{-}\eta - R, \\
+\frac{d\eta}{d\phi}
+  &= -\xi &&- \frac{dA}{d\phi} \sin\phi &&+ \frac{dB}{d\phi} \cos\phi
+  &&= -\xi;
+\end{alignat*}
+%% -----File: 183.png---Folio 175-------
+d'où
+\[
+ \frac{dA}{d\phi} \cos\phi + \frac{dB}{d\phi} \sin\phi = -R, \quad
+-\frac{dA}{d\phi} \sin\phi + \frac{dB}{d\phi} \cos\phi = 0;
+\]
+d'où
+\[
+\frac{dA}{d\phi} = - R \cos\phi, \qquad
+\frac{dB}{d\phi} = - R \sin\phi;
+\]
+ou, en réintroduisant $s$ d'après la formule~\Eq{(2)},
+\[
+\frac{dA}{ds} = -\cos\phi, \qquad
+\frac{dB}{ds} = -\sin\phi;
+\]
+et
+\[
+A = -\int \cos\phi · ds, \qquad
+B = -\int \sin\phi · ds.
+\]
+Posons
+\[
+x_{0} = \int \cos\phi · ds, \qquad
+y_{0} = \int \sin\phi · ds;
+\]
+alors
+\[
+A = -x_{0}, \qquad
+B = -y_{0}.
+\]
+Nous avons donc une solution particulière
+\[
+\xi = -x_{0} \cos\phi - y_{0} \sin\phi, \qquad
+\eta = x_{0} \sin\phi - y_{0} \cos\phi;
+\]
+et la solution générale est, $x_{1}\Add{,} y_{1}$ désignant \Card{2} constantes arbitraires,
+\[
+\Tag{(5)}
+\left\{
+\begin{alignedat}{2}%[** TN: Set on one line in original; added brace]
+\xi  &= \phantom{-} (x_{1} - x_{0}) \cos\phi &&+ (y_{1} - y_{0}) \sin\phi, \\
+\eta &= -(x_{1} - x_{0}) \sin\phi &&+ (y_{1} - y_{0}) \cos\phi.
+\end{alignedat}
+\right.
+\]
+Nous avons \Card{3} quadratures à effectuer. Interprétons géométriquement
+ces résultats:
+
+Les formules précédentes, résolues en $\Err{x, y}{x_{1}, y_{1}}$, donnent
+\[
+\Tag{(6)}
+x_{1} = x_{0} + \xi \cos\phi - \eta \sin\phi, \quad
+y_{1} = y_{0} + \xi \sin\phi + \eta \cos\phi.
+\]
+
+\Illustration[2.25in]{183a}
+Prenons dans le plan~$P$ deux axes
+fixes $0_{1}x_{1}\Add{,} \DPtypo{}{0_{1}}y_{1}$, et construisons par
+rapport à ces axes la courbe~$(R)$ lieu
+du point $(x_{0}\Add{,} y_{0})$. La courbe~$(R)$ est
+la courbe du plan~$P$ qui a même rayon
+de courbure que l'arête de rebroussement~$(A)$.
+Pour chaque valeur de~$s$,
+le point $(x_{0}\Add{,} y_{0})$ occupa une position~$\omega$
+%% -----File: 184.png---Folio 176-------
+sur la courbe~$(R)$, et $\phi$~est l'angle de la tangente à~$R$ en~$\omega$
+avec~$\Err{0,x}{0x_{1}}$. Considérons un système d'axes $\omega \xi \eta$, où l'axe $\omega \xi$
+est la tangente à~$(R)$ correspondant au sens dans lequel se
+déplace~$\omega$; $\phi$~est l'angle de~$\omega \xi$ avec~$\DPtypo{o}{O}_{1}x_{1}$; $\xi, \eta$ fonctions de~$s$,
+sont les coordonnées d'un point~$M$ fixe par rapport au système~$x_{1} \DPtypo{o}{O}_{1} y_{1}$,
+prises par rapport aux axes~$\xi \omega \eta$, et~$x_{1}\Add{,} y_{1}$ sont
+les coordonnées de ce même point par rapport aux axes~$x_{1} \DPtypo{o}{O}_{1} y_{1}$.
+Pour avoir la trajectoire orthogonale, il suffit de porter le
+plan~$P$ dans l'espace, sur le plan osculateur à la courbe~$(A)$,
+$\omega \xi$~et~$\omega \eta$ coïncidant respectivement avec $\omega \xi$~et~$\omega \eta$; dans ce
+mouvement, les courbes $(R)$~et~$(A)$ coïncideront successivement
+en tous leurs points; les rayons de courbure étant les mêmes
+en grandeur et en signe, les centres de courbure seront confondus.
+Si $S$~varie, la courbe~$(R)$ va rouler sur la courbe~$(A)$,
+et un point quelconque~$M$ invariablement lié à la courbe~$(R)$
+décrira la trajectoire orthogonale. \emph{Le mouvement du plan~$P$
+s'obtiendra donc en faisant rouler la courbe plane $(R)$ sur la
+courbe~$(A)$ de façon que le plan~$P$ \DPtypo{coincide}{coïncide} à chaque instant
+avec le plan osculateur à la courbe~$(A)$.} On peut dire que \emph{le
+plan~$P$ roule sur la développable qu'il enveloppe}, comme nous
+allons l'expliquer.
+
+Considérons l'arête de rebroussement~$(A)$ et une tangente~$\omega \xi$;
+pour développer cette courbe
+sur un plan, il faut construire la
+courbe plane dont le rayon de courbure
+en chaque point ait même expression
+en fonction de l'arc que celui
+de la courbe~$(A)$, c'est précisément
+\Figure[1.5in]{184a}%****
+%% -----File: 185.png---Folio 177-------
+la courbe~$(R)$. La position d'un point~$P$ sur la développable
+est définie par l'arc~$S$, qui fixe le point~$\omega$ sur~$(A)$ et par
+le segment $\omega P = u$. Le point~$P'$ qui correspond à~$P$ dans le développement
+est déterminé par les mêmes valeurs de~$s, u$. Les
+génératrices de la développable viennent se développer suivant
+les tangentes à la courbe~$(R)$. Considérons une courbe~$(\Gamma)$ sur
+la développable, et la courbe correspondante~$(\Gamma_{1})$ dans le plan:
+les arcs homologues sur ces \Card{2} courbes sont égaux, de sorte
+que toute courbe tracée sur le plan roule sur la courbe correspondante
+de la développable. \emph{On peut imaginer que l'on ait
+enroulé sur la développable une feuille plane déformable; le
+mouvement du plan~$P$ consistera alors à dérouler cette feuille
+de façon qu'elle reste constamment tendue.} Un point quelconque
+de la feuille décrira une trajectoire orthogonale des plans
+tangents à la développable. Nous obtenons ainsi en quelque
+sorte la \emph{surface développante d'une développable} par la généralisation
+du procédé qui donne les développantes d'une courbe
+plane.
+
+Nous pouvons enfin examiner le mouvement du plan~$P$ au
+point de vue cinématique. Nous avons
+\[
+\frac{dX}{ds} = -\frac{a''}{T} \eta, \qquad
+\frac{dY}{ds} = -\frac{b''}{T} \eta, \qquad
+\frac{dZ}{ds} = -\frac{c''}{T} \zeta;
+\]
+et par suite les projections de la vitesse sur les axes~$\xi \eta \zeta$
+invariablement liés au plan~$P$ sont
+\[
+V_{\xi} = \sum a · \frac{dX}{ds} = 0, \quad
+V_{\eta} = \sum a' \frac{dY}{ds} = 0, \quad
+V_{\zeta} = \sum a'' \frac{dZ}{ds} = - \frac{1}{T} \eta.
+\]
+\DPtypo{le}{Le} mouvement instantané du plan~$P$ est une rotation autour de~$\omega \xi$
+tangente à~$(A)$, la rotation instantanée étant~$-\dfrac{1}{T}$. \emph{Le
+plan \DPtypo{osoulateur}{osculateur}~$P$ roule sur la courbe~$(A)$ en tournant autour
+de la tangente avec une vitesse de rotation égale à~$-\dfrac{1}{T}$.}
+%% -----File: 186.png---Folio 178-------
+
+La surface~$(S)$ engendrée par le mouvement précédent est
+une \emph{surface moulure}, ou \emph{surface de Monge}. Considérons dans le
+plan~$P$ une courbe~$(c)$ invariablement liée au système d'axes~$\omega \xi \eta$
+et sa développée~$(K)$. La \Ord{2}{e} nappe de la surface focale
+sera engendrée par cette développée~$(K)$ dans le mouvement du
+plan~$P$. C'est une surface moulure. Ainsi \emph{une des nappes de la
+développée d'une surface moulure est une développable, l'autre
+est une surface moulure}.
+
+\MarginNote{Cas
+particuliers.}
+Examinons le cas particulier où la développable enveloppe
+du plan~$P$ est un cylindre ou un cône.
+
+\ParItem{\Primo.} Si \emph{le plan~$P$ enveloppe un cylindre}, les tangentes
+aux trajectoires orthogonales sont parallèles aux plans de
+section droite, les trajectoires sont les développantes des
+sections droites; ce sont des lignes planes; \emph{les \Card{2} systèmes
+de lignes de courbure de la surface sont des courbes planes.
+Le plan~$P$ roule sur le cylindre de façon que son intersection
+avec le plan d'une section droite roule sur cette section
+droite. On peut encore engendrer la surface en considérant
+dans un plan une famille de courbes parallèles (qui sont ici
+les développantes de la section droite du cylindre), et en
+déplaçant chacune de ces courbes d'un mouvement de translation
+perpendiculaire au plan}.
+
+\ParItem{\Secundo.} Si \emph{le plan~$P$ enveloppe un cône} de sommet~$A$, considérons
+une trajectoire orthogonale rencontrant le plan~$P$ en~$M$,
+la tangente en~$M$ est perpendiculaire à~$A M$, donc la trajectoire
+orthogonale est une courbe tracée sur une sphère de centre~$A$\Add{.}
+Coupons alors le cône par une sphère de centre~$A$ et de
+%% -----File: 187.png---Folio 179-------
+rayon~$R$, soit~$(c)$ l'intersection, et considérons dans le plan~$P$
+le cercle~$(S)$ de centre~$A$ et de rayon~$R$. \emph{Le plan~$F$ roule
+sur le cône de façon que le cercle~$(S)$ roule sur la courbe~$(c)$\Add{.}}
+
+\MarginNote{Autres hypothèses.}
+Cherchons maintenant si les deux nappes de la développée
+d'une surface peuvent être des développables. La surface
+est alors surface moulure de \Card{2} manières; les \Card{2} systèmes
+de lignes de courbure sont des courbes planes. Les trajectoires
+orthogonales des plans~$P$, qui enveloppent l'une des nappes
+de la développée, constituant un des systèmes de lignes
+de courbure, doivent être planes. Soit $P'$~le plan de l'une
+d'elles. Les plans~$P$ sont tous normaux à une courbe située
+dans~$P'$; ils sont donc tous perpendiculaires à $P'$. Si donc
+les plans~$P$ ne sont pas parallèles, les plans~$P'$ le sont tous;
+les plans~$P$ enveloppent un cylindre, et les plans~$P'$ sont perpendiculaires
+aux génératrices de ce cylindre, ainsi que les
+normales à la surface; le \DPtypo{profit}{profil} situé dans un plan~$P$ et qui
+engendre la surface moulure est une parallèle aux génératrices
+du cylindre. Les surfaces obtenues sont donc des cylindres;
+la seconde nappe de la développée est une droite rejetée à
+l'infini.
+
+Si les plans~$P$ sont parallèles, on arrive à la même conclusion,
+car les plans~$P'$ enveloppent un cylindre.
+
+Le cas supposé est donc impossible.
+
+Supposons qu'une des nappes de la développée soit une
+développable, l'autre étant une courbe. La surface est une
+surface moulure qui s'obtient par le mouvement d'un profil
+situé dans le plan~$P$ qui enveloppe la développable. La \Ord{2}{e} nappe
+de la développée est engendrée dans ce mouvement par
+%% -----File: 188.png---Folio 180-------
+la développée du profil; pour que ce soit une courbe, il faut
+que la développée du profil soit un point, donc que ce profil
+soit un cercle; imaginons alors la sphère qui a ce profil
+pour grand cercle; elle est inscrite dans la surface; \emph{la surface
+est une enveloppe de sphères de rayon constant}. C'est une
+surface canal.
+
+\emph{Réciproquement toute enveloppe d'une famille de sphères
+égales satisfait à la condition précédente.} Soit la sphère
+\[
+\sum (x - a)^{2} - r^{2} = 0,
+\]
+la caractéristique a pour \Ord{2}{e} équation
+\[
+\sum (x - a)\, da = 0\DPtypo{;}{.}
+\]
+C'est un grand cercle de la sphère; les normales à la surface
+enveloppe sont dans le plan de ce cercle. L'une des nappes de
+la développée sera l'enveloppe des plans de ce cercle. Si
+nous considérons le lieu du centre de la sphère, le plan du
+grand cercle lui est constamment normal; \emph{la surface est engendrée
+par un cercle de rayon constant dont le centre décrit une
+courbe, et dont le plan reste constamment normal à cette
+courbe}.
+
+Enfin comme cas singulier, nous avons encore celui où
+l'une des nappes de la développée est une droite. La surface
+est alors de révolution autour de cette droite.
+
+
+\ExSection{VII}
+
+\begin{Exercises}
+\item[32.] \DPchg{Etudier}{Étudier} la congruence formée des droites tangentes à une
+sphère et normales à une même surface; étudier les surfaces
+normales à ces droites, et leurs lignes de courbure.
+
+\item[33.] \DPchg{Etudier}{Étudier} la congruence formée des droites normales à une surface
+dont une famille de lignes de courbure est située sur
+des sphères concentriques.
+
+\item[34.] Montrer que les surfaces moulures, dans le cas où l'une des
+nappes de la développée est un cylindre ou un cône, peuvent
+être \DPtypo{definies}{définies} par le mouvement d'un profil plan, de forme
+invariable, dont le plan reste constamment normal à un \DPtypo{cylindr}{cylindre}
+ou à un cône. \DPtypo{Prèciser}{Préciser} le mouvement de ce profil. Chercher si
+l'on peut dire quelque chose d'analogue pour les surfaces
+moulures générales.
+
+\item[35.] Montrer que les droites tangentes à deux quadriques homofocales
+constituent une congruence de normales. Si on fait réfléchir
+toutes ces droites, considérées comme des rayons lumineux,
+sur une autre quadrique homofocale aux deux premières,
+quelles seront les multiplicités focales de cette seconde
+congruence?
+
+\item[36.] \DPchg{Etant}{Étant} données deux surfaces homofocales du second \DPtypo{degre}{degré} et un
+plan~$P$, si on mène par les droites du plan~$P$ des plans tangents
+aux deux surfaces, les droites qui joignent les points
+de contact correspondants sont normales à une famille de surfaces
+parallèles. Soit~$(\delta)$ la droite qui contient les pôles
+du plan~$P$ par rapport aux deux quadriques homofocales, et
+$(d')$~la droite du plan~$P$ qui correspond à une droite~$(d)$ de
+la congruence de normales considérée. Le plan mené par~$(\delta)$
+perpendiculairement à~$(d')$ coupe~$(d)$ en un point~$m$. Le lieu
+du point~$m$ est l'une des surfaces cherchées: c'est une cyclide.
+Les développables de la congruence découpent sur les
+surfaces homofocales des réseaux conjugués.
+
+\item[37.] On considère la congruence des droites de l'espace sur lesquelles
+trois plans formant un trièdre trirectangle déterminent
+des segments invariables. Démontrer que c'est une congruence
+de normales et déterminer les surfaces normales aux
+droites de la congruence. Déterminer les points focaux sur
+une quelconque de ces droites. Déterminer les cônes directeurs
+des développables de la congruence.
+
+\item[38.] \DPtypo{Demontrer}{Démontrer} qu'il existe des congruences (isogonales) telles
+que les plans focaux forment un dièdre constant. Quelle est
+la propriété des arêtes de rebroussement des développables de
+la congruence par rapport aux nappes de la surface focale qui
+les contiennent? Chercher l'équation différentielle de ces
+courbes sur la surface focale supposée donnée. Que peut-on
+dire du cas où l'une des nappes de la multiplicité focale est
+une développable, une courbe, une sphère?
+
+\item[39.] Si on considère une famille de sphères dont le lieu des
+centres~$\omega$ est une courbe plane~$C$, et dont les rayons sont
+proportionnels aux distances des centres~$\omega$ à une droite fixe~$\Delta$
+du plan de la courbe~$C$, démontrer que l'enveloppe de ces
+sphères a toutes ses lignes de courbure planes. Que peut-on
+dire des plans de ces lignes de courbure? Réciproquement,
+comment peut-on obtenir toutes les surfaces canaux dont toutes
+les lignes de courbure sont planes?
+\end{Exercises}
+%% -----File: 189.png---Folio 181-------
+
+
+\Chapitre{VIII}{Les Congruences de Droites et les Correspondances Entre
+Deux Surfaces.}
+
+\Section{Nouvelle représentation des congruences.}
+{1\Add{.}}{} Dans ce qui précède, nous avons défini une congruence
+par son support, et en donnant la direction de la
+droite ou des droites~$(D)$ qui passent par chaque point
+du support. On peut plus généralement, et ce sera préférable
+au point de vue projectif, considérer \Card{2} surfaces supports se
+correspondant point par point, les droites de la congruence
+étant celles qui joignent les points homologues des deux surfaces.
+En réalité, les \Card{2} surfaces se correspondront élément
+de contact à élément de contact, et en même temps que la congruence
+des droites joignant les points homologues, on pourra
+considérer celle des intersections des plans tangents homologues.
+
+Il est naturel alors d'employer des coordonnées homogènes\Add{.}
+Soient $M(x\Add{,}y\Add{,}z\Add{,}t)$ et $\DPtypo{M}{M_{1}}(x_{1}\Add{,} y_{1}\Add{,} z_{1}\Add{,} t_{1})$ les points homologues sur les
+\Card{2} surfaces; on pourra définir la congruence par les équations
+\[
+X = x + \rho x_{1}, \qquad
+Y = y + \rho y_{1}, \qquad
+Z = z + \rho z_{1}, \qquad
+T = t + \rho t_{1}.
+\]
+Soient de même $u, v, w, r$ les coordonnées tangentielles d'un plan
+tangent à la \Ord{1}{ère} surface, $u_{1}, v_{1}, w_{1}, r_{1}$ celles du plan tangent
+homologue à la \Ord{2}{e} surface. La congruence pourra être définie
+au point de vue tangentiel par les équations
+\[
+U = u + \rho u_{1}, \qquad
+V = v + \rho v_{1}, \qquad
+W = w + \rho w_{1}, \qquad
+R = r + \rho r_{1}.
+\]
+
+Soient $(S), (S_{1})$ les \Card{2} surfaces supports; les systèmes conjugués
+%% -----File: 190.png---Folio 182-------
+sur ces surfaces étant invariants, d'après leur définition
+même, par toute transformation projective, nous sommes
+conduits à étudier leurs relations. Soient
+\begin{align*}
+\Tag{(S)}
+x &= f(\lambda,\mu), &
+y &= g(\lambda,\mu), &
+z &= h(\lambda,\mu), &
+t &= \DPtypo{h}{k}(\lambda,\mu); \\
+\Tag{(S_{1})}
+x_{1} &= f_{1}(\lambda,\mu), &
+y_{1} &= g_{1}(\lambda,\mu), &
+z_{1} &= h_{1}(\lambda,\mu), &
+t_{1} &= \DPtypo{k}{h}_{1}(\lambda,\mu);
+\end{align*}
+les équations des deux surfaces.
+
+Le choix des paramètres $\lambda\Add{,} \mu$ est fixé par le Théorème suivant:
+\emph{Quand \Card{2} surfaces $(S)\Add{,} (S_{1})$ se correspondent point par point, il
+existe sur~$(S)$ un réseau conjugué qui correspond à un réseau
+conjugué de~$S_{1}$, et en général il n'en existe qu'un.} Soient $d\lambda,
+d\mu$ et~$\delta\lambda, \delta\mu$ les paramètres définissant \Card{2} directions conjuguées
+sur~$(S)$, elles sont conjuguées harmoniques par rapport aux
+directions
+\[
+E'\, d\lambda^{2} + 2F'\, d\lambda · d\mu + G'\, d\mu^{2} = 0.
+\]
+De même sur~$(S_{1})$, \Card{2} directions conjuguées sont conjuguées harmoniques
+par rapport aux directions
+\[
+E'_{1}\, d\lambda^{2} + 2F'_{1}\, d\lambda · d\mu + G'_{1}\, d\mu^{2} = 0.
+\]
+Chercher un système conjugué commun revient donc à chercher un
+couple de points conjugués par rapport à \Card{2} couples donnés par
+\Card{2} équations quadratiques; si les \Card{2} formes quadratiques n'ont
+pas de facteur commun, il y a un couple et un seul répondant
+à la question. Or\Add{,} les \Card{2} équations précédentes définissent les
+lignes asymptotiques des \Card{2} surfaces; si donc \Card{2} surfaces se
+correspondent point par point d'une façon telle qu'il n'y ait
+pas sur~$(S)$ une famille d'asymptotiques correspondant à une
+famille d'asymptotiques de~$(S_{1})$, il existe un système conjugué
+%% -----File: 191.png---Folio 183-------
+de~$(S)$ et un seul qui correspond à un système conjugué de~$(S_{1})$\Add{.}
+Il est défini par l'équation:
+\[
+\begin{vmatrix}
+E'\, d\lambda + F'\, d\mu    & F'\, d\lambda + G'\, d\mu \\
+E_1'\,d\lambda + F_1'\, d\mu & F_1'\, d\lambda + G_1'\, d\mu
+\end{vmatrix} = 0\Add{.}
+\]
+Il y aura impossibilité si les formes ont un facteur commun,
+et indétermination si les \Card{2} facteurs sont communs, c'est-à-dire,
+si les lignes asymptotiques se correspondent sur les
+deux surfaces. \DPchg{Ecartant}{Écartant} ces cas d'exception, nous supposerons
+que les paramètres $\lambda\Add{,} \mu$ correspondent à ce système conjugué
+commun.
+
+\Section{Emploi des coordonnées homogènes.}
+{2.}{} Nous allons reprendre les formules usuelles et
+voir ce qu'elles deviennent en coordonnées homogènes.
+
+%[** TN: Removed several ". ---" start-of-paragraph markers]
+Une \emph{courbe} en coordonnées homogènes est définie
+par \Card{4} équations
+\[
+x = f(\lambda), \qquad
+y = g(\lambda), \qquad
+z = h(\lambda), \qquad
+t = k(\lambda).
+\]
+La tangente au point $M(x,y,z,t)$ joint le point~$M$ au point
+\[%[** TN: Two large expressions are in-line in original]
+M'\left(\dfrac{dx}{d\lambda}, \dfrac{dy}{d\lambda}, \dfrac{dz}{d\lambda}, \dfrac{dt}{d\lambda}\right).
+\]
+Le plan osculateur passe par la droite~$MM'$ et
+par le point
+\[
+M''\left(\dfrac{d^{2}x}{d\lambda^{2}}, \dfrac{d^{2}y}{d\lambda^{2}}, \dfrac{d^{2}z}{d\lambda^{2}}, \dfrac{d^{2}t}{d\lambda^{2}}\right)\Add{.}
+\]
+
+Corrélativement une \emph{développable} sera l'enveloppe du
+plan~$P$
+\[
+u = f(\lambda), \qquad
+v = g(\lambda), \qquad
+w = h(\lambda), \qquad
+r = k(\lambda).
+\]
+La caractéristique (génératrice) sera l'intersection du plan~$P$
+et du plan
+$P'\left(\dfrac{du}{d\lambda},
+         \dfrac{dv}{d\lambda},
+         \dfrac{dw}{d\lambda},
+         \dfrac{dr}{d\lambda}\right)$.
+Le point de contact avec l'arête
+de rebroussement sera en outre dans le plan
+$P''\left(\dfrac{d^{2}u}{d\lambda^{2}},
+          \dfrac{d^{2}v}{d\lambda^{2}},
+          \dfrac{d^{2}w}{d\lambda^{2}},
+          \dfrac{d^{2}r}{d\lambda^{2}}\right)$.
+%% -----File: 192.png---Folio 184-------
+
+Une \emph{surface} quelconque peut se définir au point de
+vue ponctuel par
+\[
+x = f(\lambda, \mu), \qquad
+y = g(\lambda, \mu), \qquad
+z = h(\lambda, \mu), \qquad
+t = k(\lambda, \mu);
+\]
+et au point de vue tangentiel par
+\[
+u = F(\lambda, \mu), \qquad
+v = G(\lambda, \mu), \qquad
+w = H(\lambda, \mu), \qquad
+r = K(\lambda, \mu).
+\]
+On peut définir le \emph{plan tangent} en fonction du point de contact
+$(x, y, z, t)$. Ce plan contient le point, donc
+\[
+\sum ux = 0;
+\]
+il contient les tangentes aux courbes $\lambda = \cte$\DPtypo{.}{,} $\mu = \cte$, donc
+les points
+$\left(\dfrac{\dd x}{\dd \mu}, \dfrac{\dd y}{\dd \mu},
+       \dfrac{\dd z}{\dd \mu}, \dfrac{\dd t}{\dd \mu}\right)$
+et
+$\left(\dfrac{\dd x}{\dd \lambda}, \dfrac{\dd y}{\dd \lambda},
+       \dfrac{\dd z}{\dd \lambda}, \dfrac{\dd t}{\dd \lambda}\right)$.
+\[
+\sum u \frac{\dd x}{\dd \lambda} = 0, \qquad
+\sum u \frac{\dd x}{\dd \mu} = 0;
+\]
+et nous avons ainsi \Card{3} équations définissant des quantités
+proportionnelles à $u, v, w, r$. On peut écrire l'équation ponctuelle
+du plan tangent au point $(x\Add{,}y\Add{,}z\Add{,}t)$
+\[
+\begin{vmatrix}
+X  &   Y  &  Z  &  T \\
+x  &   y  &  z  &  t \\
+\mfrac{\dd x}{\dd \lambda} &
+\mfrac{\dd y}{\dd \lambda} &
+\mfrac{\dd z}{\dd \lambda} &
+\mfrac{\dd t}{\dd \lambda} \\
+%
+\mfrac{\dd x}{\dd \mu} &
+\mfrac{\dd y}{\dd \mu} &
+\mfrac{\dd z}{\dd \mu} &
+\mfrac{\dd t}{\dd \mu}
+\end{vmatrix} = 0.
+\]
+
+\emph{Corrélativement} on définira un \emph{point} de la surface en
+fonction du plan tangent en ce point, au moyen des équations:
+\[
+\sum ux = 0, \qquad
+\sum x \frac{\dd u}{\dd \lambda} = 0, \qquad
+\sum x \frac{\dd u}{\dd \mu} = 0;
+\]
+de sorte qu'en définitive, on peut définir l'un des éléments
+point, plan tangent, en fonction de l'autre au moyen des formules
+\[
+\sum ux = 0, \qquad
+\sum u\, dx = 0, \qquad
+\sum x\, du = 0.
+\]
+
+Proposons-nous maintenant d'\emph{exprimer que \Card{2} directions
+%% -----File: 193.png---Folio 185-------
+$MT(d\lambda, d\mu)$ et $MS(\delta \lambda\DPtypo{.}{,} \delta \mu)$ sont conjuguées}. Ces \Card{2} directions
+sont conjuguées si, le point de contact du plan tangent se déplaçant
+dans la direction $MT, MS$ est la caractéristique de ce
+plan tangent. Or\Add{,} cette caractéristique est
+\[
+\sum uX = 0, \qquad
+\sum du·X = 0;
+\]
+la droite~$MS$ est définie par le point $(x, y, z, t)$ et le point
+$(\delta x, \delta y, \delta z, \delta t)$. Pour exprimer que $MS$~est la caractéristique, il
+faut exprimer que les \Card{2} points précédents sont sur la caractéristique,
+ce qui donne:
+\begin{alignat*}{2}
+&\sum ux = 0, \qquad
+&&\sum du·x = 0; \\
+&\sum u · \delta x = 0, \qquad
+&&\sum du · \delta x = 0;
+\end{alignat*}
+les \Card{3} \Ord{1}{ères} équations sont vérifiées, nous avons donc la condition
+unique
+\[
+\sum du · \delta x = 0,
+\]
+ou la condition symétrique
+\[
+\sum \delta u · dx = 0.
+\]
+En particulier nous trouvons la condition pour qu'une direction
+soit conjuguée d'elle-même, c'est-à-dire soit direction
+asymptotique
+\[
+\sum du · dx = 0.
+\]
+
+Exprimons alors que les directions $\lambda = \cte$, $\mu = \cte$ forment
+un réseau conjugué. Nous avons
+\[
+\Tag{(1)}
+\sum · \frac{\dd u}{\dd \lambda} · \frac{\dd x}{\dd \mu}  = 0.
+\]
+Cette condition peut se transformer: l'équation
+\[
+\sum u \frac{\dd x}{\dd \mu} = 0
+\]
+différentiée par rapport à $\lambda$ donne
+\[
+\sum \frac{\dd u}{\dd \lambda} · \frac{\dd x}{\dd \mu}
+  + \sum u \frac{\dd^{2} x}{\dd \lambda\, \dd \mu} = 0;
+\]
+%% -----File: 194.png---Folio 186-------
+et \Eq{(1)}~s'écrit
+\[
+\Tag{(2)}
+\sum u \frac{\dd^{2} x}{\dd \lambda·\dd \mu} = 0.
+\]
+De même l'équation
+\[
+\sum u \frac{\dd x}{\dd \lambda} = 0
+\]
+différentiée par rapport à~$\mu$ donne
+\[
+\sum \frac{\dd u}{\dd \mu} · \frac{\dd x}{\dd \lambda}
+  + \sum u \frac{\dd^{2} x}{\dd \lambda\, \DPtypo{d}{\dd}\mu} = 0,
+\]
+et \Eq{(1)}~peut s'écrire
+\[
+\Tag{(3)}
+\sum \frac{\dd u}{\dd \mu} · \frac{\dd x}{\dd \lambda} = 0.
+\]
+En partant de l'une des relations
+\[
+\sum x \frac{\dd u}{\dd \lambda} = 0, \qquad
+\sum x \frac{\dd u}{\dd \mu} = 0,
+\]
+on obtiendrait la relation
+\[
+\Tag{(4)}
+\sum x \frac{\dd^{2} u}{\dd \lambda · \dd \mu} = 0.
+\]
+Ces \Card{4} équations \Eq{(1)}\Add{,} \Eq{(2)}\Add{,} \Eq{(3)}\Add{,} \Eq{(4)} dépendent simultanément des éléments
+ponctuel et tangentiel. En exprimant $u, v, w, r$ en fonction
+de~$x, y, z, t$, on obtient la condition en coordonnées ponctuelles:
+\[
+\Tag{(5)}
+\begin{vmatrix}
+x & \mfrac{\dd x}{\dd \lambda}
+  & \mfrac{\dd x}{\dd \mu}
+  & \mfrac{\dd^{2} x }{\dd\lambda\, \dd\mu}
+\end{vmatrix} = 0.
+\]
+Dans cette relation~\Eq{(5)}, le premier membre représente, par
+abréviation, le déterminant dont la première ligne serait la
+ligne écrite entre les deux traits verticaux, et dont les
+trois autres lignes se déduiraient de celle-là en~$y$ remplaçant~$x$
+par~$y, z, t$. Cette notation sera employée couramment dans la
+suite.
+
+Lorsque $t = \cte$, la condition~\Eq{(5)} se réduit à la condition
+connue
+\[
+\begin{vmatrix}
+\mfrac{\dd x}{\dd \lambda} &
+\mfrac{\dd x}{\dd \mu} &
+\mfrac{\dd^{2} x}{\dd \lambda · \dd \mu}
+\end{vmatrix} = F' = 0.
+\]
+
+La condition~\Eq{(5)} peut s'interpréter autrement: il existe
+%% -----File: 195.png---Folio 187-------
+une même relation linéaire et homogène entre les éléments correspondants
+des lignes
+\begin{align*}%[** TN: Filled in last three equations]
+\frac{\dd^2 x}{\dd \lambda · \dd \mu}
+  &= L \frac{\dd x}{\dd \lambda} + M \frac{\dd x}{\dd \mu} + N x\Add{,} \\
+\frac{\dd^2 y}{\dd \lambda · \dd \mu}
+  &= L \frac{\dd y}{\dd \lambda} + M \frac{\dd y}{\dd \mu} + N y\Add{,} \\
+\frac{\dd^2 z}{\dd \lambda · \dd \mu}
+  &= L \frac{\dd z}{\dd \lambda} + M \frac{\dd z}{\dd \mu} + N z\Add{,} \\
+\frac{\dd^2 t}{\dd \lambda · \dd \mu}
+  &= L \frac{\dd t}{\dd \lambda} + M \frac{\dd t}{\dd \mu} + N t\Add{,}
+\end{align*}
+c'est-à-dire: \emph{les \Card{4} coordonnées homogènes~$x,y,z,t$ satisfont à
+une même équation linéaire aux dérivées partielles de la forme}:
+\[
+\frac{\dd^2 f}{\dd \lambda· \dd \mu}
+  = L \frac{\dd f}{\dd \lambda} + M \frac{\dd f}{\dd \mu} + Nf.
+\]
+En \DPtypo{operant}{opérant} au point de vue tangentiel, on verrait de même que
+\emph{la condition cherchée est que $u,v,w,r$~soient intégrales d'une
+même équation}:
+\[
+\frac{\dd^2f}{\dd \lambda\, \dd \mu}
+  = P \frac{\dd f}{\dd \lambda} + Q \frac{\dd f}{\dd \mu} + Rf.
+\]
+On montrerait sans peine que si $x,y,z,t$\DPtypo{,}{} ou $u,v,w,r$ satisfont
+à une équation de la forme précédente, elles ne satisfont qu'à
+une seule.
+
+\Paragraph{Remarque.} En coordonnées cartésiennes, $t = 1$, $r = 1$, et on a
+$R = N = 0$.
+
+Considérons une \emph{surface réglée}; les équations d'une
+génératrice, joignant le point $M(x\Add{,}y\Add{,}z\Add{,}t)$ au point $M_1 (x_1, y_1, z_1, t_1)$
+sont:
+\[
+X = x + \rho x_1, \qquad
+Y = y + \rho y_1, \qquad
+Z = z + \rho z_1, \qquad
+T = t + \rho t_1.
+\]
+Supposons la surface \emph{développable}; les plans tangents aux points
+$(x,y,z,t)$ et $(x_1, y_1, z_1, t_1)$ sont les mêmes. Or\Add{,} le plan tangent en~$M$
+%% -----File: 196.png---Folio 188------
+passant par la génératrice et par la tangente à la courbe
+$\rho = \Err{\cte}{0}$ contient le point $(dx, dy, dz, dt)$. De même le plan tangent
+en~$M_1$, contient le point $(dx_1, dy_1, dz_1, dt_1)$. La condition
+pour que les plans soient confondus est donc
+\[
+\begin{vmatrix}
+x  &  x_1  &  dx  &  dx_1
+\end{vmatrix} = 0.
+\]
+
+Si nous définissions la surface en coordonnées tangentielles,
+nous arriverions de même à la condition
+\[
+\begin{vmatrix}
+u  &  u_1  &  du  &  du_1
+\end{vmatrix} = 0.
+\]
+
+Voyons enfin une \emph{congruence}: nous pouvons encore la
+représenter par les équations
+\[
+X = x + \rho x_1, \qquad
+Y = y + \rho y_1, \qquad
+Z = z + \rho z_1, \qquad
+T = t + \rho t_1;
+\]
+mais ici $x, y, z, t$ et $x_1, y_1, z_1, t_1$, sont fonctions de deux paramètres
+arbitraires $(\lambda, \mu)$. Cherchons les \emph{éléments focaux}. Soit $F$ un
+foyer d'une droite~$D(\lambda, \mu)$. Soit $\rho$ la valeur du paramètre qui
+correspond à ce point. Toutes les surfaces réglées de la congruence
+qui contiennent la droite~$D$ ont en ce point~$F$ même
+plan tangent. Considérons en particulier les surfaces $\lambda = \cte$
+et $\mu = \cte[]$. Les plans tangents à ces surfaces contiennent respectivement
+les points $(x, y, z, t)$\Add{,} $(x_1, y_1, z_1, t_1)$\Add{,}
+$\left(\dfrac{\dd x}{\dd \mu} + \rho \dfrac{\dd x_1}{\dd \mu},\dots\right)$
+et $(x, y, z, t)$\Add{,} $(x_1, y_1, z_1, t_1)$\Add{,}
+$\left(\dfrac{\dd x}{\dd \lambda} + \rho \dfrac{\dd x\Add{_1}}{\dd \lambda},\dots\right)$. La condition pour
+que ces plans \DPtypo{coincident}{coïncident}, c'est-à-dire l'\emph{équation aux points
+focaux}, est donc
+\[
+\begin{vmatrix}
+x  &  x_1  &
+\mfrac{\dd x}{\dd \lambda} + \rho\mfrac{\dd x_1}{\dd \lambda} &
+\mfrac{\dd x}{\dd \mu} + \rho\mfrac{\dd x_1}{\dd \mu}
+\end{vmatrix} = 0;
+\]
+On trouvera de même l'\emph{équation aux plans focaux}:
+\[
+\begin{vmatrix}
+u & u_1 &
+\mfrac{\dd u}{\dd \lambda} + \rho \mfrac{\dd u_1}{\dd \lambda} &
+\mfrac{\dd u}{\dd \mu} + \rho \mfrac{\dd \DPtypo{u}{u_{1}}}{\dd \mu}
+\end{vmatrix} = 0.
+\]
+%% -----File: 197.png---Folio 189-------
+
+\Section{Correspondances spéciales.}
+{3.}{} Nous allons étudier la \emph{correspondance entre \Card{2} points
+$M\Add{,} \Err{M}{M_{1}}$, de \Card{2} surfaces telle que les développables de la congruence
+des droites $M\Err{M}{M_{1}}$ coupent les \Card{2} surfaces suivant les
+\Card{2} réseaux conjugués qui se correspondent}. Nous supposerons que
+les paramètres $\lambda\Add{,} \mu$ qui fixent la position d'un point sur chacune
+des surfaces sont précisément tels quo les courbes conjuguées
+homologues soient $\lambda = \cte$ et $\mu = \cte[]$. Les courbes
+$\lambda = \cte\Add{,} \mu = \cte$ sont conjuguées sur la \Ord{1}{ère} surface~$(S)$ donc
+$x, y, z, t$ satisfont à une même équation différentielle
+\[
+\Tag{(1)}
+\frac{\dd^2 f}{\dd \lambda\, \dd \mu}
+  = P \frac{\dd f}{\dd \lambda} + Q \frac{\dd f}{\dd \mu} + R f;
+\]
+de même les courbes $\lambda = \cte$ et $\mu = \cte$ étant conjuguées sur la
+\Ord{2}{e} surface~$(S_1)$, $x_1, y_1, z_1, t_1$\DPtypo{,}{} satisfont à une même équation différentielle
+\[
+\Tag{(2)}
+\frac{\dd^2 f}{\dd \lambda\, \dd \mu}
+  = P_1 \frac{\dd f}{\dd \lambda} + Q_1 \frac{\dd f}{\dd \mu} + R_1 f.
+\]
+Exprimons maintenant que les développables de la congruence
+correspondent à $\lambda = \cte$ et $\mu = \cte[]$. Si nous représentons la
+congruence par les équations
+\[
+X = x + \rho x, \qquad
+Y = y + \rho y, \qquad
+Z = z + \rho z, \qquad
+T = t + \rho t,
+\]
+les développables sont données par l'équation
+\[
+\begin{vmatrix}
+x  &  x_1  &  dx  &  dx_1
+\end{vmatrix} = 0.
+\]
+Or,
+\begin{alignat*}{5}
+dx &= \frac{\dd x}{\dd \lambda}\, d\lambda
+   &&+ \frac{\dd x}{\dd \mu}\, d\mu,\qquad &
+dy &= \dots\dots,\qquad &
+dz &= \dots\dots,\qquad &
+dt &= \dots\dots, \\
+%
+dx_1 &= \frac{\dd x_1}{\dd\lambda}\, d\lambda
+   &&+ \frac{\dd x_1}{\dd\mu}\, d\mu,\qquad &
+dy_1 &= \dots\dots,\qquad &
+dz_1 &= \dots\dots,\qquad &
+dt_1 &= \dots\dots;
+\end{alignat*}
+\iffalse%%%%%[** TN: Code follows for three sets of equations above]
+dy &= \frac{\dd y}{\dd \lambda}\, d\lambda + \frac{\dd y}{\dd \mu}\, d\mu, &
+dz &= \frac{\dd z}{\dd \lambda}\, d\lambda + \frac{\dd z}{\dd \mu}\, d\mu, &
+dt &= \frac{\dd t}{\dd \lambda}\, d\lambda + \frac{\dd t}{\dd \mu}\, d\mu, \\
+%
+dy_1 &= \frac{\dd y_1}{\dd\lambda}\, d\lambda + \frac{\dd y_1}{\dd\mu}\, d\mu, &
+dz_1 &= \frac{\dd z_1}{\dd\lambda}\, d\lambda + \frac{\dd z_1}{\dd\mu}\, d\mu, &
+dt_1 &= \frac{\dd t_1}{\dd\lambda}\, d\lambda + \frac{\dd t_1}{\dd\mu}\, d\mu,
+\fi %%%% End of code for elided equations
+et l'équation précédente devant être vérifiée pour $d\lambda = 0$,
+$d\mu = 0$, nous avons les conditions
+%% -----File: 198.png---Folio 190-------
+\begin{align}
+\Tag{(3)}
+\begin{vmatrix}
+x  &  x_1  &  \mfrac{\dd x}{\dd \lambda}  &  \mfrac{\dd x_1}{\dd \lambda}
+\end{vmatrix} = 0, \\
+\Tag{(4)}
+\begin{vmatrix}
+x  &  x_1  &  \mfrac{\dd x}{\dd \mu}  &  \mfrac{\dd x_1}{\dd \mu}
+\end{vmatrix} = 0.
+\end{align}
+Il existe une même relation linéaire et homogène entre les
+éléments des colonnes, donc
+\begin{alignat*}{2}
+\Tag{(5)}
+Ax + B \frac{\dd x}{\dd \lambda}
+  &= A_1 x_1 + B_1 \frac{\dd x_1}{\dd \lambda},\qquad
+  && \text{et les analogues}\quad\dots\dots\Add{,}
+ \\
+\Tag{(6)}
+Cx + D \frac{\dd x}{\dd \mu}
+  &= C_1 x_1 + D_1 \frac{\dd x_1}{\dd \mu},\qquad
+  && \text{et les analogues}\quad\dots\dots\Add{.}
+\end{alignat*}
+
+\Paragraph{\DPchg{\Ord{1}{er} Cas}{Premier Cas}.} Voyons d'abord ce qui arrive si l'un des \Card{4} coefficients
+$B\Add{,} B_1\Add{,} D\Add{,} D_1$ est nul. Soit $B_1 = 0$. Alors les équations~\Eq{(5)}
+expriment que le point $M_1(x_1\Add{,} y_1\Add{,} z_1\Add{,} t_1)$ est sur la droite qui
+joint les points $M(x\Add{,}y\Add{,}z\Add{,}t)$ et
+\[%[** TN: Large in-line expression in original]
+M\left(\dfrac{\dd x}{\dd \lambda}, \dfrac{\dd y}{\dd \lambda}, \dfrac{\dd z}{\dd \lambda}, \dfrac{\dd t}{\dd \lambda}\right).
+\]
+La droite~$MM_1$
+est tangente à la courbe~$\mu = \cte$ tracée sur la surface~$(S)$.
+Toutes les droites~$MM_1$ sont ainsi tangentes à la surface~$(S)$
+qui est une des nappes de la surface focale de la congruence.
+Sur la surface~$(S)$ les courbes $\mu = \cte$ correspondent à une
+famille de développables, et par suite les courbes $\lambda = \cte$
+conjuguées des précédentes correspondent à la \Ord{2}{e} famille. Il
+nous faut alors chercher comment on peut définir~$(S_1)$ pour
+que cette surface soit coupée suivant un réseau conjugué par
+les développables de la congruence. Les équations~\Eq{(5)} peuvent
+s'écrire dans le cas considéré
+\[
+x_1 = Ax + B \frac{\dd x}{\dd \lambda}, \quad
+y_1 = Ay + B \frac{\dd y}{\dd \lambda}, \quad
+z_1 = Az + B \frac{\dd z}{\dd \lambda}, \quad
+t_1 = At + B \frac{\dd t}{\dd \lambda}\Add{.}
+\]
+Posons
+\[
+x = \theta X, \qquad
+y = \theta Y, \qquad
+z = \theta Z, \qquad
+t = \theta T;
+\]
+nous avons alors
+\[
+x_1 = A \theta X + B \left(\theta \frac{\dd x}{\dd \lambda}
+  + X \frac{\dd \theta}{\dd \lambda}\right), \quad
+y_1 = \dots\dots, \quad
+z_1 = \dots\dots, \quad
+t_1 = \dots\dots;
+\]
+déterminons la fonction~$\theta$ par la relation
+\[
+A \theta + B \frac{d \theta}{d \lambda} = 0,
+\]
+%% -----File: 199.png---Folio 191-------
+ce qui est toujours possible. Nous avons
+\[
+x_1 = A \frac{\dd x}{\dd \lambda}, \qquad
+y_1 = A \frac{\dd y}{\dd \lambda}, \qquad
+z_1 = A \frac{\dd z}{\dd \lambda}, \qquad
+t_1 = A \frac{\dd t}{\dd \lambda};
+\]
+et comme les coordonnées homogènes ne sont définies qu'à un
+facteur près, nous pouvons écrire
+\[
+\Tag{(7)}
+x_1 = \frac{\dd x}{\dd \lambda}, \qquad
+y_1 = \frac{\dd y}{\dd \lambda}, \qquad
+z_1 = \frac{\dd z}{\dd \lambda}, \qquad
+t_1 = \frac{\dd t}{\dd \lambda}.
+\]
+Alors, d'après ces relations, l'équation différentielle~\Eq{(1)}
+s'écrit
+\[
+\Tag{(8)}
+\frac{\dd x_1}{\dd \mu} = Px_1 + Q \frac{\dd x}{\dd \mu} + Rx,
+\]
+condition de la forme~\Eq{(6)}. Les équations \Eq{(3)}~et~\Eq{(4)} sont alors
+vérifiées. Différentions la relation~\Eq{(8)} par rapport à~$\lambda$
+\[
+\frac{\dd^2 x_1}{\dd \lambda\, \dd \mu}
+  = \frac{\dd P}{\dd \lambda} x_1 + P \frac{\dd x_1}{\dd \lambda}
+  + \frac{\dd Q}{\dd \lambda} · \frac{\dd x}{\dd \mu}
+  + Q \frac{\dd^2 x}{\dd \lambda\, \dd \mu}
+  + \frac{\dd R}{\dd \lambda x} + R \frac{\dd x}{\dd \lambda}.
+\]
+Mais, $\Err{x}{x_{1}}$~vérifiant l'équation~\Eq{(2)}, nous avons
+\[
+\frac{\dd^2 x_1}{\dd \lambda\, \dd \mu}
+  = P_1 \frac{\dd x_1}{\dd \lambda} + Q_1 \frac{\dd x_1}{\dd \mu} + R_1 x_1,
+\]
+et nous obtenons ainsi
+\[
+\Tag{(9)}
+P_1 \frac{\dd x_1}{\dd \lambda} + Q_1 \frac{\dd x_1}{\dd \mu} + R_1 x_1
+  = \frac{\dd P}{\dd \lambda} x_1
+  + P \frac{\dd x_1}{\dd \lambda}
+  + \frac{\dd Q}{\dd \lambda}\Err{}{\, \frac{dx}{d\mu}}
+  + Q \frac{\dd^2 x}{\dd \lambda\, \dd \mu}
+  + \frac{\dd R}{\dd \lambda} · x
+  + R \frac{\dd x}{\dd \lambda};
+\]
+\Eq{(8)}\Add{,}~\Eq{(9)} sont \Card{2} équations en $x$~et~$\dfrac{\dd x}{\dd \mu}$. Si on peut les résoudre, on
+en peut tirer $x$ en particulier, en fonction linéaire de~$x_1$,
+$\dfrac{\dd x_1}{\dd \lambda}$\Add{,}~et~$\dfrac{\dd \Err{x}{x_1}}{\dd \mu}$;
+car~$\dfrac{\dd^2 x}{\dd \lambda\, \dd \mu}\Err{}{=\dfrac{\dd x_1}{\dd \mu}}$\Err{}{ et $\dfrac{\dd x}{\dd\lambda} = x_1$} s'exprime en fonction linéaire de ces \Card{3} quantités;
+donc le point $M(x, y, z, t)$ se trouve dans le plan des \Card{3} points
+$(x_1, y_1, z_1, t_1)$\Add{,}
+%[** TN: Added elided components in next two points]
+$\left(\dfrac{\dd x_{1}}{\dd\lambda}, \dfrac{\dd y_{1}}{\dd\lambda},
+       \dfrac{\dd z_{1}}{\dd\lambda}, \dfrac{\dd t_{1}}{\dd\lambda}\right)$\Add{,}
+$\left(\dfrac{\dd x_{1}}{\dd \mu}, \dfrac{\dd y_{1}}{\dd \mu},
+       \dfrac{\dd z_{1}}{\dd \mu}, \dfrac{\dd t_{1}}{\dd \mu}\right)$,
+c'est-à-dire dans le
+plan tangent en~$M_1$, à la surface~$(S_1)$. La droite~$MM_1$ est donc aussi
+tangente à~$(S_1)$, et $(S_1)$~est la \Ord{2}{e} nappe de la surface focale.
+Nous avons ainsi établi une \emph{correspondance point par point entre
+les \Card{2} nappes de la surface focale d'une congruence}.
+%% -----File: 200.png---Folio 192-------
+
+\emph{\DPchg{Ecartons}{Écartons} ce cas}; il faut alors supposer que les équations
+\Eq{(8)}\Add{,}~\Eq{(9)} ne sont pas résolubles en $x$~et~$\dfrac{\dd x}{\dd \mu}$; ce qui exige que
+l'on ait
+\[
+\begin{vmatrix}
+Q                          & R  \\
+\mfrac{\dd Q}{\dd \lambda} & \mfrac{\dd R}{\dd \lambda}
+\end{vmatrix} = 0,
+\]
+ou
+\[
+Q \frac{\dd R}{\dd \lambda} - R \frac{\dd Q}{\dd \lambda} = 0;
+\]
+ce qui exprime que $\dfrac{Q}{R}$ est fonction de $\mu$ seulement
+\[
+R = Q \psi(\mu).
+\]
+Reprenons alors la relation~\Eq{(8)}, et multiplions les \Card{4} coordonnées
+$x\Add{,}y\Add{,}z\Add{,}t$ par un facteur fonction de~$\mu$ de façon à simplifier
+la relation~\Eq{(8)}, qui s'écrit
+\[
+\frac{\dd x_1}{\dd \mu}
+  = Px_1 + Q \left[ \frac{\dd x}{\dd \mu} + x \psi(\mu)\right].
+\]
+On peut multiplier~$x$ par un facteur~$\omega$ tel que l'expression entre
+crochets se réduise~$\theta\omega \dfrac{\dd x}{\dd \mu}$; comme ce facteur~$\omega$ ne dépend pas de~$\lambda$,
+les équations~\Eq{(7)} subsistent, et nous avons des relations de la forme
+\[
+\frac{\dd x_1}{\dd \mu} = Px_1 + Q \frac{\dd x}{\dd \mu}, \quad
+\frac{\dd y_1}{\dd \mu} = \dots\dots, \quad
+\frac{\dd z_1}{\dd \mu} = \dots\dots, \quad
+\frac{\dd t_1}{\dd \mu} = \dots\dots.
+\]
+Ceci revient à supposer $R = 0$ dans les équations~\Eq{(1)}; ce qui
+donne enfin
+\[
+\Tag{(10)}
+\frac{\dd^2 x}{\dd \lambda\, \dd \mu}
+  = P \frac{\dd x}{\dd \lambda} + Q \frac{\dd x}{\dd \mu}, \quad
+\frac{\dd^2 y}{\dd \lambda\, \dd \mu} = \dots, \quad
+\frac{\dd^2 z}{\dd \lambda\, \dd \mu} = \dots, \quad
+\frac{\dd^2 t}{\dd \lambda\, \dd \mu} = \dots.
+\]
+Il est facile de voir que si $x\Add{,}y\Add{,}z\Add{,}t$ satisfont à~\Eq{(10)}, les conditions
+\Eq{(1)}\Add{,} \Eq{(2)}\Add{,} \Eq{(3)}\Add{,} \Eq{(4)} sont satisfaites. Tout d'abord \Eq{(3)}~et~\Eq{(4)}
+le sont, ainsi que~\Eq{(1)}. Voyons alors~\Eq{(2)}. Les équations~\Eq{(10)}
+peuvent s'écrire
+\[
+\frac{\dd x_1}{\dd \mu} = Px_1 + Q \frac{\dd x}{\dd \mu}\Add{,}
+\]
+d'où
+\[
+\frac{\dd^2 x_1}{\dd \lambda\, \dd \mu}
+  = \frac{\dd P}{\dd \lambda} x_1 + P \frac{\dd x_1}{\dd \lambda}
+  + Q \frac{\dd x_1}{\dd \mu}
+  + \frac{1}{Q} \left( \frac{\dd x_1}{\dd \mu} - Px_1\right)
+      \frac{\dd Q}{\dd \lambda}\Add{,}
+\]
+ce qui est bien une équation de la forme~\Eq{(2)}.
+%% -----File: 201.png---Folio 193-------
+
+\Paragraph{\DPchg{\Ord{2}{ème} Cas}{Deuxième Cas}.} Nous supposons maintenant $B, B_1, D, D_1, \neq 0$. Reprenons
+les équations~\Eq{(5)}\Add{,}~\Eq{(6)}\Add{.} En multipliant $x, y, z, t$ et $x_1, y_1, z_1, t_1$,
+par des facteurs convenables, on peut faire \DPchg{disparaitre}{disparaître} dans~\Err{\Eq{(S)}}{\Eq{(5)}}
+le terme en~$x$ et le terme en~$x_1$, de sorte que nous pouvons
+écrire
+\[
+\Tag{(11)}
+\frac{\dd x_1}{\dd \lambda} = L \frac{\dd x}{\dd \lambda}, \qquad
+\frac{\dd y_1}{\dd \lambda} = L \frac{\dd y}{\dd \lambda}, \qquad
+\frac{\dd z_1}{\dd \lambda} = L \frac{\dd z}{\dd \lambda}, \qquad
+\frac{\dd t_1}{\dd \lambda} = L \frac{\dd t}{\dd \lambda}.
+\]
+L'équation~\Eq{(6)} peut s'écrire
+\[
+\Tag{(12)}
+\frac{\dd x_1}{\dd \mu} = M \frac{\dd x}{\dd \mu} + N x + S x_1;
+\]
+différentions par rapport à~$\lambda$ en tenant compte de~\Eq{(11)}, nous
+avons
+\[
+\frac{\dd}{\dd \mu} \left(L \frac{\dd x}{\dd \lambda}\right)
+  = \frac{\dd}{\dd \lambda} \left(M \frac{\dd x}{\dd \mu}\right)
+  + \frac{\dd}{\dd \lambda} (N x) + \frac{\dd}{\dd \lambda} (S x_1);
+\]
+$\dfrac{\dd^2 x}{\dd \lambda\, \dd \mu}$ peut d'après~\Eq{(1)} s'exprimer en fonction de~$\Err{x_1}{x}$, $\dfrac{\dd x}{\dd \lambda}$ et $\dfrac{\dd x}{\dd \mu}$,
+et la relation précédente s'écrit
+\[
+\frac{\dd}{\dd \lambda} (S x_1)
+  = F \left(x, \frac{\dd x}{\dd \lambda}, \frac{\dd x}{\dd \mu}\right),
+\]
+$F$~étant une fonction linéaire, ce qu'on peut écrire encore
+\[
+\frac{\dd S}{\dd \lambda} x_1 + S L \frac{\dd x}{\dd \lambda}
+  = F \left(x, \frac{\dd x}{\dd \lambda}, \frac{\dd x}{\dd \mu}\right).
+\]
+Si $\dfrac{\dd S}{\dd \lambda} \neq 0$, $x_1$ est fonction linéaire de~$x$, $\dfrac{\dd x}{\dd \lambda}$, $\dfrac{\dd x}{\dd \mu}$. Le point~$M$
+est dans le plan tangent en~$M$ à la surface~$(S)$, qui est alors
+une des nappes de la surface focale, cas qui a été précédemment
+examiné. Il faut donc supposer $\dfrac{\dd S}{\dd \lambda} = 0$, $S$~n'est fonction
+que de~$\mu$. Alors si nous reprenons l'équation~\Eq{(12)}, nous pouvons
+multiplier $x_1, y_1, z_1, t_1$ par une fonction de~$\mu$ telle que le
+terme en~$x_1$ \DPchg{disparaisse}{disparaîsse}, les relations~\Eq{(11)} subsistant. Et
+nous ramènerons~\Eq{(12)} à la forme
+\[
+\frac{\dd x_1}{\dd \mu} = H \frac{\dd x}{\dd \mu} + K x.
+\]
+%% -----File: 202.png---Folio 194-------
+Le même raisonnement montrera que $K$~est indépendant de~$\lambda$ et
+que par suite on peut faire \DPchg{disparaitre}{disparaître} le terme en~$x$; finalement
+on a
+\[
+\Tag{(13)}
+\frac{\dd x_1}{\dd \mu} = M \frac{\dd x}{\dd \mu},  \qquad
+\frac{\dd y_1}{\dd \mu} = M \frac{\dd y}{\dd \mu},  \qquad
+\frac{\dd z_1}{\dd \mu} = M \frac{\dd z}{\dd \mu},  \qquad
+\frac{\dd t\Add{_1}}{\dd \mu} = M \frac{\dd t}{\dd \mu}.
+\]
+Les relations \Eq{(11)}~et~\Eq{(13)} sont d'ailleurs suffisantes, car
+on en conclut
+\begin{align*}
+\frac{\dd^2 x_1}{\dd \lambda\, \dd \mu}
+  &= \frac{\dd}{\dd \mu} \left(L \frac{\dd x}{\dd \lambda}\right), \\
+\frac{\dd^2 x_1}{\dd \lambda\, \dd \mu}
+  &= \frac{\dd}{\dd \lambda} \left(M \frac{\dd x}{\dd \mu}\right); \\
+\intertext{d'où}
+\Tag{(14)}
+\frac{\dd}{\dd \lambda} \left(M \frac{\dd x}{\dd \mu}\right)
+  &= \frac{\dd}{\dd \mu} \left(L \frac{\dd x}{\dd \lambda}\right),
+\end{align*}
+équation de la forme~\Eq{(1)}, où $R = 0$; on obtiendrait de même
+\[
+\Tag{(15)}
+\frac{\dd}{\dd \lambda} \left(\frac{1}{M} · \frac{\dd x_1}{\dd \mu}\right)
+  = \frac{\dd}{\dd \mu} \left(\frac{1}{M}\, \frac{\dd x_1}{\dd \lambda}\right),
+\]
+équation de la forme~\Eq{(2)} où $R_1=0$.
+
+\MarginNote{Conclusions.}
+Dans le \emph{\Ord{1}{er} cas}, nous avons été ramenés à faire \DPchg{disparaitre}{disparaître}
+le terme en~$x$ dans l'équation
+\[
+\Tag{(1)}
+\frac{\dd^2 x}{\dd \lambda\, \dd \mu}
+  = P \frac{\dd x}{\dd \lambda} + Q \frac{\dd x}{\dd \mu} + R x
+\]
+au moyen de la substitution
+\[
+x = \omega X
+\]
+on trouve immédiatement la condition
+\[
+\frac{\dd^2 \omega}{\dd \lambda\, \dd \mu}
+  = P \frac{\dd \omega}{\dd \lambda}
+  + Q \frac{\dd \omega}{\dd \mu} + R \omega\Add{,}
+\]
+et on peut dire alors que \emph{la surface~$(S_1)$ est définie par les
+équations}
+\[
+x_1 = \frac{\dd}{\dd \lambda} \left(\frac{x}{\omega}\right)\Add{,} \qquad
+y_1 = \frac{\dd}{\dd \lambda} \left(\frac{y}{\omega}\right)\Add{,} \qquad
+z_1 = \frac{\dd}{\dd \lambda} \left(\frac{z}{\omega}\right)\Add{,} \qquad
+t_1 = \frac{\dd}{\dd \lambda} \left(\frac{t}{\omega}\right)\Add{,}
+\]
+\emph{$\omega$~étant une solution de l'équation~\Eq{(1)}}.
+%% -----File: 203.png---Folio 195-------
+
+Passons au \emph{\Ord{2}{e} cas}: il faut encore faire \DPtypo{disparaitre}{disparaître} le
+terme en~$x$ de l'équation~\Eq{(1)}, ce qui revient à chercher une
+intégrale de cette équation. L'équation prend alors la forme
+\[
+\Tag{(2)}
+\frac{\dd^2 x}{\dd \lambda\, \dd \mu}
+  = P \frac{\dd x}{\dd \lambda} + Q \frac{\dd x}{\dd \mu}\Add{.}
+\]
+Identifions avec l'équation~\Eq{(14)} précédemment obtenue. Nous
+avons
+\[
+\frac{\dd L}{\dd \mu} = P(M - L), \qquad
+\frac{\dd M}{\dd \lambda} = Q(L - M);
+\]
+posons alors
+\[
+L - M = \theta,
+\]
+et nous aurons
+\[
+\frac{\dd L}{\dd \mu} = -P\theta, \qquad
+\frac{\dd M}{\dd \lambda} = Q\theta;
+\]
+et l'on voit immédiatement que $\theta$~doit être intégrale de l'équation
+\[
+\Tag{(3)}
+\frac{\dd^2 \theta}{\dd \lambda\, \dd \mu}
+  + \frac{\dd (P \theta)}{\dd \lambda}
+  + \frac{\dd (Q \theta)}{\dd \mu} = 0\Add{,}
+\]
+qui est ce qu'on appelle \emph{l'adjointe} de~\Eq{(2)}. Ayant~$\theta$, on détermine
+par quadratures $L$~et~$M$; car
+\[
+L = -\int P\theta · d\mu, \qquad
+M =  \int Q\theta · d\lambda\Add{.}
+\]
+
+\MarginNote{Propriétés de
+la correspondance
+précédente.}
+Soient $M(x, y, z, t)$, $M_1(x_1, y_1, z_1, t_1)$; soit maintenant $P$~le
+\Figure{203a}
+point de coordonnées $\left(\dfrac{\dd x}{\dd \lambda},\dots\right)$ ou
+$\left(\dfrac{\dd x_1}{\dd \lambda},\dots\right)$ et $Q$~le point $\left(\dfrac{\dd x}{\dd \mu},\dots\right)$ ou
+$\left(\dfrac{\dd x_1}{\dd \mu},\dots\right)$, de sorte que la droite~$PM$
+est tangente à la courbe $\mu = \cte$ sur
+la surface~$(S)$ et~$PM_1$ à la courbe $\mu =
+\cte$, sur la surface~$(S_1)$, et de même
+%% -----File: 204.png---Folio 196-------
+la droite~$QM$ est tangente à la courbe~$\lambda = \cte$ sur la surface~$(S)$,
+et~$QM_1$ à la courbe~$\lambda = \cte$ sur la surface~$(S_1)$. Les plans
+tangents aux \Card{2} surfaces $(S), (S_1)$ aux points $M, M_1$ se coupent suivant
+la droite~$PQ$. Considérons la congruence de ces droites~$PQ$.
+On peut la définir par les équations
+\[
+X = \frac{\dd x}{\dd \lambda} + \rho \frac{\dd x}{\dd \mu},  \quad
+Y = \frac{\dd y}{\dd \lambda} + \rho \frac{\dd y}{\dd \mu},  \quad
+Z = \frac{\dd z}{\dd \lambda} + \rho \frac{\dd z}{\dd \mu},  \quad
+T = \frac{\dd t}{\dd \lambda} + \rho \frac{\dd t}{\dd \mu}.
+\]
+Les développables de cette congruence sont définies par l'équation
+\[
+\begin{vmatrix}
+  \mfrac{\dd x}{\dd \lambda} &
+  \mfrac{\dd x}{\dd \mu} &
+   \mfrac{\dd^2 x}{\dd \lambda^2} · d \lambda
+ + \mfrac{\dd^2 x}{\dd \lambda\, \dd \mu} d\mu &
+   \mfrac{\dd^2 x}{\dd \lambda\, \dd \mu} · d\lambda
+ + \mfrac{\dd^2 x}{\dd \mu^2} d\mu
+\end{vmatrix} = 0;
+\]
+mais on a
+\[
+\frac{\dd^2 x}{\dd \lambda\, \dd \mu}
+  = P \frac{\dd x}{\dd \lambda} + Q \frac{\dd x}{\dd \mu},
+\]
+de sorte que l'équation précédente se réduit à
+\[
+d\lambda · d\mu = 0.
+\]
+\emph{Les développables de la congruence des droites~$PQ$ correspondent
+donc aux développables de la congruence des droites~$MM\Add{_1}$,
+c'est-à-dire encore aux systèmes conjugués homologues.}
+
+Cherchons maintenant les points focaux. Ils sont donnés
+par l'équation
+\[
+\begin{vmatrix}
+  \mfrac{\dd x}{\dd \lambda} &
+  \mfrac{\dd x}{\dd \mu} &
+  \mfrac{\dd^2 x}{\dd \lambda^2} + \rho \mfrac{\dd^2 x}{\dd \lambda\, \dd \mu} &
+  \mfrac{\dd^2 x}{\dd \lambda\, \dd \mu} + \mfrac{\rho \dd^2 x}{\dd \mu^2}
+\end{vmatrix} = 0,
+\]
+équation qui, à cause de la même condition que précédemment,
+se réduit à $\rho = 0$; une racine est nulle, l'autre infinie;
+\emph{les points focaux ne sont autres que les points~$P, Q$. Ils sont
+dans les plans focaux de la congruence~$MM_1$}.
+
+Considérons le point~$P$, et supposons que l'on fasse $\lambda =
+\cte[]$. La direction de la tangente à la trajectoire du point~$P$
+est définie par un \Ord{2}{e} point, dont les coordonnées sont
+\[
+\frac{\dd}{\dd \mu} \left(\frac{\dd x}{\dd \lambda}\right)
+  = P \frac{\dd x}{\dd \lambda} + Q \frac{\dd x}{\dd \mu},
+\]
+et les analogues.
+%% -----File: 205.png---Folio 197-------
+C'est un point de~$PQ$. Le point~$P$ décrit une courbe tangente à~$PQ$,
+arête de rebroussement de la développable correspondant à
+$\lambda = \cte[]$. Le point~$Q$ décrira de même l'arête de rebroussement
+de la développable correspondant à~$\mu = \cte[]$.
+
+Les propriétés de la correspondance que nous venons d'étudier
+se transforment en elles-mêmes par dualité. En choisissant
+convenablement les coordonnées tangentielles homogènes,
+on aurait donc
+\begin{align*}
+\frac{\dd u_1}{\dd \lambda} &= H \frac{\dd u}{\dd \lambda},
+  \quad\text{et les analogues;}\quad\dots\dots, \\
+\frac{\dd u_1}{\dd \mu} &= K \frac{\dd u}{\dd \mu},
+  \quad\text{et les analogues.}\quad\dots\dots.
+\end{align*}
+Appelons alors congruence~$(K)$ celle des droites~$MM_1$, congruence~$(K')$
+celle des droites~$PQ$. \emph{Si les développables de la congruence~$(K)$
+coupent les surfaces $(S)\Add{,} (S_1)$ suivant deux réseaux
+conjugués, les développables de la congruence~$(K')$ sont circonscrites
+à ces surfaces suivant les mêmes réseaux, et réciproquement.
+Les points focaux de~$(K')$ sont dans les plans focaux
+de~$(K)$, chaque point focal se trouvant dans le plan focal
+qui ne lui correspond pas.}
+
+\Section{Correspondance par plans tangents parallèles.}
+{4.}{} Soit sur la surface~$(S)$ l'une des courbes~$(c)$ du réseau
+conjugué qui correspond à un réseau conjugué sur~$(S_1)$
+et soit~$(c_1)$ la courbe correspondante sur~$(S_1)$. Supposons
+qu'en deux points homologues les plans tangents aux surfaces
+$(S)\Add{,} (S_1)$ soient parallèles; leurs caractéristiques le sont aussi;
+donc \emph{les directions conjuguées homologues sont parallèles.}
+%% -----File: 206.png---Folio 198-------
+Faisons $t = 1$ et~$t_1 = 1$, nous avons
+\begin{alignat*}{3}
+\Tag{(1)}
+\frac{\dd x_1}{\dd \lambda} &= L \frac{\dd x}{\dd \lambda},  \qquad &
+\frac{\dd y_1}{\dd \lambda} &= L \frac{\dd y}{\dd \lambda},  \qquad &
+\frac{\dd z_1}{\dd \lambda} &= L \frac{\dd z}{\dd \lambda}; \\
+\Tag{(2)}
+\frac{\dd x_1}{\dd \mu} &= M \frac{\dd x}{\dd \mu}  \qquad &
+\frac{\dd y_1}{\dd \mu} &= M \frac{\dd y}{\dd \mu}  \qquad &
+\frac{\dd z_1}{\dd \mu} &= M \frac{\dd z}{\dd \mu}.
+\end{alignat*}
+Nous pouvons donc appliquer les résultats précédemment obtenus\Add{.}
+Les plans tangents en~$M\Add{,} M_1$ étant parallèles, la droite~$PQ$ est à
+l'infini. Les droites de la congruence~$(K')$ sont les droites
+du plan de l'infini. Sur chacune de ces droites, les points~$P\Add{,} Q$
+sont les points où elles sont rencontrées par les tangentes
+conjuguées homologues sur $(S)\Add{,} (S_1)$, et le lieu des points~$P\Add{,}Q$
+est tangent à chaque droite~$PQ$ aux points~$P, Q$.
+
+\MarginNote{Cas particulier.}
+En particulier, supposons que, la surface~$(S)$ étant quelconque,
+la surface~$(S_1)$ soit une sphère. La congruence des
+droites~$MM_1$ a des développables qui découpent sur $(S)\Add{,} (S_1)$ des
+réseaux conjugués, les tangentes homologues étant parallèles.
+Or\Add{,} sur une sphère, un réseau conjugué est un réseau orthogonal;
+donc le réseau conjugué de~$(S)$ est aussi un réseau orthogonal,
+ce sont les \emph{lignes de courbure}, dont la recherche est
+ainsi ramenée à celle des développables d'une congruence. En
+particulier, supposons la surface~$(S)$ du \Ord{2}{e} degré, et considérons
+la congruence des droites~$PQ$ du plan de l'infini. Le
+plan de l'infini coupe $(S),(S_1)$ suivant deux coniques $(\Gamma),(\Gamma_1)$.
+Considérons leurs points d'intersection avec une droite~$PQ$;
+les points d'intersection avec~$(\Gamma)$ correspondent aux directions
+des génératrices de~$(S)$ qui passent par~$M$, et qui sont
+les tangentes asymptotiques; les points~$P\Add{,} Q$ qui correspondent
+aux directions principales sont donc conjugués par rapport à
+ces points d'intersection, c'est-à-dire conjugués par rapport
+%% -----File: 207.png---Folio 199-------
+à la conique~$(\Gamma)$. Ils sont de même conjugués par rapport à~$(\Gamma_1)$.
+Les points~$P,Q$ sont les points doubles de l'involution
+déterminée sur la droite~$PQ$ par le faisceau de coniques ayant
+pour bases $(\Gamma),(\Gamma_1)$. La droite~$PQ$ est tangente en~$P\Add{,} Q$ aux \Card{2} coniques
+de ce faisceau qui lui sont tangentes; de sorte que la
+détermination des développables de la congruence~$(K)$, c'est-à-dire
+des lignes de courbure de la quadrique~$(S)$, revenant à
+celle d'un faisceau de coniques, peut se faire algébriquement.
+Si on prend pour paramètres ceux des génératrices rectilignes
+qui passent par un point de~$(S)$, on obtient ainsi l'intégration
+de l'\emph{équation d'Euler}.
+
+\Paragraph{Remarque.} Au lieu du plan de l'infini, on pourrait considérer
+un plan fixe quelconque~$(\Pi)$. La correspondance serait
+telle que les plans tangents en \Card{2} points homologues de $(S), (S_1)$
+se coupent dans le plan~$\Pi$.\DPnote{** TN: [sic], no ()} Les résultats seraient analogues;
+et de même si, corrélativement, on établissait entre les \Card{2} surfaces
+une correspondance telle que la droite~$MM_1$ passe par
+un point fixe.
+
+Considérons deux surfaces $(S), (S_1)$ qui se correspondent
+par plans tangents parallèles. Prenons dans l'espace un point
+fixe~$O$, et substituons à~$(S_1)$ une de ses homothétiques par
+rapport à~$O,(S'_1)$. A tout réseau conjugué sur~$(S_1)$ correspond
+sur~$(S_1\Add{'})$ un réseau homothétique qui est aussi conjugué, et le
+réseau conjugué de~$(S)$ qui correspond à un réseau conjugué
+sur~$(\Err{S_1}{S'_1})$ correspond aussi à un réseau conjugué sur~$(S'_1)$. Imaginons
+que le rapport d'homothétie croisse indéfiniment le
+point~$M'_1$ homothétique de~$\Err{M}{M_1}$ s'éloigne à l'infini, la droite~$MM_1$
+est la parallèle menée par~$M$ au rayon~$OM$. Donc, \emph{si l'on a
+%% -----File: 208.png---Folio 200-------
+\Card{2} surfaces $(S),(S_1)$ se correspondant par plans tangents parallèles,
+si on prend dans l'espace un point fixe~$O$, et si par
+le point~$M$ de~$(S)$ on mène la parallèle~$MN$ au rayon~$\Err{OM}{OM_1}$, les
+développables de la congruence des droites~$MN$ découpent sur~$(S)$
+le réseau conjugué qui correspond à un réseau conjugué
+sur~$(S_1)$}. Si en particulier nous prenons pour~$(S_1)$ une sphère,
+pour~$O$ son centre, $\Err{OM}{OM_1}$~est perpendiculaire au plan tangent à~$(\Err{S}{S_1})$,
+et par conséquent au plan tangent à~$(S)$; $MN$~qui lui est
+parallèle est la normale à~$(S)$. \emph{La congruence des normales à
+une surface a des développables qui déterminent sur cette
+surface un réseau conjugué orthogonal.} On retrouve donc la
+propriété fondamentale des lignes de courbure de la surface~$(S)$.
+
+Remarquons encore que si le rayon de la sphère~$(S_1)$ est
+égal à~$1, x, y, z,$ sont les cosinus directeurs de la normale,
+et les formules \Eq{(1)}\Add{,}~\Eq{(2)} ne sont autres que les formules
+d'Olinde Rodrigues.
+
+
+\ExSection{VIII}
+
+\begin{Exercises}
+\item[40.] On donne deux courbes $C, C_{1}$. Trouver toutes les surfaces~$S$ sur
+lesquelles les courbes de contact des cônes circonscrits à~$S$,
+ayant leurs sommets sur $C$ et~$C_{1}$, forment un réseau conjugué.
+En définissant $C$~et~$C_{1}$ par les équations
+\begin{alignat*}{4}
+x &= f(\lambda), \qquad &
+y &= g(\lambda), \qquad &
+z &= h(\lambda), \qquad &
+t &= k(\lambda); \\
+%
+x &= \phi(\mu), &
+y &= \psi(\mu), &
+z &= \chi(\mu), &
+t &= \theta(\mu),
+\end{alignat*}
+la surface la plus générale répondant à la question est définie
+par les équations
+\begin{align*}
+x &= \int A(\lambda)f(\lambda)\, d\lambda + \int B(\mu)\phi(\mu)\, d\mu, \\
+y &= \int A(\lambda)g(\lambda)\, d\lambda + \int B(\mu)\psi(\mu)\, d\mu, \\
+z &= \int A(\lambda)h(\lambda)\, d\lambda + \int B(\mu)\chi(\mu)\, d\mu, \\
+t &= \int A(\lambda)k(\lambda)\, d\lambda + \int B(\mu)\theta(\mu)\, d\mu.
+\end{align*}
+
+Interpréter géométriquement les formules obtenues de
+façon à trouver une définition géométrique de ces surfaces.
+Transformer par dualité les divers résultats obtenus.
+
+\item[41.] Soit $\Sigma$ la sphère de centre~$O$ et de rayon égal à un; soit $S$
+une surface quelconque et $S'$~sa polaire \DPtypo{reciproque}{réciproque} par rapport
+à~$\Sigma$. Soit $M$ un point quelconque de~$S$ et~$P$ le plan tangent en
+ce point; soient $M'$~et~$P'$ le point et le plan tangent de~$S'$ qui
+correspondent à~$P$ et~$M$ par polaires réciproques. On considère
+la congruence~$K$ des droites~$MM'$ et la congruence~$K'$ des intersections
+des plans $P$~et~$P'$. Montrer que leurs développables
+se correspondent, et que les développables de~$K$ découpent
+sur~$S$ et~$S'$ des réseaux conjugués. Comment les développables
+de~$K$ coupent-elles~$\Sigma?$ Chercher à déterminer~$S$ de manière
+que $K$~soit une congruence de normales; que peut-on dire alors
+des développables de~$K$ et de la surface~$S$?
+
+\item[42.] \DPchg{Etant}{Étant} \DPtypo{donnee}{donnée} une courbe gauche~$C$, par un point fixe~$O$ on
+mène des segments~$OM$ équipollents aux diverses cordes de~$C$.
+Le lieu des points~$M$ est une surface~$S_{0}$. Par chaque point~$M$
+de cette surface on mène la parallèle~$\Delta$ à l'intersection
+des plans osculateurs de $C$~\DPtypo{menes}{menés} aux points $P$~et~$P_{1}$ de~$C$ tels
+que $PP_{1}$~est équipollent à~$OM$. Soient $S_{1}$~et~$S_{2}$ les deux nappes de
+la surface focale de la congruence des droites~$\Delta$:
+
+%[** TN: Regularized formatting of parts]
+\Primo. déterminer
+$S_{1}$~et~$S_{2}$, leur~$ds^2$, leur~$\sum l\,d^{2}x$. Montrer que les
+asymptotiques se correspondent sur $S_{1}$~et~$S_{2}$. Quelles sont les
+courbes de $S_{0}$ qui leur correspondent?
+
+\Secundo. Condition nécessaire
+et suffisante que doit remplir~$C$ pour que la congruence
+des droites~$\Delta$ soit une congruence de normales. Trouver alors
+l'une des surfaces normales. Montrer que les rayons de courbure
+de~$\Sigma$ sont fonctions l'un de l'autre.
+
+\Tertio. En restant
+dans ce cas, rapporter le~$ds^{2}$, de $S_{1}$~aux géodésiques tangentes
+aux droites~$\Delta$ et à leurs trajectoires orthogonales. En conclure
+que $S_{1}$~est applicable sur un \DPchg{paraboloide}{paraboloïde} de révolution.
+
+\textsc{Nota}. Les deux dernières parties de cet exercice se
+rattachent à la fin du chapitre~XIII\@.
+\end{Exercises}
+%% -----File: 209.png---Folio 201-------
+
+
+\Chapitre{IX}{Complexes de Droites.}
+
+\Section{\DPchg{Eléments}{Éléments} fondamentaux d'un complexe de droites.}
+{1.}{} On appelle \emph{complexe} un système de $\infty^{3}$~droites,
+c'est-à-dire une famille de droites dépendant de \Card{3} paramètres.
+
+Soit $A$ un point de l'espace, toutes les droites~$(D)$ du
+complexe qui passent par ce point sont au nombre de~$\infty^{1}$, et
+constituent le \emph{cône du complexe} attaché au point~$A$: nous
+l'appellerons le cône~$(K)$.
+
+Corrélativement: soit un plan~$P$, toutes les droites~$(D)$
+du complexe situées dans ce plan sont au nombre de~$\infty^{1}$, et enveloppent
+une courbe~$(c)$ qui est la \emph{courbe du complexe} associée
+à~$P$. La tangente en tout point de cette courbe est une
+droite du complexe.
+
+Plus généralement nous appellerons \emph{courbe du complexe}
+une courbe~$(c)$ dont toutes les tangentes appartiennent au complexe.
+Considérons sur une telle courbe un point~$A$, et le
+cône du complexe~$(K)$ associé au point~$A$. Ce cône est tangent
+à la courbe~$(c)$. \emph{Une courbe du complexe est tangente en chacun
+de ses points au cône du complexe associé à ce point.}
+
+\Illustration[2.25in]{210a}%[** TN: Moved to top of paragraph]
+Considérons un plan~$P$, et un point~$A$ de ce plan; cherchons
+les droites du complexe situées dans le plan~$P$ et passant
+par~$A$. Considérons le cône du complexe associé au point~$A$,
+%% -----File: 210.png---Folio 202-------
+les droites cherchées sont les
+génératrices de ce cône situées dans
+le plan~$P$: si nous considérons la
+courbe du complexe associée au plan~$P$,
+les droites cherchées sont aussi
+les tangentes issues de~$A$ à cette
+courbe. Cherchons dans le plan~$P$ le
+lieu des points~$A$ tels que deux des
+droites du complexe situées dans le plan~$P$ et passant par~$A$
+soient confondues; les points~$A$ correspondants seront, d'après
+ce qui précède, tels que le cône du complexe correspondant
+soit tangent au plan~$P$: ils doivent aussi être sur la
+courbe du complexe: les droites du complexe confondues \DPtypo{coincident}{coïncident}
+avec la génératrice de contact du cône du complexe, ou
+avec la tangente à la courbe du complexe. Ainsi l'on peut définir
+la courbe du complexe située dans un plan comme étant
+le lieu des points de ce plan pour lesquels le cône du complexe
+est tangent au plan, et la génératrice de contact n'est
+autre que la tangente en ce point à la courbe. La courbe du
+complexe est ainsi définie par points et par tangentes.
+
+\begin{wrapfigure}[14]{O}{1.625in}
+\Input[1.5in]{210b}
+\end{wrapfigure}
+Considérons alors une droite~$(D)$ du complexe; prenons sur
+cette droite un point~$A$, et considérons le cône~$(K)$ du complexe
+associé au point~$A$; soit $P$ le plan
+tangent à ce cône le long de la génératrice~$(D)$.
+A chaque point~$A$ de la droite
+correspond ainsi un plan~$P$. Considérons
+maintenant la courbe~$(c)$ du complexe située
+dans le plan~$P$, elle est tangente à
+%% -----File: 211.png---Folio 203-------
+la droite~$(D)$ précisément au point~$A$, de sorte qu'à chaque
+plan~$P$ passant par la droite correspond un point de cette
+droite. \emph{Il y a une correspondance homographique entre les
+points et les plans d'une droite du complexe.}
+
+Précisons la nature de cette homographie. Une droite
+quelconque peut être représentée par \Card{2} équations de la forme
+\[
+\Tag{(1)}
+X = a Z + f, \qquad
+Y = b Z + g.
+\]
+Pour qu'elle appartienne à un complexe, il faut et il suffit
+qu'il existe une relation entre les paramètres~$a\Add{,}b\Add{,}f\Add{,}g$:
+\[
+\Tag{(2)}
+\phi (a,b,f,g) = 0.
+\]
+Cherchons alors toutes les droites du complexe infiniment
+voisines de la droite~\Eq{(1)} et rencontrant cette droite. Une
+telle droite peut être représentée par les équations
+\[
+\Tag{(3)}
+X = (a + da) Z + (f + df), \qquad
+Y = (b + db) Z + (g + dg).
+\]
+Exprimons qu'elle rencontre la droite~\Eq{(1)}. Les équations
+\[
+\Tag{(4)}
+Z\, da + df = 0, \qquad
+Z\, db + dg = 0,
+\]
+doivent avoir une solution commune en~$Z$, ce qui donne la condition
+\[
+\Tag{(5)}
+da · dg - db · df = 0.
+\]
+Le point d'intersection~$M$ des \Card{2} droites infiniment voisines
+aura alors pour cote
+\[
+Z = - \frac{df}{da}.
+\]
+Différentions la relation~\Eq{(2)}, nous avons
+\[
+\Tag{(6)}
+\frac{\dd \phi}{\dd a}\, da +
+\frac{\dd \phi}{\dd b}\, db +
+\frac{\dd \phi}{\dd f}\, df +
+\frac{\dd \phi}{\dd g}\, dg = 0.
+\]
+Supposons connu le point~$M$, nous avons les relations~\Eq{(4)} dans
+lesquelles $Z$ est connu, et qui par conséquent déterminent les
+rapports des différentielles. Cherchons alors le plan passant
+%% -----File: 212.png---Folio 204-------
+par les deux droites infiniment voisines. Il suffit de multiplier~\Eq{(3)}
+respectivement par $db$ et~$-da$, et d'ajouter, il
+vient, en tenant compte de~\Eq{(5)}
+\[
+\Tag{(7)}
+(X - aZ - f)\, db - (Y - bZ - g)\, da = 0\Add{.}
+\]
+Telle est l'équation du plan cherché: il ne dépend que du rapport~$\dfrac{db}{da}$.
+Nous en concluons que \emph{toutes les droites du complexe
+infiniment voisines de la droite~$D$ et rencontrant cette droite
+en un point~$M$ donné sont dans un même plan, et inversement
+toutes les droites du complexe infiniment voisines de la droite~$D$
+et situées dans un même plan passant par~$D$ rencontrent $D$
+au même point}. Posons
+\[
+\lambda = \frac{da}{db}\Add{,}
+\]
+l'équation~\Eq{(7)} s'écrit
+\[
+\Tag{(8)}
+X - a Z - f -\lambda (Y - b Z - g) = 0\Err{\lambda}{}\Add{.}
+\]
+Démontrons qu'il y a une relation homographique entre~$\lambda, Z$:
+tirons en effet $df, dg$ des équations~\Eq{(4)} et portons dans~\Eq{(6)},
+nous avons
+\[
+\left(\frac{\dd \phi}{\dd a} - Z \frac{\dd \phi}{\dd f}\right) da +
+\left(\frac{\dd \phi}{\dd b} - Z \frac{\dd \phi}{\dd g}\right) db = 0
+\]
+et la relation d'homographie est
+\[
+\lambda \left(\frac{\dd \phi}{\dd a} - Z \frac{\dd \phi}{\dd f}\right)
+            + \frac{\dd \phi}{\dd b} - Z \frac{\dd \phi}{\dd g} = 0.
+\]
+
+Considérons en particulier le cône du complexe de sommet~$M$;
+la génératrice infiniment voisine est une droite du complexe
+rencontrant $D$ en~$M$: le plan de ces \Card{2} droites est le plan
+tangent au cône du complexe, et nous avons l'homographie
+précédemment définie.
+
+Considérons encore une courbe du complexe quelconque
+%% -----File: 213.png---Folio 205-------
+tangente à la droite~$D$ au point~$A$. Considérons une tangente
+infiniment voisine à cette courbe; à la limite cette tangente
+rencontre $D$ au point~$A$, et le plan de ces \Card{2} droites n'est autre
+que le plan osculateur à la courbe au point~$A$, et ce plan
+osculateur est associé au point~$A$ dans l'homographie précédente.
+Donc \emph{toutes les courbes du complexe tangentes à une
+droite~$D$ en un même point~$A$ ont même plan osculateur en ce
+point: c'est le plan tangent au cône du complexe associé au
+point~$A$}.
+
+Considérons enfin une congruence de droites appartenant
+au complexe; prenons dans cette congruence une droite~$D$, et
+sur cette droite un point focal~$A$; le point~$A$ appartient à
+une des nappes de la surface focale de la congruence; il appartient
+aussi à l'arête de rebroussement d'une des développables
+de la congruence, et cette arête de rebroussement, enveloppe
+de droites~$D$ appartenant au complexe, est une courbe du
+complexe. Son plan osculateur en~$A$ est le \Ord{2}{e} plan focal de la
+congruence; d'après ce qui précède, \emph{toutes les congruences du
+complexe passant par la droite~$D$ et ayant un foyer en~$A$ ont
+même \Ord{2}{e} plan focal relatif à la droite~$D$;} il y a correspondance
+homographique entre ce \Ord{2}{e} plan focal et le point~$A$.
+
+\Section{Surfaces du complexe.}
+{2.}{} Cherchons si dans un complexe il y a des congruences
+ayant une surface focale double. Sur une telle surface~$(\Phi)$
+les arêtes de rebroussement des développables sont des lignes
+asymptotiques; or\Add{,} ce sont des courbes du complexe.
+Il s'agit donc de trouver des surfaces telles qu'une famille
+%% -----File: 214.png---Folio 206-------
+de lignes asymptotiques soit formée de courbes du complexe.
+Considérons une telle asymptotique~$(c)$ et un de ses points~$A$.
+Le plan osculateur à la courbe~$(c)$ en~$A$ est le plan tangent
+au cône~$(K)$ du complexe associé au point~$A$, et ce plan osculateur
+est tangent à la surface~$(\Phi)$. Les surfaces cherchées
+sont donc tangentes en chacun de leurs points au cône du complexe
+associé à ce point. Réciproquement soit~$(\Phi)$ une telle
+surface; considérons en chacun de ses points la génératrice
+de contact~$(D)$ du cône du complexe avec le plan tangent. Nous
+déterminons ainsi sur la surface~$(\Phi)$ une famille de courbes
+tangentes en chaque point aux droites~$(D)$; ces courbes~$(\DPtypo{C}{c})$
+sont des courbes du complexe; leur plan osculateur est le
+plan tangent au cône du complexe le long de la droite~$(D)$,
+c'est le plan tangent à la surface~$(\Phi)$ et les courbes~$(c)$ sont
+des asymptotiques de cette surface. De telles surfaces sont
+appelées \emph{surfaces du complexe}.
+
+Considérons les équations d'une droite du complexe
+\[
+\Tag{(1)}
+x = az + f, \qquad
+y = bz + g,
+\]
+$a\Add{,} b\Add{,} f\Add{,} g$ étant liés par l'équation
+\[
+\Tag{(2)}
+\phi(a, b, f, g) = 0.
+\]
+Transportons l'origine au point~$(x, y, z)$ et appelons $X, Y, Z$ les
+nouvelles coordonnées. $X, Y, Z$~sont alors les coefficients de
+direction d'une droite du complexe
+\[
+a = \frac{X}{Z}, \qquad
+b = \frac{Y}{Z},
+\]
+et l'équation du cône du complexe associé au point $(x\Add{,}y\Add{,}z)$ est
+\[
+\phi\left(\frac{X}{Z},
+          \frac{Y}{Z},
+          x - \frac{X}{Z} z,
+          y - \frac{Y}{Z} z\right) = 0,
+\]
+%% -----File: 215.png---Folio 207-------
+ou, en rendant homogène
+\[
+\Psi(X, Y, Z, xZ - zX, yZ - zY) = 0;
+\]
+les courbes du complexe sont alors définies par l'équation
+différentielle, homogène en~$dx, dy, dz$,
+\[
+\Psi(dx · dy · dz, x\,dz - z\,dx, y\,dz - z\,dy) = 0.
+\]
+Une telle équation s'appelle une \emph{équation de Monge}, et \emph{équation
+de Pfaff} si elle est du \Ord{1}{er} degré.
+
+Prenons maintenant l'équation tangentielle du cône du
+complexe
+\[
+F(x, y, z, U, V, W) = 0;
+\]
+la condition pour qu'une surface $z = G(x, y)$ soit tangente à
+ce cône en chacun de ses points, est que l'équation soit vérifiée
+par $U = \dfrac{\dd G}{\dd x} = p$, $V = \dfrac{\dd G}{\dd y} = q$, $W = - 1$; les surfaces du complexe
+sont donc définies par l'équation aux dérivées partielles
+\[
+F(x, y, z, p, q, -1) = 0,
+\]
+qui est de la forme:
+\[
+\Tag{(3)}
+f(x, y, z, p, q) = 0.
+\]
+Nous obtenons une équation aux dérivées partielles du \Ord{1}{er} ordre.
+Inversement, avec les notations précédentes, toute
+équation aux dérivées partielles du \Ord{1}{er} ordre pouvant se
+mettre sous la forme
+\[
+\Tag{(4)}
+f\left(x, y, z, \Err{}{-}\frac{U}{W}, \Err{}{-}\frac{V}{W}\right) = 0
+\]
+exprime que le plan tangent à une surface intégrale est tangent
+à un certain cône associé au point de contact, mais les
+génératrices de tous ces $\infty^{3}$~cônes remplissent en général tout
+l'espace, et ne forment un complexe qu'exceptionnellement.
+
+Pour pouvoir mieux préciser ce cas d'exception, rappelons
+%% -----File: 216.png---Folio 208-------
+les points essentiels de la théorie des équations générales
+aux dérivées partielles du premier ordre, c'est-à-dire
+de la forme~\Eq{(3)}.
+
+Un \emph{élément de contact intégral} est un élément de contact
+dont les coordonnées $(x, y, z, p, q)$ satisfont à l'équation donnée~\Eq{(3)}.
+
+Le \emph{cône élémentaire} associé au point $(x, y, z)$ est l'enveloppe
+des éléments de contact intégraux appartenant à ce point
+son équation tangentielle est précisément l'équation~\Eq{(4)}.
+Tout élément linéaire formé d'un point et d'une génératrice du
+cône élémentaire associé à ce point s'appelle un \emph{élément linéaire
+intégral}. Si $dx, dy, dz$~sont les coefficients de direction
+d'une telle génératrice, l'équation qui caractérise les
+éléments linéaires intégraux s'obtient en éliminant $p$~et~$q$
+entre les équations:
+\[
+\Tag{(5)}
+f(x, y, z, p, q) = 0, \quad
+dz - p\, dx - q\, dy = 0, \quad
+\frac{\dd f}{\dd p} dy - \frac{\dd f}{\dd q} dx = 0.
+\]
+
+\Illustration[1.75in]{217a}
+\noindent L'équation obtenue est une équation de Monge:
+\[
+\Tag{(6)}
+G(x, y, z, dx, dy, dz) = 0.
+\]
+
+Les \emph{courbes intégrales} sont les courbes dont tous les
+éléments linéaires (points-tangentes) sont intégraux. Elles
+sont définies par l'équation~\Eq{(6)}.
+
+Une \emph{bande intégrale} est un lieu d'éléments de contact
+appartenant à une même courbe (points-plans tangents), et
+qui soient tous des éléments de contact intégraux. C'est donc
+un ensemble de $\infty^{1}$~éléments de contact satisfaisant aux équations
+\[
+\Tag{(7)}
+f(x, y, z, p, q) = 0, \qquad
+dz - p\,dx - q\,dy = 0.
+\]
+%% -----File: 217.png---Folio 209-------
+Si on prend une courbe quelconque et si par chacune de ses
+tangentes on mène un plan tangent au cône élémentaire associé
+au point de contact, on obtient une bande intégrale. Par une
+courbe quelconque passent donc, si l'équation~\Eq{(3)}
+est algébrique en~$p, q$, un nombre
+limité de bandes intégrales. Ce nombre
+se réduit de un dans le cas où la courbe
+est une courbe intégrale.
+
+\Paragraph{Par une bande intégrale passe en général
+une surface intégrale et une seule.}
+Les bandes intégrales qui font exception s'appellent \emph{bandes
+caractéristiques}. Les courbes qui leur servent de supports
+sont des courbes intégrales particulières, qu'on appelle
+\emph{caractéristiques}.
+
+Les bandes caractéristiques sont définies par les équations
+\begin{gather*}
+f(x, y, z, p, q) = 0, \\
+%
+\Tag{(8)}
+\frac{dx}{\dfrac{\dd f}{\dd p}}
+  = \frac{dy}{\dfrac{\dd f}{\dd q}}
+  = \frac{dz}{p \dfrac{\dd f}{\dd p} + q \dfrac{\dd f}{\dd q}}
+  = \frac{dp}{- \dfrac{\dd f}{\dd x} - p \dfrac{\dd f}{\dd z}}
+  = \frac{dq}{- \dfrac{\dd f}{\dd y} - q \dfrac{\dd f}{\dd z}}.
+\end{gather*}
+
+On les obtient donc en intégrant un système d'équations
+différentielles ordinaires. \emph{Par un élément de contact intégral
+passe une bande caractéristique et une seule.}
+
+\emph{La surface intégrale qui passe par une bande intégrale
+non caractéristique donnée est engendrée par les bandes caractéristiques
+passant par les divers éléments de contact intégraux
+de cette bande intégrale.}
+
+Sur une surface intégrale il y a au plus une courbe
+%% -----File: 218.png---Folio 210-------
+intégrale qui ne soit pas une caractéristique.
+
+\Paragraph{Toute courbe intégrale est l'enveloppe d'une famille
+de $\infty^{1}$~courbes caractéristiques.} Ces caractéristiques engendrent
+une surface intégrale.
+
+\emph{Réciproquement}: si une famille de $\infty^{1}$~caractéristiques
+a une enveloppe, cette enveloppe est une courbe intégrale.
+
+L'intégration du système~\Eq{(8)} suffit donc pour l'intégration
+de l'équation~\Eq{(3)} et de l'équation de Monge~\Eq{(6)}, qui lui
+est associée.
+
+Ces trois intégrations s'achèvent enfin immédiatement si
+on a une \emph{intégrale complète}, c'est-à-dire une équation où
+figure deux constantes arbitraires
+\[
+H(x, y, z, a, b) = 0
+\]
+définissant des surfaces intégrales, pour toutes les valeurs
+de ces constantes.
+
+Les \emph{courbes caractéristiques} sont alors définies par les
+équations
+\[
+H = 0,\qquad
+\frac{\dd H}{\dd a} + c \frac{\dd H}{\dd b} = 0,
+\]
+où $c$~est une nouvelle constante arbitraire.
+
+Une \emph{surface intégrale quelconque} s'obtient en prenant
+l'enveloppe de $\infty^{1}$~surfaces, faisant partie de l'intégrale
+complète, c'est-à-dire en éliminant~$a$ entre les équations\DPtypo{.}{}
+\[
+H(x, y, z, a, b) = 0,\qquad
+\frac{\dd H}{\dd a}\, da + \frac{\dd H}{\dd b}\, db = 0,
+\]
+après y avoir remplacé~$b$ par une fonction arbitraire de~$a$.
+
+Les caractéristiques tracées sur une telle surface ont
+nécessairement une enveloppe; par suite on obtient une \emph{courbe
+%% -----File: 219.png---Folio 211-------
+intégrale quelconque}, en éliminant~$a$ entre les équations
+\[
+H = 0,\quad
+\frac{\dd H}{\dd a}\, da + \frac{\dd H}{\dd b}\, db = 0,\quad
+\frac{\dd^2 H}{\dd a^2}\, da^2 + 2 \frac{\dd^2 H}{\dd a\, \dd b}\, da\, db +
+\frac{\dd^2 H}{\dd b^2}\, db^2 + \frac{\dd H}{\dd b}\, d^2 b = 0,
+\]
+après y avoir remplacé~$b$ par une fonction arbitraire de~$a$.
+
+Si nous revenons maintenant au cas particulier où l'équation~\Eq{(3)}
+est celle qui définit les surfaces d'un complexe,
+nous voyons que les courbes intégrales sont les courbes du
+complexe, et que les caractéristiques situées sur une surface
+intégrale constituent la famille de $\infty^{1}$~courbes du complexe
+qui sont les lignes asymptotiques de cette surface. Il en résulte
+que les équations~\Eq{(8)} ont alors pour conséquence
+\[
+dp\, dx + dq\, dy = 0,
+\]
+c'est-à-dire que l'équation~\Eq{(3)} a elle-même pour conséquence
+\[
+\frac{\dd f}{\dd p} \left(\frac{\dd f}{\dd x}
+  + p\, \frac{\dd f}{\dd z}\right) +
+\frac{\dd f}{\dd q} \left(\frac{\dd f}{\dd y}
+  + q\, \frac{\dd f}{\dd z}\right) = 0.
+\]
+On démontre que, réciproquement, les seules équations~\Eq{(3)}
+pour lesquelles les caractéristiques sont les lignes asymptotiques
+des surfaces intégrales, sont, (si on excepte les
+équations linéaires), les équations dont les cônes élémentaires
+sont les cônes des complexes de droites.
+
+\Paragraph{Remarque.} Si le cône du complexe se réduit à un plan, le complexe
+est appelé un \emph{complexe linéaire}. Le cône n'a alors pas
+d'équation tangentielle, et la théorie précédente ne s'applique
+plus.
+
+Le cas des complexes linéaires sera étudié dans le chapitre
+suivant.
+
+
+\Section{Complexes spéciaux.}
+{3.}{} Nous dirons qu'un complexe est \emph{spécial} quand l'homographie
+qui existe entre les points et les plans d'une droite
+%% -----File: 220.png---Folio 212-------
+du complexe est spéciale. A un élément d'un système correspond
+toujours le même élément dans le système associé, sauf
+pour un seul élément du \Ord{1}{er} système, dont le correspondant
+est indéterminé. L'équation de l'homographie étant
+\[
+\lambda \left(\frac{\dd \phi}{\dd a} - z\, \frac{\dd \phi}{\dd f}\right)
+            + \frac{\dd \phi}{\dd b} - z\, \frac{\dd \phi}{\dd b} = 0\Add{,}
+\]
+la condition pour qu'on ait une homographie spéciale est
+\[
+\Tag{(1)}
+\frac{\dd \phi}{\dd a} · \frac{\dd \phi}{\dd g} -
+\frac{\dd \phi}{\dd b} · \frac{\dd \phi}{\dd f} = 0.
+\]
+Considérons le \emph{complexe des droites tangentes à une surface};
+considérons une congruence de ce complexe; \Err{les développables}{ces développables de l'une des familles}
+de la congruence seront circonscrites à la surface, l'un des
+plans focaux sera indépendant de la \Err{développable}{congruence} que l'on
+considère. Même résultat si on considère le \emph{complexe des
+droites rencontrant une courbe donnée}. On obtient donc ainsi
+des complexes spéciaux. Nous allons montrer qu'il n'y en a
+pas d'autres. Prenons l'équation d'un complexe sous la forme
+\[
+\Phi = g - \phi(a, b, f) = 0;
+\]
+\Eq{(1)}~s'écrit
+\[
+%[** TN: Original has a leading +; suspect artifact from typist]
+\Tag{(2)}
+\frac{\dd \phi}{\dd a} + \frac{\dd \phi}{\dd b} · \frac{\dd \phi}{\dd f} = 0.
+\]
+Cette relation ne contient plus~$g$, elle doit être une identité
+par rapport à~$a, b, f$. Considérons une droite~$D$ du complexe, et
+les droites infiniment voisines qui la rencontrent; on a la
+condition
+\[
+da · d\phi - db · df = 0,
+\]
+ou
+\[
+db · df - da \left(
+  \frac{\dd \phi}{\dd a}\, da
++ \frac{\dd \phi}{\dd b}\, db
++ \frac{\dd \phi}{\dd f}\, df\right) = 0;
+\]
+remplaçons $\dfrac{\dd \phi}{\dd a}$ par sa valeur tirée de~\Eq{(2)}, il vient
+\[
+\frac{\dd \phi}{\dd b} · \frac{\dd \phi}{\dd f}\, da^2
+  - \frac{\dd \phi}{\dd b}\, da · db
+  - \frac{\dd \phi}{\dd f}\, da · df + db · df = 0,
+\]
+ou
+\[
+\Tag{(3)}
+\left(\frac{\dd \phi}{\dd b}\, da - df\right)
+\left(\frac{\dd \phi}{\dd f}\, da - db\right) = 0.
+\]
+%% -----File: 221.png---Folio 213-------
+le point de rencontre de la droite~$D$ avec les droites infiniment
+voisines est
+\[
+\Tag{(4)}
+z = - \frac{df}{da} = - \frac{\dd \phi}{\dd b},
+\]
+de sorte qu'à tout plan passant par~$D$ correspond toujours le
+même point~$F$:
+\[
+x = az + f,\qquad
+y = bz + \phi,\qquad
+z = - \frac{\dd \phi}{\dd b}.
+\Tag{(5)}
+\]
+Différentions $x, y$
+\[
+dx = a\, dz + z\, da + df,\qquad
+dy = b\, dz + z\, db + d \phi,
+\]
+d'où, en remplaçant $z$ par sa valeur
+\[
+dx - a\, dz = - \frac{\dd \phi}{\dd b}\, da + df,\qquad
+dy - b\, dz =   \frac{\dd \phi}{\dd a}\, da + \frac{\dd \phi}{\dd f}\, df;
+\]
+d'où la relation
+\[
+\Tag{(6)}
+-\frac{\dd \phi}{\dd f} (dx - a\, dz) + dy - b\, dz = 0.
+\]
+Les différentielles $dx, dy, dz$ sont liées par une relation linéaire
+et homogène; les fonctions~$x, y, z$ sont liées au moins
+par une relation.
+
+Si on n'a qu'une relation, le lieu des points~$F$ est une
+surface, et \Eq{(6)}~exprime que la droite~$D$ est tangente à cette
+surface. Si on a 2~relations, le lieu des points~$F$ est une
+courbe et la droite~$D$ rencontre cette courbe. Tels sont les
+2~seuls cas possibles pour les complexes spéciaux.
+
+\Paragraph{Remarques. \1.} Dans l'équation~\Eq{(3)} nous avons jusqu'à
+présent considéré le seul facteur $\left(\dfrac{\dd \phi}{\dd b}\, da - df\right)$. Annulant l'autre
+facteur
+\[
+\frac{db}{da} = \frac{\dd \phi}{\dd f},
+\]
+nous aurions alors des droites du complexe qui seraient
+%% -----File: 222.png---Folio 214-------
+toutes situées dans un même plan avec~$D$, ce plan serait le
+plan singulier de l'homographie, et précisément le plan tangent
+à la surface lieu des points~$F$. On voit ainsi qu'en prenant
+l'un ou l'autre des facteurs, on définit la même surface
+par points et par plans tangents.
+
+\Paragraph{\2.} Si l'équation du complexe ne contient ni~$f$ ni~$g$, on
+a une relation entre les coefficients de direction de la droite~$D$,
+on a le complexe des droites rencontrant une même courbe
+à l'infini.
+
+\Paragraph{\3.} Le calcul précédent peut s'interpréter dans le cas
+d'un complexe quelconque. L'équation~\Eq{(1)}, qui n'est plus alors
+conséquence de l'équation du complexe, jointe à cette équation
+du complexe, définit une congruence des droites du complexe
+sur lesquelles l'homographie est spéciale. Ce sont les
+\emph{droites singulières} du complexe. Alors \emph{toutes les surfaces
+réglées du complexe passant par une droite singulière ont même
+plan tangent au point~$F$ de cette droite défini précédemment},
+ce plan tangent étant parallèle au plan
+\[
+-\frac{\dd \phi}{\dd f} (x - az) + y - bz = 0.
+\]
+Si le lieu des points singuliers est une surface, \Eq{(6)}~montre
+que cette surface est aussi l'enveloppe des plans singuliers,
+et les droites singulières lui sont tangentes. \emph{La surface des
+singularités est une des nappes de la surface focale de la
+congruence des droites singulières, les points et les plans
+singuliers sont des éléments focaux de cette congruence non
+associés entre eux. Si le lieu des points singuliers est une
+%% -----File: 223.png---Folio 215-------
+courbe, les plans singuliers sont \(d'après~\Eq{(6)}\) tangents à
+cette courbe, qui est une courbe focale de la congruence des
+droites singulières.}
+
+\Paragraph{\4.} Considérons en particulier le cas des \emph{complexes du
+\Ord{2}{e} degré}. En un point quelconque, le plan associé est tangent
+au cône du complexe; il est unique et bien déterminé. Il ne
+peut y avoir indétermination que si le cône du complexe associé
+à ce point se décompose. \emph{La surface des singularités est
+donc le lieu des points où le cône du complexe se décompose;
+c'est aussi l'enveloppe des plans pour lesquels la courbe du
+complexe se décompose}, comme le verrait par un raisonnement
+analogue.
+
+\MarginNote{Surfaces et
+courbes des
+complexes spéciaux.}
+Revenons aux complexes spéciaux: considérons d'abord le
+cas du complexe des tangentes à une surface~$(\Phi)$. Les cônes
+du complexe sont les cônes circonscrits à cette surface.
+Les plans tangents à~$(\Phi)$ constituent une intégrale complète.
+Une intégrale quelconque est donc l'enveloppe de $\infty^{1}$~plans tangents
+à~$(\Phi)$, c'est-à-dire une développable quelconque circonscrite
+à~$(\Phi)$. Les caractéristiques, qui sont en général les
+courbes de contact de la surface intégrale avec les surfaces,
+faisant partie de l'intégrale complète, qu'elle enveloppe,
+sont les génératrices rectilignes de ces développables, c'est-à-dire
+les droites même du complexe. Enfin on obtiendra les
+courbes intégrales en prenant l'enveloppe des caractéristiques
+sur les surfaces intégrales; ce sont précisément les
+arêtes de rebroussement des développables qui sont les courbes
+%% -----File: 224.png---Folio 216-------
+du complexe.
+
+Considérons maintenant le complexe des droites rencontrant
+une courbe; on voit de même que les surfaces du complexe
+sont les développables passant par la courbe, les caractéristiques
+sont les droites du complexe, et les courbes
+du complexe sont les arêtes de rebroussement.
+
+\emph{Dans les complexes spéciaux, l'équation aux dérivées partielles
+du \Ord{1}{e} ordre dont dépend la recherche des surfaces du
+complexe a pour caractéristiques les droites du complexe.
+Réciproquement toute équation aux dérivées partielles du \Ord{1}{er} ordre
+dont les caractéristiques sont des droites est associée
+à un complexe spécial.}
+
+Soit en effet l'équation aux dérivées partielles
+\[
+f(x, y, z, p, q) = 0
+\]
+dont les caractéristiques sont des droites. On obtient les
+surfaces intégrales en prenant une courbe intégrale et en menant
+les caractéristiques tangentes: donc les surfaces intégrales
+sont des développables, et le plan tangent est le même
+le long de chaque caractéristique, c'est-à-dire que $dp = 0$,
+$dq = 0$ doivent être conséquences de l'équation des caractéristiques,
+ce qui revient à dire que $f = 0$ doit \DPtypo{entrainer}{entraîner} comme
+conséquence les équations
+\[
+\frac{\dd f}{\dd x} + p\, \frac{\dd f}{\dd z} = 0,\qquad
+\frac{\dd f}{\dd y} + q\, \frac{\dd f}{\dd z} = 0.
+\]
+Supposons alors que $z$~figure dans l'équation aux dérivées
+partielles et posons
+\[
+f = z - \Phi(x, y, p, q);
+\]
+%% -----File: 225.png---Folio 217-------
+les conditions précédentes s'écriront
+\[
+\frac{\dd \Phi}{\dd x} - p = 0,\qquad
+\frac{\dd \Phi}{\dd y} - q = 0,
+\]
+d'où il résulte que $\Phi$~est de la forme
+\[
+\Phi = px + qy + \Psi(p, q),
+\]
+et l'équation aux dérivées partielles est
+\[
+z - px - qy = \Psi(p, q).
+\]
+Le plan tangent à une quelconque des surfaces intégrales est
+donc
+\[
+pX + qY - Z + \Psi(p\DPtypo{.}{,} q) = 0\Add{.}
+\]
+L'ensemble de tous ces plans a donc une enveloppe, surface ou
+courbe. Le cône élémentaire associé à un point quelconque est
+le cône circonscrit à cette surface ou à cette courbe, et
+l'équation aux dérivées partielles est bien associée à un
+complexe spécial.
+
+\Paragraph{Remarque.} Nous avons dû supposer que $z$~figurait dans
+l'équation aux dérivées partielles; s'il n'en est pas ainsi,
+cette équation s'écrit
+\[
+\Phi(x, y, p, q) = 0
+\]
+et les conditions obtenues plus haut s'écrivent
+\[
+\frac{\dd \Phi}{\dd x} = 0,\qquad
+\frac{\dd \Phi}{\dd y} = 0;
+\]
+$\Phi$ doit être indépendant de~$x\Add{,} y$, et l'équation aux dérivées partielles
+prend la forme
+\[
+\Phi(p, q) = 0.
+\]
+On a alors le complexe des droites rencontrant une courbe à
+l'infini.
+
+Considérons par exemple l'équation
+\[
+1 + p^2 + q^2 = 0
+\]
+%% -----File: 226.png---Folio 218-------
+elle définit le \emph{complexe des droites isotropes}; les courbes
+du complexe sont les courbes minima, et on les obtient sans
+intégration comme arêtes de rebroussement des développables
+isotropes.
+
+\Section{Surfaces normales aux droites du complexe\Add{.}}
+{4.}{} Proposons-nous maintenant de chercher les \emph{surfaces
+dont les normales appartiennent au complexe} défini par l'équation
+\[
+\Phi(a, b, f, g) = 0.
+\]
+Une normale à une surface du complexe est définie par les
+équations
+\[
+\frac{X - x}{p} = \frac{Y - y}{q} = -(Z - z)
+\]
+ou
+\[
+X = -pZ + x + pz,\qquad
+Y = -qZ + y + qz;
+\]
+de sorte que les surfaces cherchées sont définies par l'équation
+aux dérivées partielles
+\[
+\Phi(-p, -q, x + pz, y + qz) = 0.
+\]
+Si une surface répond à la question, il est évident que toutes
+les surfaces parallèles répondent aussi à la question.
+Si le complexe est spécial, le problème revient à la recherche
+d'une congruence de normales, connaissant une des multiplicités
+focales. Pour le cas d'un complexe quelconque, nous
+allons chercher les congruences de normales appartenant au
+complexe: on obtiendra ensuite les surfaces au moyen d'une
+quadrature. Pour que  $\infty^{2}$~droites:
+\[
+\frac{x - f}{a} = \frac{y - g}{b} = \frac{z - 0}{1}
+\]
+%% -----File: 227.png---Folio 219-------
+soient les normales d'une même surface, la condition est, en
+posant
+\[
+\alpha = \frac{a}{\sqrt{a^2 + b^2 + 1}},\qquad
+\beta  = \frac{b}{\sqrt{a^2 + b^2 + 1}},\qquad
+\gamma = \frac{1}{\sqrt{a^2 + b^2 + 1}},
+\]
+que $\alpha\, df + \beta\, dg$ soit une différentielle exacte. Or\Add{,} l'équation
+du complexe, résolue par rapport à~$\beta$ peut s'écrire
+\[
+\beta = \phi(\alpha, f, g)\Add{,}
+\]
+et $\alpha\, df + \phi(\alpha, f, g)\, dg$ doit être une différentielle exacte par
+rapport à deux variables indépendantes. Déterminons~$\alpha$ par
+exemple en fonction de $f, g$, nous aurons la condition
+\[
+\frac{\dd \alpha}{\dd g}
+  = \frac{\dd \phi}{\dd \alpha} · \frac{\dd \alpha}{\dd f}
+  + \frac{\dd \phi}{\dd f}.
+\]
+Cherchons une solution de la forme
+\[
+F(\alpha, f, g) = \cte,
+\]
+nous avons, pour déterminer~$F$,
+\[
+\frac{\dd F}{\dd f} + \frac{\dd \alpha}{\dd f} · \frac{\dd F}{\dd \alpha} = 0,
+\qquad
+\frac{\dd F}{\dd g} + \frac{\dd \alpha}{\dd g} · \frac{\dd F}{\dd \alpha} = 0.
+\]
+On est ramené à l'équation
+\[
+\frac{\dd F}{\dd g}
+  - \frac{\dd \phi}{\dd \alpha} · \frac{\dd F}{\dd f}
+  + \frac{\dd \phi}{\dd f} · \frac{\dd F}{\dd \alpha} = 0,
+\]
+qui se ramène au système d'équations différentielles ordinaires
+\[
+dg = \frac{\ -df\ }{\dfrac{\dd \phi}{\dd \alpha}}
+   = \frac{\ d \alpha\ }{\dfrac{\dd \alpha}{\dd f}}.
+\]
+
+Remarquons encore que \emph{les développées des surfaces cherchées
+sont les surfaces pour lesquelles $\infty^{1}$~géodésiques sont
+des courbes du complexe}. Ce sont les surfaces focales des congruences
+considérées.
+
+
+\ExSection{IX}
+
+\begin{Exercises}
+\item[43.] On considère deux plans rectangulaires, et toutes les droites
+telles que le segment intercepté sur chacune d'elles par les
+plans précédents ait une longueur constante. Trouver les congruences
+de normales du complexe de ces droites.
+
+\item[44.] On considère trois plans formant un trièdre trirectangle et
+les droites telles que le rapport des segments déterminés par
+ces trois plans sur chacune d'elles soit constant. Trouver
+les surfaces dont les normales appartiennent au complexe de
+ces droites. Il y a parmi ces surfaces une infinité de surfaces
+du \Ord{2}{e} ordre admettant les \Card{3} plans donnés comme plans de
+symétrie. Le complexe précédent est celui des normales à une
+famille de quadriques homofocales, ou homothétiques par rapport
+à leur centre.
+\end{Exercises}
+%% -----File: 228.png---Folio 220-------
+
+
+\Chapitre{X}{Complexes Linéaires.}
+
+\Section{Généralités sur les complexes algébriques.}
+{1.}{} Soit une droite
+\[
+\Tag{(1)}
+x = az + f,\qquad
+y = bz + g;
+\]
+un \emph{complexe algébrique} sera défini par une relation algébrique
+entre $a, b, f, g$:
+\[
+\Phi(a, b, f, g) = 0.
+\]
+Si on considère les droites du complexe passant par un point~$A$,
+et situées dans un plan~$P$ passant par ce point, ce sont
+les génératrices d'intersection du plan~$P$ avec le cône du
+complexe associé au point~$A$, ou bien les tangentes issues de~$A$
+à la courbe du complexe située dans le plan~$P$; si le complexe
+est algébrique, le cône et la courbe sont algébriques,
+et on voit que \emph{le degré du cône du complexe est égal à l'ordre
+de la courbe plane du complexe}; leur valeur commune s'appelle
+le \emph{degré du complexe}, c'est le nombre de droites du
+complexe situées dans un plan et passant par un point de ce
+plan.
+
+Si ce nombre est égal à~$1$, on a ce qu'on appelle un
+\emph{complexe linéaire}; le cône du complexe associé au point~$A$ est
+un plan qu'on appelle \emph{plan focal} ou \emph{plan polaire} du point~$A$.
+La courbe du complexe située dans un plan~$P$ se réduit à un
+point, qu'on appelle \emph{foyer} ou \emph{pôle} du plan~$P$; si le plan~$P$ est
+%% -----File: 229.png---Folio 221-------
+le plan polaire du point~$A$, le point~$A$ est le pôle du plan~$P$;
+\emph{il y a réciprocité entre un pôle et son plan polaire}.
+
+\Section{Coordonnées homogènes.}
+{2.}{} Pour l'étude des complexes algébriques il y a avantage
+à remplacer $a, b, f, g$ par les coordonnées homogènes de
+droites.
+
+\Paragraph{Coordonnées de Plücker.} Considérons les équations d'une
+droite en coordonnées cartésiennes
+\[
+\Tag{(2)}
+\frac{X - f}{a} = \frac{Y - g}{b} = \frac{Z - h}{c},
+\]
+équations qui contiennent comme cas particulier les équations~\Eq{(1)}.
+Nous prendrons pour coordonnées pluckériennes de la
+droite les \Card{6} quantités
+\[
+a,\quad b,\quad c,\qquad
+p = gc - hb,\qquad
+q = ha - fc,\qquad
+r = fb - ga\Add{.}
+\]
+Ces \Card{6} coordonnées sont, comme on le voit immédiatement, liées
+par la relation homogène
+\[
+\Tag{(3)}
+pa + qb + rc = 0.
+\]
+Ces \Card{6} paramètres liés par une relation homogène se réduisent
+à~\Card{4} en réalité; $a, b, c$~sont les projections sur les axes d'un
+certain segment porté par la droite; $p, q, r$~sont les moments de
+ce segment par rapport aux axes (en coordonnées rectangulaires).
+
+Voyons ce que devient l'équation du complexe. De~\Eq{(2)} on
+tire
+\[
+X = \frac{a}{c}\, Z - \frac{q}{c},\qquad
+Y = \frac{b}{c}\, Z + \frac{p}{c},
+\]
+et l'équation
+\[
+\Phi(a, b, f, g) = 0
+\]
+devient
+\[
+\Phi \left(\frac{a}{c}, \frac{b}{c}, -\frac{q}{c}, \frac{p}{c}\right) = 0.
+\]
+%% -----File: 230.png---Folio 222-------
+Cette équation peut être rendue homogène, et prend la forme
+\[
+\Psi(a, b, c, p, q) = 0
+\]
+on peut y introduire~$r$ en vertu de l'équation~\Eq{(3)}, et on obtient
+finalement, pour définir le complexe, une équation homogène
+entre les coordonnées pluckériennes:
+\[
+\chi(a, b, c, p, q, r) = 0.
+\]
+Réciproquement, toute équation de la forme précédente peut
+être ramenée à la forme
+\[
+\Psi \left(a, b, c, p, -q, -\frac{pa + qb}{c}\right) = 0,
+\]
+et par suite à la forme primitive de l'équation du complexe.
+
+Cherchons le \emph{cône du complexe} de sommet $(x, y, z)$. Nous
+avons, $X, Y, Z$~étant les coordonnées courantes,
+\begin{alignat*}{3}%[** TN: Set on one line in original]
+a &= X - x, &
+b &= Y - y, &
+c &= Z - z, \\
+\Err{}{\intertext{ou encore}}
+p &= \Err{cY - bZ}{yZ - zY},\qquad &
+q &= \Err{aZ - cX}{zX - xZ},\qquad &
+r &= \Err{bX - aY}{xY - yX};
+\end{alignat*}
+l'équation du cône du complexe s'obtiendra en remplaçant $a, b, c$\Add{,}
+$p, q, r$ par les valeurs précédentes dans l'équation du complexe.
+C'est donc:
+\[
+\chi (X - x, Y - y, Z - z,
+\Err{cY - bZ, aZ - cX, bX - aY}{yZ - zY, zX - xZ, xY - yX}) = 0.
+\]
+
+Si on veut une \emph{courbe du complexe}, on prendra
+\begin{alignat*}{3}%[** TN: Set on one line in original]
+a &= dx, &
+b &= dy, &
+c &= dz, \\
+p &= y\, dz - z\, dy,\qquad &
+q &= z\, dx - x\, dz,\qquad &
+r &= x\, dy - y\, dx,
+\end{alignat*}
+et on a l'équation différentielle des courbes du complexe
+\[
+\chi(dx, dy, dz, y\, dz - z\, dy, z\, dx - x\, dz, x\, dy - y\, dx) = 0.
+\]
+
+La condition pour qu'un complexe soit spécial est
+\[
+\frac{\dd \Phi}{\dd a} · \frac{\dd \Phi}{\dd g}
+  - \frac{\dd \Phi}{\dd b} · \frac{\dd \Phi}{\dd f} = 0;
+\]
+elle devient ici
+\[
+\frac{\dd \chi}{\dd a} · \frac{\dd \chi}{\dd p}
+  + \frac{\dd \chi}{\dd b} · \frac{\dd \chi}{\dd q}
+  + \frac{\dd \chi}{\dd c} · \frac{\dd \chi}{\dd r} = 0;
+\]
+%% -----File: 231.png---Folio 223-------
+dans le cas d'un complexe algébrique quelconque, cette équation,
+jointe à celle du complexe définit \emph{la congruence des
+droites singulières}.
+
+Reprenons l'homographie entre droites et plans d'une
+droite du complexe; les coefficients de cette homographie
+sont $\dfrac{\dd \Phi}{\dd a}$, $\dfrac{\dd \Phi}{\dd b}$, $\dfrac{\dd \Phi}{\dd f}$, $\dfrac{\dd \Phi}{\dd g}$, et par suite en coordonnées homogènes, ce
+sont $\dfrac{\dd \chi}{\dd a}, \dots, \dfrac{\dd \chi}{\dd r}$. Considérons la droite $(a_{0}, b_{0}, c_{0}, p_{0}, q_{0}, r_{0})$.
+L'équation
+\[
+\sum a\, \frac{\dd \chi}{\dd a_{0}} + \sum p\, \frac{\dd \chi}{\dd p_{0}} = 0
+\]
+définit un complexe linéaire contenant la droite considérée,
+et sur cette droite, l'homographie pour ce complexe linéaire
+est précisément la même que pour le complexe primitif. Ce
+complexe linéaire est dit \emph{tangent} au complexe donné.
+
+\Paragraph{Remarques.} Si nous définissons la droite par \Card{2} points
+$(x, y, z)$ et $(x', y', z')$ nous avons
+\begin{alignat*}{3}%[** TN: Set on one line in original]
+a &= x' - x, &
+b &= y' - y, &
+c &= z' - z, \\
+p &= yz' - z\Err{y}{y'},\qquad &
+q &= zx' - x\Err{z}{z'},\qquad &
+r &= xy' - yx';
+\end{alignat*}
+d'où l'équation du cône du complexe
+\[
+\chi (x' - x, y' - y, \DPtypo{z}{z'} - z, yz' - zy', zx' - xz', xy' - yx') = 0;
+\]
+
+Corrélativement, définissons la droite par \Card{2} plans
+$(u, v, w, s)$\Add{,} $(u', v', w', s')$. On trouve facilement
+\begin{alignat*}{3}%[** TN: Set on one line in original]
+a &= vw' - wv', &
+b &= wu' - uw', &
+c &= uv' - vu', \\
+p &= su' - us',\qquad &
+q &= sv' - vs',\qquad &
+r &= sw' - ws';
+\end{alignat*}
+on obtient alors l'équation tangentielle d'une courbe plane
+du complexe
+\[
+\chi (vw' - wv', \dots, su' - us', \dots) = 0,
+\]
+%% -----File: 232.png---Folio 224-------
+et on voit bien ainsi que la classe de cette courbe est égale
+à l'ordre du cône du complexe.
+
+\Paragraph{Coordonnées générales de Grassmann et Klein.} Plus généralement
+prenons un tétraèdre de référence quelconque, et
+soient $x_{1}, x_{2}, x_{3}, x_{4}$ les coordonnées d'un point; $u_{1}, u_{2}, u_{3}, u_{4}$ les
+coordonnées d'un plan. Considérons la droite comme définie
+par \Card{2} points~$(x)\Add{,} (y)$. Nous prendrons comme coordonnées de
+cette droite les quantités
+\[
+p_{ik} = \begin{vmatrix}
+\Err{x}{x_{i}} & x_{k} \\
+\Err{y}{y_{i}} & y_{k}
+\end{vmatrix}
+\qquad
+(i, k = 1, 2, 3, 4);
+\]
+remarquons que l'on a $p_{ii} = 0$ et $p_{ki} = -p_{ik}$, de sorte que
+l'on n'obtient ainsi que \Card{6} coordonnées $p_{12}, p_{13}, p_{14}$, $p_{23}, p_{24}, p_{34}$\Add{.}
+Ce sont les moments par rapport au segment des \Card{2} points~$(x)\Add{,} (y)$
+des segments égaux à~$1$ pris sur les \Card{6} arêtes du tétraèdre,
+ou du moins des quantités proportionnelles à ces moments.
+
+Si on a \Card{2} droites $(p_{ik})$~et~$(p'_{ik})$, leur moment relatif~$M$
+est donné par la formule
+\[
+\rho M = \sum p_{ik} p'_{hl}.
+\]
+Si ce moment est nul, les \Card{2} droites se rencontrent. Or\Add{,} considérons
+le déterminant
+\[
+\Theta = \begin{vmatrix}
+x_{1} & x_{2} & x_{3} & x_{4} \\
+y_{1} & y_{2} & y_{3} & y_{4} \\
+x_{1} & x_{2} & x_{3} & x_{4} \\
+y_{1} & y_{2} & y_{3} & y_{4}
+\end{vmatrix} = 0.
+\]
+Développons d'après la règle de Laplace, nous avons
+\[
+\Theta = 2 (p_{12} p_{34} + p_{13} p_{42} + p_{14} p_{\DPtypo{24}{23}})
+  = 2 \Phi (p_{ik}) = 0;
+\]
+de sorte que la condition de rencontre des \Card{2} droites est
+%% -----File: 233.png---Folio 225-------
+\[
+\sum p'_{ik}\, \frac{\dd \Phi}{\dd p_{ik}} = 0.
+\]
+
+Si nous définissons la droite par \Card{2} plans $(u)\Add{,} (v)$, nous
+prendrons pour coordonnées
+\[
+q_{ik} = \begin{vmatrix}
+u_{i} & u_{k} \\
+v_{i} & v_{k}
+\end{vmatrix}\Add{.}
+\]
+Cherchons les relations entre les~$p, q$. La droite étant l'intersection
+des plans $(u)\Add{,} (v)$, un point de cette droite sera
+l'intersection des plans $(u)\Add{,} (v)\Add{,} (w)$. On aura donc
+\[
+\left\{
+\begin{alignedat}{5}
+&u_{1} x_{1} &&+ u_{2} x_{2} &&+ u_{3} x_{3} &&+ u_{4} x_{4} &&= 0, \\
+&v_{1} x_{1} &&+ v_{2} x_{2} &&+ v_{3} x_{3} &&+ v_{4} x_{4} &&= 0, \\
+&w_{1} x_{1} &&+ w_{2} x_{2} &&+ w_{3} x_{3} &&+ w_{4} x_{4} &&= 0.
+\end{alignedat}
+\right.
+\]
+Considérons le déterminant
+\[
+\Omega = \begin{vmatrix}
+u_{1} & u_{2} & u_{3} & u_{4} \\
+v_{1} & v_{2} & v_{3} & v_{4} \\
+w_{1} & w_{2} & w_{3} & w_{4} \\
+s_{1} & s_{2} & s_{3} & s_{4}
+\end{vmatrix}\Add{;}
+\]
+la coordonnée~$x_{i}$\DPtypo{;}{} est égale au coefficient~$S_{i}$ de~$s_{i}$. Pour avoir
+un \Ord{2}{e} point de la droite, nous le définirons par les \Card{3} plans
+$(u)\Add{,} (v)\Add{,} (s)$, et alors $y_{i} = W_{i}$. Considérons l'adjoint de~$\Omega$
+\[
+\begin{vmatrix}
+U_{1} & U_{2} & U_{3} & U_{4} \\
+V_{1} & V_{2} & V_{3} & V_{4} \\
+W_{1} & W_{2} & W_{3} & W_{4} \\
+S_{1} & S_{2} & S_{3} & S_{4}
+\end{vmatrix},
+\]
+nous avons, en associant à chaque mineur du \Ord{2}{e} ordre de~$\Omega$ le
+mineur complémentaire de l'adjoint
+\[
+p_{ik} = \Omega\, \frac{\dd \Phi}{\dd q_{ik}};
+\]
+on peut prendre~$\Omega$ comme arbitraire, et écrire
+\[
+p_{ik} = \frac{\dd \Phi}{\dd q_{ik}}.
+\]
+%% -----File: 234.png---Folio 226-------
+et de même
+\[
+q_{ik} = \frac{\dd \Phi}{\dd p_{ik}}.
+\]
+
+L'équation du complexe sera alors $F (p_{ik}) = 0$ ou $F (q_{hl})
+= 0$, d'où les équations du cône ou de la courbe du complexe.
+La condition pour que le complexe soit spécial est
+\[
+\frac{\dd F}{\dd p_{12}} · \frac{\dd F}{\dd p_{34}} +
+\frac{\dd F}{\dd p_{13}} · \frac{\dd F}{\dd p_{24}} +
+\frac{\dd F}{\dd p_{14}} · \frac{\dd F}{\dd p_{23}} = 0\Add{.}
+\]
+
+\Section{Complexe linéaire.}
+{3.}{} \DPchg{Etudions}{Étudions} plus spécialement le complexe linéaire. Son
+équation s'écrit
+\[
+\sum A_{hl} p_{ik} = 0;
+\]
+le complexe est spécial s'il satisfait à la relation
+\[
+A_{12} A_{34} + A_{13} A_{42} + A_{14} A_{23} = 0,
+\]
+et cette équation exprime que les~$A$ sont les coordonnées d'une
+droite; l'équation du complexe exprime que toute droite du
+complexe rencontre cette droite. \emph{Un complexe linéaire spécial
+est constitué par les droites rencontrant une droite fixe},
+qu'on appelle \emph{directrice du complexe}.
+
+Si on a une droite du complexe, un point~$A$ de cette droite et
+son plan polaire~$P$, le cône du complexe se réduisant ici au
+plan~$P$, l'homographie du complexe est celle des plans de la
+droite~$D$ associés à leurs pôles.
+
+\Section{Faisceau de complexes.}
+{4.}{} Soient \Card{2} complexes linéaires
+\[
+\sum A_{hl} p_{ik} = 0, \qquad
+\sum B_{hl} p_{ik} = 0;
+\]
+l'équation
+\[
+\sum (A_{hl} + \lambda B_{hl}) p_{ik} = 0
+\]
+représentera un \emph{faisceau de complexes}. Cherchons dans ce
+faisceau les complexes spéciaux. Ils sont définis par
+%% -----File: 235.png---Folio 227-------
+l'équation
+\begin{align*}%[** TN: Rebroken]
+     &(A_{14} + \lambda B_{14})(A_{23} + \lambda B_{23}) \\
+{}+{}&(A_{12} + \lambda B_{12})(A_{34} + \lambda B_{34}) \\
+{}+{}&(A_{13} + \lambda B_{13})(A_{24} + \lambda B_{24}) = 0,
+\end{align*}
+équation du \Ord{2}{e} degré. \emph{Dans un faisceau de complexes linéaires
+il y a donc \Card{2} complexes \DPtypo{speciaux}{spéciaux}.} Cherchons a quelles conditions
+il y a racine double. Supposons que $\lambda = 0$ soit racine,
+on a
+\[
+\sum A_{12} A_{34} = 0,
+\]
+et l'équation précédente se réduit à
+\[
+A_{12} B_{34} + A_{34} B_{12} + \dots + \lambda (B_{12} B_{34} + \dots ) = 0.
+\]
+Nous appellerons \emph{invariant du complexe} la quantité
+\[
+\Delta_{A} = A_{12} A_{34} + A_{13} A_{\DPtypo{42}{24}} + A_{14} A_{23},
+\]
+et \emph{invariant relatif} la quantité
+\[
+\Delta_{AB} = \sum  B_{ik}\, \frac{\dd \Delta_{A}}{\dd A_{ik}};
+\]
+l'équation devient alors
+\[
+\Delta_{AB} + \lambda \Delta_{B} = 0;
+\]
+pour que $\lambda = 0$ soit racine double, il faut que $\Delta_{AB} = 0$. Or\Add{,} les
+$A_{ik}$ sont des coordonnées de droite, la condition $\Delta_{AB} = 0$ exprime
+que cette droite appartient au \Ord{2}{e} complexe qui définit le
+faisceau. Elle appartient évidemment au \Ord{1}{er}. Donc \emph{pour que
+l'un des complexes spéciaux soit double, il faut et il suffit
+que sa directrice appartienne à tous les complexes du faisceau}\Add{.}
+Pour que l'équation se réduise à une identité, c'est-à-dire
+pour que tous les complexes du faisceau soient spéciaux, il
+faut encore que $\Delta_{B} = 0$; il faut donc que les \Card{2} complexes
+soient spéciaux, et que leurs directrices se rencontrent.
+
+Nous appellerons \emph{congruence linéaire} l'ensemble des
+droites communes à \Card{2} complexes linéaires. Par tout point de
+%% -----File: 236.png---Folio 228-------
+l'espace passe une droite de cette congruence, et dans tout
+plan il y a une droite. Considérons le faisceau déterminé par
+les \Card{2} complexes qui définissent la congruence. Si ce faisceau
+a \Card{2} complexes spéciaux distincts, toutes les droites de la
+congruence appartiennent à ces complexes spéciaux, et par
+suite rencontrent \Card{2} directrices fixes. \emph{Une congruence linéaire
+est formée en général des droites rencontrant \Card{2} directrices
+fixes.} Si les complexes spéciaux sont confondus, soit $\Delta$ leur
+directrice commune; considérons un complexe quelconque~$(c)$ du
+faisceau. $\Delta$~est une droite du complexe~$(c)$; à chaque point~$A$
+de~$\Delta$ correspond son plan polaire par rapport au complexe~$(c)$;
+les droites de la congruence passant par~$A$ et appartenant au
+complexe~$(c)$ sont dans ce plan polaire. Or\Add{,} les points de~$\Delta$
+ont même plan polaire par rapport à tous les complexes du
+faisceau. Les droites de la congruence rencontrent la droite~$\Delta$,
+et pour chaque point de cette droite sont situées dans le
+plan polaire correspondant.
+
+\Section{Complexes en involution.}
+{5.}{} Reprenons le faisceau de complexes précédent. Les \Card{2} complexes
+de base sont dits \emph{en involution} si on a $\Delta_{AB} = 0$.
+Considérons une droite~$D$ commune aux \Card{2} complexes. A un point~$A$
+de cette droite correspond son plan polaire dans chacun des
+complexes, soient $P, Q$ ces plans; il en résulte une correspondance
+homographique entre les plans $P, Q$ de la droite.
+De même, en partant d'un plan de la droite, on verrait qu'il
+existe une homographie entre les points de la droite. Cherchons
+les plans doubles de cette homographie. Considérons une des
+directrices~$\Delta$ de la congruence linéaire définie par les \Card{2} complexes,
+%% -----File: 237.png---Folio 229-------
+et le plan~$D \Delta$; le pôle de ce plan est l'intersection~$A'$
+de~$D$ avec la \Ord{2}{e} directrice~$\Delta'$,
+car toutes les droites passant par~$A'$
+et rencontrant $\Delta$ appartiennent à la
+congruence, et par suite aux \Card{2} complexes.
+Ainsi $A'$~est foyer du plan~$D \Delta$;
+il est aussi évidemment foyer
+du plan~$D \Delta'$; et l'on voit facilement
+que ces \Card{2} plans sont les plans doubles cherchés. Maintenant
+pour que l'homographie entre les plans~$P, Q$ soit une involution,
+il faut que les plans $P\Add{,}Q$ soient conjugués par rapport à
+ces plans doubles. L'équation du plan polaire d'un point par
+rapport à un complexe quelconque du faisceau est
+\[
+\sum (A_{hl} + \lambda B_{hl})
+\begin{vmatrix}
+X_{i}   &   X_{k}  \\
+x_{i}   &   x_{k}
+\end{vmatrix} = 0,
+\]
+
+\Illustration[1.25in]{237a}
+\noindent équation de la forme
+\[
+P + \lambda Q = 0.
+\]
+Considérons alors \Card{4} complexes quelconques du faisceau, le
+rapport anharmonique des \Card{4} plans polaires d'un même point
+dans ces \Card{4} complexes est égal au rapport anharmonique des \Card{4} quantités~$\lambda$
+correspondantes. Or\Add{,} prenons en particulier les \Card{2} complexes
+de base et les complexes spéciaux. Les valeurs de
+correspondantes sont~$0, \infty$, et les racines de l'équation
+\[
+\sum (A_{14} + \lambda B_{14}) (A_{23} + \lambda B_{23}) = 0;
+\]
+et la condition pour que les \Card{2} \Ord{1}{ères} soient conjuguées harmoniques
+par rapport aux \Card{2} autres est
+\[
+\lambda_{1} + \lambda_{2} = 0
+\]
+ou $\Delta_{AB} = 0$. \emph{Ainsi donc si \Card{2} complexes sont en involution, les
+%% -----File: 238.png---Folio 230-------
+plans polaires d'un point dans ces \Card{2} complexes sont conjugués
+harmoniques par rapport aux plans passant par ce point et
+par les directrices de la congruence commune aux \Card{2} complexes.
+Et réciproquement.}
+
+\Paragraph{Application.} On peut généraliser encore les coordonnées
+de droites. Reprenons la relation fondamentale
+\[
+ap + bq + cr = 0;
+\]
+elle est homogène et du \Ord{2}{e} degré. Or\Add{,} il existe un type remarquable
+d'équations du \Ord{2}{e} degré, celui où ne figurent que les
+carrés. Posons
+\begin{alignat*}{3}
+a + ip &= t_{1},        & b + iq &= t_{3},        & c + ir &= t_{5}, \\
+a - ip &= it_{2},\qquad & b - iq &= it_{4},\qquad & c - ir &= it_{6};
+\end{alignat*}
+la condition précédente devient
+\[
+t_{1}^{2} + t_{3}^{2} + t_{5}^{2} + t_{2}^{2} + t_{4}^{2} + t_{6}^{2} = 0.
+\]
+On introduit comme coordonnées homogènes les $t$, qui sont des
+fonctions linéaires homogènes des coordonnées pluckériennes.
+En égalant ces \Card{6} coordonnées à~$0$, on a les équations de \Card{6} complexes
+qui sont \Card{2} à \Card{2} en involution, car on voit facilement
+que la condition pour que les \Card{2} complexes
+\[
+\sum A_{i} t_{i} = 0, \qquad
+\sum B_{i} t_{i} = 0,
+\]
+soient en involution est
+\[
+\sum A_{i} B_{i} = 0.
+\]
+
+\Section{Droites conjuguées.}
+{6.}{} Considérons un complexe~$(c)$ et une droite~$\Delta$ n'appartenant
+pas à ce complexe\DPtypo{;}{.} Considérons la congruence commune
+à~$(c)$ et au complexe spécial de directrice~$\Delta$\DPtypo{;}{.} Cette
+congruence a une \Ord{2}{e} directrice $\Delta'$ qui est dite la \emph{droite conjuguée} %[**TN: Partial hyphenated word in original]
+%% -----File: 239.png---Folio 231-------
+de~$\Delta$. Il y a évidemment réciprocité entre ces \Card{2} droites.
+\emph{Toutes les droites du complexe~$(c)$ qui rencontrent
+la droite~$\Delta$ rencontrent sa conjuguée~$\Delta'$}, puisque ce sont des
+droites de la congruence, et inversement \emph{toute droite rencontrant
+à la fois les \Card{2} droites conjuguées~$\Delta\Add{,} \Delta'$ appartient à la
+congruence et par suite au complexe}. Si on considère un point~$A$
+de~$\Delta$, son plan polaire passe par~$\Delta'$, puisque toutes les
+droites passant par~$A$ et rencontrant~$\Delta'$ appartiennent au complexe.
+\emph{$\Delta'$~est donc l'enveloppe des plans polaires des points
+de sa conjuguée~$\Delta$}. On voit de même \emph{que $\Delta'$~est le lieu des pôles
+des plans passant par sa conjuguée~$\Delta$}. Si la droite~$\Delta$ appartient
+au complexe~$(c)$, la congruence précédente a ses \Card{2} directrices
+confondues. \emph{Les droites du complexe sont à elles-mêmes
+leurs conjuguées.}
+
+Supposons l'équation du complexe
+\[
+F (a, b, c, p, q, r) = Pa + Qb + Rc + Ap + Bq + Cr = 0\DPtypo{;}{.}
+\]
+Cherchons les coordonnées $(a', b', c', p', q', r')$ de la conjuguée
+d'une droite $(a, b, c, p, q, r)$. Il suffit d'exprimer que le complexe
+donné, et les complexes spéciaux ayant pour directrices
+les droites $(a, b, c, p, q, r)$\Add{,} $(a', b', c', p', q', r')$ appartiennent à un
+même faisceau, ce qui donne
+\[
+P + \lambda p + \lambda' p' = 0, \quad\text{et les analogues}\dots.
+\]
+Multiplions respectivement par $a, b, c, p, q, r$ et ajoutons membre
+à membre, le coefficient de~$\lambda$ \DPchg{disparait}{disparaît} et nous avons
+\[
+F(a, b, c, p, q, r) + \lambda' \sum (ap' + pa') = 0;
+\]
+posons pour abréger
+\[
+\sum (ap' + pa') = \sigma,
+\]
+%% -----File: 240.png---Folio 232-------
+nous avons
+\[
+\Tag{(1)}
+F(a, b, c, p, q, r) + \lambda' \sigma = 0.
+\]
+Si nous multiplions par $a', b', c', p', q', r'$ et si nous ajoutons, c'est
+le coefficient de~$\lambda'$ qui \DPchg{disparaitra}{disparaîtra} et nous aurons
+\[
+\Tag{(2)}
+F(a', b', c', p', q', r') + \lambda \sigma = 0.
+\]
+Enfin si nous multiplions par $A, B, C, P, Q, R$, nous obtenons, en
+posant
+\begin{gather*}
+\Delta = AP + BQ + \Err{c}{C}R, \\
+2\Delta + \lambda F (a, b, c, p, q, r)
+  + \lambda' F(a', b', c', p', q', r') = 0\Add{,}
+\end{gather*}
+ce qui peut s'écrire, en tenant compte de~\Eq{(1)}\Add{,}~\Eq{(2)}
+\[
+\Delta = \lambda \lambda' \sigma,
+\]
+d'où
+\[
+\lambda = \frac{\Delta}{\lambda' \sigma}
+      = - \frac{\Delta}{F(a, b,c, p, q, r)};
+\]
+et nous pouvons prendre pour coordonnées de la droite conjuguée
+\[
+a = A - \frac{\Delta}{F(a\Add{,} \dots)}\, a, \quad\text{et les analogues}, \dots
+\]
+ou
+\[
+a' = AF(a, b, c, p, q, r) - \Delta a, \quad\text{et les analogues}, \dots.
+\]
+
+Supposons qu'on prenne \Card{2} droites conjuguées pour arêtes
+opposées du tétraèdre de référence. Si nous appelons $x, y, z, t$
+les coordonnées tétraédriques, nous aurons
+\begin{alignat*}{3}
+a &= xt' - tx', &
+b &= yt' - ty', &
+c &= zt'- tz', \\
+p &= yz' - zy', \qquad &
+q &= zx' - xz', \qquad &
+r &= xy' - yx'.
+\end{alignat*}
+Supposons qu'on prenne pour droites conjuguées les droites
+$(x = 0, y = 0)$ et $(z = 0, t = 0)$. Leurs coordonnées sont
+\begin{align*}
+&a = 0,     && b = 0,    && c,         && p = 0, && q = 0, && r = 0; \\
+&a' = 0, && b' = 0, && c' = 0, && p' = 0, && q' = 0, && r'.
+\end{align*}
+Exprimons que ces droites sont conjuguées. D'après les conditions
+%% -----File: 241.png---Folio 233-------
+trouvées précédemment, nous avons
+\begin{alignat*}{3}
+0 &= AF (a\Add{,} \dots),\qquad &
+0 &= BF (a\Add{,} \dots),\qquad &
+0 &= CF - \Delta c, \\
+%
+0 &= PF, &
+0 &= QF, &
+r' &= RF.
+\end{alignat*}
+Or,
+\[
+F(a, b, c, p, q, r) = F(0, 0, c, 0, 0, 0) = Rc;
+\]
+il en résulte que $A = 0, B = 0, P = 0, Q = 0$. Alors
+\[
+\Delta = RC,
+\]
+et l'équation du complexe devient
+\[
+Cr + Rc = 0,
+\]
+ou
+\[
+r = kc.
+\]
+
+En particulier cherchons à effectuer cette réduction en
+axes cartésiens. Nous supposerons que $\Delta$ soit l'axe~$Oz$ et que
+$\Delta'$ soit rejetée à l'infini dans le plan des~$x\Add{,}y$. Il faut
+d'abord montrer qu'il y a des droites dont la conjuguée peut
+être rejetée à l'infini. Pour qu'une droite $(a, b, c, p, q, r)$
+soit à l'infini, il faut que $a = 0$, $b = 0$, $c = 0$; et d'après
+les formules précédemment trouvées, nous avons pour les conjuguées
+de ces droites
+\[
+\frac{a'}{A} = \frac{b'}{B} = \frac{c'}{C}
+  = \frac{F(0, 0, 0, p, q, r)}{\Delta};
+\]
+$a', b', c'$ sont donc proportionnels à des quantités fixes. \emph{Les
+conjuguées des droites de l'infini sont parallèles à une
+même direction. Ces droites sont les lieux des pôles des plans
+parallèles à un plan fixe.} On les appelle \emph{diamètres}. En rapportant
+donc un complexe à un diamètre et au plan conjugué,
+on peut mettre l'équation du complexe sous la forme
+\[
+r = kc.
+\]
+
+On peut obtenir cette réduction en axes rectangulaires.
+Il existe en effet une infinité de droites perpendiculaires
+à leurs conjuguées. Elles sont définies par la relation
+%% -----File: 242.png---Folio 234-------
+\[
+aa' + bb' + cc' = 0,
+\]
+ou
+\[
+(Aa + Bb + Cc) F(a, b, c, p, q, r) - \Delta (a^{2} + b^{2} + c^{2}) = 0.
+\]
+Ces droites constituent donc un complexe du \Ord{2}{e} degré. Prenons
+un diamètre quelconque $(a, b, c, p, q, r)$. Le plan conjugué, passant
+par l'origine a pour équation
+\[
+p'X + q'Y + r'Z = 0;
+\]
+la condition pour qu'il soit perpendiculaire au diamètre est
+\[
+\frac{a}{p'} = \frac{b}{q'} = \frac{c}{r'},
+\]
+ou
+\[
+\frac{a}{PF - \Delta p} = \frac{b}{QF - \Delta q} = \frac{c}{RF - \Delta r};
+\]
+la droite conjuguée du diamètre étant à l'infini, on peut remplacer
+$a, b, c$ par $A, B, C$, ce qui donne
+\[
+\frac{A}{PF - \Delta p} = \frac{B}{QF - \Delta q} = \frac{C}{RF - \Delta r}.
+\]
+On a
+\[
+ap + bq + cr = 0,
+\]
+donc ici
+\[
+Ap + Bq + Cr = 0;
+\]
+et
+\[
+F(a, b, c, p, q, r) = Pa + Qb + Rc\Add{.}
+\]
+Multiplions alors les \Card{2} termes des rapports précédents respectivement
+par $A, B, C$ et ajoutons, nous obtenons le rapport égal~$\dfrac{\sum A^{2}}{\Delta F}$;
+nous pouvons alors prendre  $a = A$, $b = B$, $c = C$ et $F =\Delta$,
+et enfin
+\[
+\frac{A}{P \Delta - p \Delta} = \frac{\sum A^{2}}{\Delta^{2}},
+  \quad\text{et les analogues\Add{,}}
+\]
+d'où
+\[
+p = P - \frac{A \Delta}{\sum A^{2}}, \qquad
+q = Q - \frac{B \Delta}{\sum B^{2}}, \qquad
+r = R - \frac{C \Delta}{\sum C^{2}}.
+\]
+Nous obtenons ainsi un diamètre perpendiculaire au plan conjugué,
+c'est \emph{l'axe du complexe} et on a l'équation réduite en
+coordonnées rectangulaires
+\[
+r - mc = 0.
+\]
+Le complexe ne dépend que d'un seul paramètre~$m$ par rapport
+%% -----File: 243.png---Folio 235-------
+au groupe des mouvements.
+
+Si $r = 0$, $c = 0$, l'équation est satisfaite; or\Add{,} $r = 0$,
+$c = 0$ sont les coordonnées des droites rencontrant $Oz$ et perpendiculaire
+à~$Oz$. \emph{Le complexe contient toutes les droites
+rencontrant l'axe et perpendiculaires à l'axe}; $c, r$~sont des
+coordonnées qui ne changent pas si on fait tourner la droite
+autour de~$Oz$; de même si on la déplace parallèlement à~$Oz$.
+Autrement dit \emph{un mouvement \DPchg{hélicoidal}{hélicoïdal} d'axe~$Oz$ laisse le complexe
+inaltéré. Il en résulte que si on a $\infty^{1}$~droites appartenant
+au complexe et ne dérivant pas les unes des autres par
+un mouvement \DPchg{hélicoidal}{hélicoïdal}, on obtiendra toutes les droites du
+complexe en faisant subir à ce système de droites les translations
+et rotations précédentes}. Considérons les droites dont
+les coordonnées~$a, p$ sont nulles, et cherchons parmi ces
+droites celles qui appartiennent au complexe; nous trouvons
+les droites
+\[
+bx = mc, \qquad
+cy - bz = 0,
+\]
+qui constituent une famille de génératrices du \DPchg{paraboloide}{paraboloïde}
+\[
+xy - mz = 0.
+\]
+Par conséquent, \emph{pour obtenir toutes les droites d'un complexe,
+il suffit de prendre un système de génératrices d'un \DPchg{paraboloide}{paraboloïde}
+et de faire subir à chacune d'elles un des mouvements
+précédents}.
+
+\Section{Réseau de complexes.}
+{7.}{}
+$\Phi = 0$, $\Phi' = 0$, $\Phi'' = 0$ étant les équations de \Card{3} complexes
+linéaires, un \emph{réseau de complexes} sera défini par l'équation
+%% -----File: 244.png---Folio 236-------
+\[
+\lambda \Phi + \lambda' \Phi' + \lambda'' \Phi'' \DPtypo{+}{} = 0.
+\]
+Considérons les droites communes à tous les complexes du réseau,
+c'est-à-dire communes aux \Card{3} complexes $\Phi = 0$, $\Phi' = 0$,
+$\Phi'' = 0$; il y en a $\infty^{1}$; elles appartiennent aux complexes spéciaux
+du réseau, on peut les définir au moyen de \Card{3} de ces complexes
+spéciaux. Or\Add{,} un complexe spécial est formé de toutes
+les droites rencontrant sa directrice; les droites précédentes
+rencontrent donc \Card{3} droites fixes, elles constituent un
+système de génératrices d'une quadrique, le \Ord{2}{e} système de génératrices
+comprenant les directrices des complexes spéciaux
+du réseau.
+
+\Paragraph{Application. On peut définir un complexe par \Card{5} droites
+n'appartenant pas à une même congruence linéaire.} Soient en
+effet les droites $1, 2, 3, 4, 5$;  donnons-nous un point~$P$ et cherchons-en
+le plan polaire; considérons les droites $1, 2, 3, 4$;
+il existe deux droites $\Delta\Add{,} \Delta'$ qui rencontrent ces \Card{4} droites, ces
+droites sont conjuguées par rapport au complexe, et alors la
+droite passant par~$P$ et s'appuyant sur~$\Delta\Add{,} \Delta'$ appartient au complexe.
+De même en considérant les droites $2, 3, 4, 5$, nous aurons
+une \Ord{2}{e} droite passant par~$P$ et appartenant au complexe;
+le plan polaire de~$P$ est alors déterminé par ces \Card{2} droites.
+
+\Section{Courbes du complexe.}
+{8.}{} Proposons-nous de déterminer les courbes du complexe
+\[
+r = kc.
+\]
+Considérons une droite passant par un point~$(x, y, z)$ et de
+coefficients directeurs~$a, b, c$; pour qu'elle appartienne au
+complexe, il faut que l'on ait
+%% -----File: 245.png---Folio 237-------
+\[
+bx - ay = kc\Add{,}
+\]
+et l'équation différentielle des courbes du complexe est alors
+\[
+\Tag{(1)}
+x\, dy - y\, dx = k · dz.
+\]
+Cette équation s'écrit
+\[
+x^{2}\, d\left(\frac{y}{x}\right) = d(kz)\Add{,}
+\]
+posons
+\[
+\Tag{(2)}
+kz = Y, \qquad \frac{y}{x} = X, \qquad x^{2} = P;
+\]
+l'équation précédente s'écrit
+\[
+dY - P\, dX = 0\Add{,}
+\]
+elle montre que $P$~est la dérivée de~$Y$ par rapport à~$X$. On a
+donc la solution générale de~\Eq{(1)}
+\[
+\Tag{(3)}
+X = \phi(t), \qquad
+Y = \Psi(t), \qquad
+P = \frac{d\Psi}{d\phi};
+\]
+d'où $x\Add{,}y\Add{,}z$ exprimées en fonction d'une \DPtypo{varable}{variable} arbitraire~$t$ au
+moyen de \Card{2} fonctions arbitraires. Si on prend pour variable
+indépendante~$X$, on aura
+\[
+Y = f(X)\Add{,} \qquad
+P = f'(X)\Add{,}
+\]
+d'où les équations de la courbe
+\[
+\Tag{(4)}
+kz = f \left(\frac{y}{x}\right), \qquad
+x^{2} = f' \left(\frac{y}{x}\right)\DPtypo{;}{.}
+\]
+On pourra poser
+\[
+\frac{y}{x} = u\Add{,}
+\]
+d'où les expressions de $x\Add{,}y\Add{,}z$ en fonction de~$u$
+\[
+\Tag{(5)}
+x =   \sqrt{f'(u)}, \qquad
+y = u \sqrt{f'(u)}, \qquad
+z = \frac{1}{k}\, f(u).
+\]
+
+Il est facile, en particularisant la forme de la fonction~$f$,
+d'obtenir des courbes remarquables du complexe.
+
+\ParItem{\Primo.} \DPtypo{on}{On} obtiendra toutes les courbes algébriques du complexe
+en prenant pour~$f$ une fonction algébrique de~$u$. Posons
+en particulier
+\[
+f(u) = \frac{u^{3}}{3}\Add{,}
+\]
+%% -----File: 246.png---Folio 238-------
+alors
+\[
+f'(u) = u^2\Add{,}
+\]
+et nous avons
+\[
+\Tag{(6)}
+x = u, \qquad
+y = u^2, \qquad
+z = \frac{u^3}{3k};
+\]
+ces équations sont celles d'une cubique gauche osculatrice au
+plan de l'infini dans le direction $x = 0$, $y = 0$. Réciproquement
+on peut par une transformation projective ramener les équations
+de toute cubique gauche à la forme précédente, d'où il
+résulte que \emph{les tangentes à toute cubique gauche appartiennent
+à un complexe linéaire}.
+
+\ParItem{\Secundo.} Les formules générales~\Eq{(5)} contiennent un radical,
+provenant de ce qu'on a posé $x^2 = P$. On fera \DPchg{disparaitre}{disparaître} le
+radical en choisissant le paramètre de façon que $P$~soit carré
+parfait. Pour cela considérons la courbe plane~$(X\Add{,}Y)$ considérée
+comme enveloppe de la droite
+\[
+Y - u^2X + 2 \phi(u) = 0,
+\]
+car on a bien alors
+\[
+\frac{dY}{dX} = u^2;
+\]
+l'enveloppe est définie par l'équation de la droite et par
+\[
+-uX + \phi'(u) = 0\Add{,}
+\]
+d'où l'on tire
+\[
+X = \frac{\phi'(u)}{u}, \qquad
+Y = u\phi'(u) - 2\phi(u);
+\]
+d'où
+\[
+\Tag{(7)}
+x = u, \qquad
+y = \phi' (u), \qquad
+z = \frac{1}{k} \bigl[u\phi'(u) - 2\phi(u)\bigr];
+\]
+et ces formules permettent de trouver toutes les courbes unicursales
+du complexe; il n'y a qu'à prendre pour~$u$ une fonction
+rationnelle d'un paramètre arbitraire, et pour~$\phi$ une
+%% -----File: 247.png---Folio 239-------
+fonction rationnelle de~$u$.
+
+\ParItem{\Tertio.} L'équation différentielle~\Eq{(1)} peut encore s'écrire
+\[
+(x^2 + y^2)\, d \left(\arctg \frac{y}{x}\right) = k\, dz;
+\]
+posons
+\[
+kz = Y, \qquad
+\arctg \frac{y}{x} = X, \qquad
+x^2 + y^2 = P = \frac{dY}{dX}.
+\]
+En prenant $X$ comme variable indépendante, on aura la solution
+générale
+\[
+\arctg \frac{y}{x} = \omega, \qquad
+kz = f(\omega), \qquad
+x^2 + y^2 = f'(\omega);
+\]
+qu'on peut encore écrire
+\[
+\Tag{(8)}
+x = \sqrt{f'(\omega)} · \cos\omega, \qquad
+y = \sqrt{f'(\omega)} \Add{·} \sin\omega, \qquad
+z = \frac{1}{k} f(\omega).
+\]
+On obtient des courbes particulières en prenant
+\[
+f(\omega) = R^2 \omega + C;
+\]
+d'où
+\[
+\Tag{(9)}
+x = R \cos\omega, \qquad
+y = R \sin\omega, \qquad
+z = \frac{R^2}{k} \omega + a;
+\]
+\DPtypo{Ce}{ce} sont des hélices tracées sur des cylindres de révolution
+autour de l'axe du complexe. Le pas de ces hélices~$\dfrac{2 \pi R^2}{k}$ est
+uniquement fonction de~$R$, donc \emph{toutes les hélices du complexe
+tracées sur un même cylindre ayant l'axe du complexe pour axe
+ont même pas}.
+
+\MarginNote{Propriétés
+générales des
+courbes du
+complexe.}
+Il résulte immédiatement de la définition des courbes
+d'un complexe que, \emph{dans un complexe linéaire, le plan polaire
+d'un point d'une courbe du complexe est le plan osculateur à
+la courbe en ce point}. Considérons alors les plan osculateurs
+à une courbe du complexe issus d'un point~$P$. Soit~$A$ l'un des
+points de contact; le plan osculateur en~$A$ étant le plan
+%% -----File: 248.png---Folio 240-------
+polaire de~$A$, la droite~$PA$ appartient au complexe, et par suite
+est dans le plan polaire de~$P$. Il en résulte que \emph{les points
+de contact des plans osculateurs issus d'un point à une courbe
+d'un complexe linéaire sont dans un même plan}. En particulier
+\emph{les points de contact des plans osculateurs issus d'un
+point à une cubique gauche sont dans un même plan passant par
+le point donné}.
+
+Prenons les formules~\Eq{(7)}. Nous trouvons
+\begin{align*}
+A &= y' z''- z' y'' = \frac{1}{k} \phi' \phi'' = \frac{y}{k} \phi''', \\
+B &= z' x'' - x' z'' = - \frac{u}{k} \phi''' = - \frac{x}{k} \phi''', \\
+C &= x' y'' - y' x'' = \phi''' = \phi''';
+\end{align*}
+et
+\[
+\begin{vmatrix}
+x'    &  y'    &  z'   \\
+x''   &  y''   &  z''  \\
+x'''  &  y'''  &  z'''
+\end{vmatrix} = \frac{1}{k}\, \phi'''{}^2.
+\]
+On voit alors que la torsion au point $(x,y,z)$ est donnée par
+\[
+T = -\frac{x^2 + y^2 + z^2}{k};
+\]
+Elle ne dépend que du point, et pas de la courbe. Donc \emph{toutes
+les courbes du complexe linéaire passant par un point ont
+même torsion en ce point \(Sophus Lie\)}.
+
+\Section{Surfaces normales du complexe.}
+{9.}{} Il n'y a pas lieu de rechercher les surfaces d'un
+complexe linéaire. Soit en effet le complexe linéaire
+\[
+ay - bx + kc = 0;
+\]
+le plan polaire du point $(x,y,z)$ est parallèle au plan
+\[
+Xy - Yx + kZ = 0,
+\]
+%% -----File: 249.png---Folio 241-------
+et pour qu'une surface soit tangente à ce plan, il faudrait
+que l'on eût
+\[
+\frac{p}{y} = \frac{q}{-x} = \frac{-1}{k},
+\]
+ou
+\[
+p = -\frac{y}{k}, \qquad
+q =  \frac{x}{k};
+\]
+et la condition d'intégrabilité
+\[
+\frac{\dd p}{\dd y} = \frac{\dd q}{\dd x}
+\]
+n'est pas réalisée. Le problème est impossible.
+
+Nous nous proposerons alors de chercher les surfaces
+dont les normales sont des droites du complexe. Nous aurons à
+intégrer l'équation aux dérivées partielles
+\[
+py - qx - k = 0,
+\]
+ce qui revient à l'intégration du système
+\[
+\frac{dx}{y} = \frac{dy}{-x} = \frac{dz}{k} = - dt,
+\]
+qui est précisément le système auquel on arrive lorsqu'on recherche
+les courbes normales aux plans polaires de leurs
+points\Add{.} Nous pouvons écrire
+\[
+dx = -y·dt, \qquad
+dy =  x·dt, \qquad
+dz = -k·dt;
+\]
+système qui s'intègre immédiatement et donne
+\[
+x = R \cos t, \qquad
+y = R \sin t, \qquad
+z = - kt + h.
+\]
+Ces trajectoires orthogonales dépendent de \Card{2} constantes arbitraires.
+Ce sont des hélices circulaires ayant toutes même
+pas, trajectoires d'un mouvement \DPchg{hélicoidal}{hélicoïdal} uniforme de pas~$-2k$.
+D'où l'interprétation cinématique du complexe linéaire:
+considérons un mouvement \DPchg{hélicoidal}{hélicoïdal} uniforme; à chaque point~$M$
+correspond la vitesse de ce point, et le plan polaire du
+point~$M$ dans le complexe est le plan perpendiculaire à cette
+%% -----File: 250.png---Folio 242-------
+vitesse. \emph{Le complexe linéaire est constitué par les normales
+aux vitesses du mouvement instantané d'un corps solide.}
+
+Les surfaces normales du complexe sont définies par les
+équations
+\[
+x = v \cos u, \qquad
+y = v \sin u, \qquad
+z = - ku + \phi(v);
+\]
+car elles sont évidemment engendrées par les hélices précédentes.
+Ce sont les \DPchg{hélicoides}{hélicoïdes} engendrés par un profil quelconque
+dans le mouvement précédent. Les \DPtypo{equations}{équations} précédentes
+\DPtypo{representent}{représentent} d'ailleurs \DPchg{l'hélicoide}{l'hélicoïde} le plus général. Il en
+résulte que \emph{les normales issues d'un point à un \DPchg{hélicoide}{hélicoïde}
+sont dans un même plan} (plan polaire de ce point).
+
+\Paragraph{Remarque.} Les hélices trajectoires orthogonales des
+plans polaires s'obtiennent en faisant $v = \cte$, et leurs
+trajectoires orthogonales sont les courbes du complexe situées
+sur les surfaces précédentes. Cherchons-les. Formons l'élément
+linéaire sur ces surfaces:
+\begin{align*}%[** TN: Rebroken]
+ds^2 = dx^2 + dy^2 + dz^2
+  &= (\cos u · dv - v · \sin u · du)^2 \\
+  &+ (\sin u · dv + v · \cos u · du)^2 + (- m · du + \phi' · dv)^2,
+\end{align*}
+\[
+ds^2 = (v^2 + m^2)\Add{·} du^2 - 2 m \phi' · du\,dv + (1 + \phi'{}^2) · dv^2;
+\]
+et les trajectoires orthogonales des hélices $v = \cte$, $dv = 0$,
+sont définies par l'équation
+\[
+(v^2 + m^2)\Add{·} du - m \phi' · dv = 0\Add{,}
+\]
+d'où
+\[
+u = \int \frac{m \phi'}{v^2 + m^2}\, dv\Add{.}
+\]
+Leur détermination dépend d'une quadrature.
+%% -----File: 251.png---Folio 243-------
+
+\Section{Surfaces réglées du complexe.}
+{10.}{} Considérons une surface réglée dont les génératrices
+appartiennent au complexe; soit~$G$ une de ses génératrices;
+elle appartient au complexe, donc à chacun de
+ses points~$M$ correspond un plan~$P$ qui en est le plan \Err{focal}{polaire};
+d'autre part au point~$M$ correspond aussi homographiquement le
+plan tangent à la surface en ce point; il en résulte qu'\emph{il y
+a correspondance homographique entre le plan polaire d'un
+point de la génératrice et le plan tangent à la surface en ce
+point}; dans cette homographie il y a \Card{2} éléments doubles, donc
+\emph{sur chaque génératrice de la surface il existe \Card{2} points $A\Add{,} B$
+tels que les plans polaires de ces points soient tangents à la
+surface}. Considérons le lieu des points~$A$ sur la surface; en
+chacun de ses points le plan tangent à la surface est le plan
+polaire de~$A$; la tangente à la courbe, qui est dans le plan
+tangent à la surface, est donc dans le plan polaire; donc \emph{le
+lieu des points~$A$, et aussi le lieu des points~$B$, qui peuvent
+d'ailleurs se confondre algébriquement, sont des courbes du
+complexe}. Le plan osculateur en chaque point est le plan polaire,
+donc il est tangent à la surface; \emph{ces courbes sont
+donc des asymptotiques de la surface réglée}; les asymptotiques
+se déterminent au moyen d'une seule quadrature.
+
+Il peut arriver que les génératrices de la surface appartiennent
+à une congruence linéaire; elles appartiennent alors
+à une infinité de complexes linéaires, et pour chaque complexe,
+on aura \Card{2} lignes asymptotiques courbes de ce complexe.
+On obtiendra ainsi toutes les asymptotiques sans aucune intégration.
+%% -----File: 252.png---Folio 244-------
+\emph{Les génératrices de la surface précédente s'appuient
+alors sur \Card{2} directrices fixes}. C'est le cas des \DPtypo{conoides}{conoïdes} à
+plan directeur et des surfaces réglées du \Ord{3}{e} ordre. Inversement
+on verrait facilement qu'une courbe quelconque du complexe
+est asymptotique d'une infinité de surfaces réglées du
+complexe; on peut donc au moyen de ces surfaces réglées trouver
+une courbe quelconque du complexe.
+
+Si les génératrices de la surface appartiennent à un
+complexe linéaire spécial, les courbes du complexe sont des
+courbes planes dont les plans contiennent la directrice du
+complexe; \emph{les surfaces normales du complexe sont de révolution
+autour de la directrice; les surfaces réglées du complexe sont
+des surfaces dont les génératrices rencontrent une droite
+fixe}; cette directrice est une asymptotique de la surface,
+et les autres asymptotiques se déterminent par \Card{2} quadratures.
+
+
+\ExSection{X}
+
+\begin{Exercises}
+\item[45.] \DPchg{Etudier}{Étudier} les asymptotiques des surfaces réglées du \Ord{3}{e} ordre.
+Montrer que ce sont des unicursales du \Ord{4}{e} ordre, et que chaque
+génératrice rencontre une asymptotique en deux points conjugués
+harmoniques par rapport aux points où la génératrice
+s'appuie sur la droite double et sur la droite singulière.
+
+\item[46.] Déterminer les asymptotiques de la surface de Steiner. Par
+quelles courbes sont-elles représentées dans la représentation
+paramétrique de la surface? \DPchg{Etudier}{Étudier} les cas de dégénérescence.
+
+\item[47.] Déterminer la surface canal la plus générale dont toutes les
+lignes de courbure soient sphériques; montrer que ces lignes
+de courbure se déterminent sans intégration.
+
+\item[48.] Que peut-on dire de la détermination des lignes de courbure
+d'une surface canal, enveloppe de $\infty^{1}$~sphères coupant une
+sphère fixe sous un angle constant?
+
+\item[49.] Déterminer les surfaces réglées d'un complexe linéaire qui
+admettent pour ligne asymptotique une courbe donnée. Montrer
+que toutes leurs asymptotiques se \DPtypo{déterminant}{déterminent} sans intégration,
+et qu'elles sont algébriques si la courbe donnée est
+algébrique.
+\end{Exercises}
+%% -----File: 253.png---Folio 245-------
+
+
+\Chapitre{XI}
+{Transformations Dualistiques. Transformation de Sophus Lie.}
+
+\Section{\DPchg{Eléments}{Éléments} et multiplicités de contact.}
+{1.}{} On appelle \emph{élément de contact} l'ensemble d'un point~$M$
+et d'un plan~$P$ passant par ce point. Un tel élément sera
+défini par ses \emph{coordonnées}, coordonnées $x, y, z$ du point, et
+coefficients de direction $p, q, -1$ de la normale au plan. Un
+élément de contact est ainsi défini par \Card{5} coordonnées.
+
+Considérons un point~$A$, les éléments de contact de ce
+point sont formés par ce point et tous les plans passant par
+ce point; les coordonnées $x, y, z$ sont fixes, et $p, q$ arbitraires.
+Un point possède $\infty^{2}$~éléments de contact.
+
+Considérons une courbe; un de ses éléments de contact
+est formé d'un point de la courbe et d'un plan tangent à la
+courbe en ce point; les coordonnées sont $x, y, z$ fonctions d'un
+paramètre arbitraire, et $p, q$ \DPtypo{lies}{liés} par la relation
+\[
+px' + qy'- z' = 0\Add{,}
+\]
+il y a donc \Card{2} paramètres arbitraires. Une courbe possède $\infty^{2}$
+éléments de contact.
+
+Considérons maintenant une surface; un de ses éléments de
+contact est formé par un point et le plan tangent en ce point;
+%% -----File: 254.png---Folio 246-------
+ses coordonnées sont $x, y, z = f (x, y)$, $p=\dfrac{\dd f}{\dd x}$, $q=\dfrac{\dd f}{\dd y}$. Il y a \Card{2} paramètres
+arbitraires, donc une surface possède $\infty^{2}$~éléments
+de contact. Remarquons que $p, q$ peuvent ne dépendre que d'un
+seul paramètre; c'est le cas des surfaces développables, qui
+possèdent ainsi $\infty^{2}$~points et $\infty^{1}$~plans tangents, et correspondent
+par dualité aux courbes, qui possèdent $\infty^{1}$~points et $\infty^{2}$~plans
+tangents.
+
+Les points, courbes et surfaces, qui sont engendrées par
+$\infty^{2}$~éléments de contact, sont appelés \emph{multiplicités~$M_2$}. Plus
+généralement on appellera \emph{multiplicité} toute famille d'éléments
+de contact dont les coordonnées vérifient la relation
+\[
+\Tag{(1)}
+dz - p\, dx - q\, dy = 0.
+\]
+Si ces coordonnées ne dépendent que d'un paramètre arbitraire,
+on aura les multiplicités~$M_1$; si elles dépendent de \Card{2} paramètres
+arbitraires, on aura les \emph{multiplicités~$M_2$}.
+
+Cherchons à déterminer toutes les multiplicités~$M_2$:
+$x, y, z$, $p, q$ sont fonctions de \Card{2} paramètres arbitraires
+\[
+x = f (u, v), \quad
+y = g (u, v), \quad
+z = h (u, v), \quad
+p = k (u, v), \quad
+q = l (u, v).
+\]
+Considérons les \Card{3} \Ord{1}{ères} relations; entre elles on peut éliminer
+$u, v$, et par suite de cette élimination on peut obtenir
+\Card[f]{1},~ou \Card{2}, ou \Card{3} relations.
+
+Supposons d'abord qu'on obtienne une relation
+\[
+F(x, y, z) = 0\Add{,}
+\]
+on peut considérer $z$ comme fonction de $x, y$; et si on écrit
+que la relation~\Eq{(1)} est satisfaite quels que soient, $x, y$, on a
+\[
+p = \frac{\dd z}{\dd x}, \qquad
+q = \frac{\dd z}{\dd y},
+\]
+%% -----File: 255.png---Folio 247-------
+et on a les éléments de contact d'une surface.
+
+Supposons qu'on obtienne \Card{2} relations
+\[
+F(x, y, z) = 0, \qquad
+G(x, y, z) = 0;
+\]
+on peut considérer $x, y$ comme fonctions de~$z$
+\[
+x = \phi(z), \qquad
+%[** TN: Appears to be \Psi in original, but using \psi for consistency]
+y = \psi(z),
+\]
+et l'équation~\Eq{(1)} devient
+\[
+dz - p \phi'(z)\, dz - q \psi'(z)\, dz = 0,
+\]
+ou
+\[
+p \phi'(z) + q \psi'(z) - 1 = 0;
+\]
+le plan de l'élément de contact est tangent à la courbe
+$x = \phi(z)$, $y = \psi (z)$, on a les éléments de contact d'une courbe.
+
+Enfin si on obtient \Card{3} relations, c'est que $x, y, z$ sont des
+constantes; l'équation~\Eq{(1)} est alors vérifiée quels que soient
+$p\Add{,} q$, qui sont alors les paramètres arbitraires, et on a les
+éléments de contact d'un point.
+
+Cherchons maintenant les multiplicités~$M$; nous avons
+\[
+x = f (t), \qquad
+y = g (t), \qquad
+z = h (t), \qquad
+p = k (t), \qquad
+q = l (t).
+\]
+Considérons les \Card{3} \Ord{1}{ères} équations, et entre elles éliminons~$t$.
+Il y a \Card{2} ou \Card{3} relations.
+
+S'il y a \Card{2} relations, le lieu des points de la multiplicité,
+qu'on appelle aussi \emph{support de la multiplicité}, est une
+courbe, et les plans ne dépendant que d'un paramètre, pour
+chaque point de la courbe il y a un plan tangent déterminé;
+on a une \emph{bande d'éléments de contact}.
+
+S'il y a \Card{3} relations, $x, y, z$ sont des constantes, le support
+est un point; on a alors une famille de plans dépendant
+d'un paramètre et passant par un point fixe; c'est ce qu'on
+appelle un \emph{cône élémentaire}.
+%% -----File: 256.png---Folio 248-------
+
+Considérons \Card{2} multiplicités~$M_2$; elles peuvent avoir en
+commun, \Card{0}\DPtypo{,}{}~ou \Card{1} élément de contact, ou une infinité.
+
+Considérons le cas d'\emph{un élément de contact commun}; si les
+multiplicités sont \Card{2} points $A\Add{,} A'$, il ne peut y avoir un élément
+de contact commun que si les \Card{2} points sont confondus, et
+alors il y a $\infty^{2}$~éléments de contact communs. Si on a un point
+et une courbe, le point est sur la courbe, et tous les plans
+tangents à la courbe en ce point appartiennent à des éléments
+de contact communs, qui sont ainsi au nombre de~$\infty^{1}$. Si on a un
+point et une surface, le point sera sur la surface, et l'élément
+de contact commun sera constitué par le point et le plan
+tangent à la surface en ce point. Considérons \Card{2} courbes; si
+elles ont un élément de contact commun, elles se rencontrent
+en un point, et si elles n'y sont pas tangentes, il n'y a
+qu'un élément de contact commun. Considérons une courbe et
+une surface; il y aura un élément de contact commun si la
+courbe est tangente à la surface. Enfin \Card{2} surfaces ont un
+élément de contact commun si elles sont tangentes en un point.
+
+Il y aura \emph{$\infty^{1}$~éléments de contact communs} pour un point
+sur une courbe, \Card{2} courbes tangentes en un point, une courbe
+sur une surface, \Card{2} surfaces circonscrites le long d'une courbe\Add{.}
+
+Considérons un \emph{point qui décrit une courbe}; on a une
+famille de $\infty^{1}$~points dont chacun donne à la courbe $\infty^{1}$~éléments
+de contact. Considérons une \emph{surface engendrée par une courbe};
+nous avons $\infty^{1}$~courbes~$(c)$ dont chacune a en commun avec la
+surface une bande, et par suite donne à la surface $\infty^{1}$~éléments
+de contact. Considérons $\infty^{1}$~surfaces; leur \emph{enveloppe} a avec
+%% -----File: 257.png---Folio 249-------
+chacune d'elles une bande commune; nous avons encore $\infty^{1}$~éléments
+générateurs d'une multiplicité~$M_2$, donnant chacun à la
+multiplicité $\infty^{1}$~éléments de contact.
+
+On pourrait considérer le cas où chaque élément générateur
+ne donne qu'un élément de contact à la multiplicité:
+$\infty^{2}$~points engendrant une surface; $\infty^{2}$~courbes formant une congruence
+de courbes (dans ce cas, comme dans celui des congruences
+de droites, il y a en général une surface focale,
+tangente à chacune de ces courbes, et ayant avec chacune un
+élément de contact commun); enfin si on considère $\infty^{2}$~surfaces,
+leur enveloppe a en commun avec chacune d'elles un élément de
+contact.
+
+\Paragraph{Remarque.} Dans les trois cas précédents, quand nous disons
+que chaque élément générateur donne un élément de contact
+à la multiplicité, il faut entendre que cette multiplicité
+peut se décomposer en nappes, et que cela s'applique alors à
+chacune des nappes séparément.
+
+Il y a un cas exceptionnel, celui de $\infty^{1}$~courbes ayant
+une enveloppe; on a alors $\infty^{1}$~courbes cédant chacune à la multiplicité
+$\infty^{1}$~éléments de contact.
+
+\Section{Transformations de contact.}
+{2.}{} On appelle \emph{transformation de contact} toute transformation
+des éléments de contact qui change une multiplicité~$M_2$
+en une multiplicité~$M_2$. On a \Card{5} équations de
+transformation
+\begin{gather*}
+x' = f (x, y, z, p, q), \quad
+y' = g (x, y, z, p, q), \quad
+z' = h (x, y, z, p, q), \\
+p' = k (x, y, z, p, q), \qquad
+q' = l (x, y, z, p, q).
+\end{gather*}
+%% -----File: 258.png---Folio 250-------
+Si l'élément de contact~$(x, y, z, p, q)$ appartient à un multiplicité,
+on a
+\[
+\Tag{(1)}
+dz - p\, dx - q\, dy = 0,
+\]
+et pour que l'élément transformé $(x', y', z', p', q')$ appartienne
+aussi à une multiplicité, il faut que l'on ait
+\[
+dz' - p'\, dx'- q'\, dy'= 0.
+\]
+Une transformation de contact laisse invariante l'équation~\Eq{(1)}\Add{.}
+Une telle transformation change \Card{2} multiplicités ayant un élément
+de contact commun en \Card{2} multiplicités ayant un élément de
+contact commun, et de même \Card{2} multiplicités ayant $\infty^{1}$~éléments
+de contact communs en \Card{2} multiplicités ayant $\infty^{1}$~éléments de
+contact communs. Une transformation de contact change les
+points, courbes et surfaces en points, courbes, ou surfaces
+indistinctement.
+
+Reprenons les équations de la transformation, et entre
+elles éliminons~$p\Add{,} q$, nous obtenons \Card[f]{1},~ou \Card{2}, ou \Card{3} relations entre
+$x, y, z$,~$x', y', z'$.
+
+Si on obtient \Card{3} relations,
+\[
+\Tag{(2)}
+x' = f(x, y, z), \qquad
+y' = g(x, y, z), \qquad
+z' = h(x, y, z)\Add{,}
+\]
+dans la transformation de contact est contenue une transformation
+ponctuelle. Une telle transformation change un point en
+point, une courbe en courbe, une surface en surface; \Card{2} courbes
+qui se rencontrent se transforment en \Card{2} courbes qui se
+rencontrent, deux surfaces tangentes en \Card{2} surfaces tangentes.
+A un élément de contact commun à \Card{2} multiplicité correspond un
+élément de contact commun aux \Card{2} multiplicités transformées.
+On obtiendra $p', q'$ en fonction de~$p, q$ en considérant $z'$ comme
+%% -----File: 259.png---Folio 251-------
+fonction de $x', y'$. On a
+\begin{alignat*}{3}%[** TN: Added elided equations]
+dx' &= \frac{\dd f}{\dd x}\, dx
+     &&+ \frac{\dd f}{\dd y}\, dy
+     &&+ \frac{\dd f}{\dd z}\, (p\, dx + q\, dy), \\
+dy' &= \frac{\dd g}{\dd x}\, dx
+     &&+ \frac{\dd g}{\dd y}\, dy
+     &&+ \frac{\dd g}{\dd z}\, (p\, dx + q\, dy), \\
+dz' &= \frac{\dd h}{\dd x}\, dx
+     &&+ \frac{\dd h}{\dd y}\, dy
+     &&+ \frac{\dd h}{\dd z}\, (p\, dx + q\, dy);
+\end{alignat*}
+éliminant $dx, dy$ entre ces \Card{3} relations, on a
+\[
+dz' = k(x,y,z,p,q)\, dx' + l(x,y,z,p,q)\, dy',
+\]
+d'où
+\[
+p' = k(x,y,z,p,q), \qquad
+q' = l(x,y,z,p,q).
+\]
+
+Supposons ensuite que l'on obtienne \Card[f]{1}~relation d'élimination
+\[
+\Tag{(3)}
+\Omega (x,y,z, x',y',z') = 0\Add{.}
+\]
+Considérons un point $A(x\Add{,}y\Add{,}z)$ du \Ord{1}{er} espace; cherchons la multiplicité
+qui lui correspond dans le \Ord{2}{e} espace; elle est engendrée
+par des éléments de contact dont les points sont liés
+au point~$A$ par l'équation~\Eq{(3)} qui représente une surface~$S_A'$.
+La multiplicité correspondant à un point est une surface. Si
+on a une courbe lieu de points~$A$, il lui correspond une famille
+de $\infty^{1}$~surfaces, et la multiplicité engendrée par ces
+surfaces, c'est-à-dire leur enveloppe, sera la transformée de
+la courbe. Enfin si on a une surface lieu de $\infty^{2}$~points~$A$, il
+leur correspondra $\infty^{2}$~surfaces dont l'enveloppe correspondra à
+la surface donnée.
+
+\MarginNote{Transformations
+dualistiques.}
+Supposons la relation~\Eq{(3)} bilinéaire en $x,y,z$,
+$x',y',z'$. A chaque point du \Ord{1}{er} espace correspond un plan
+du \Ord{2}{e} espace et réciproquement. A $\infty^{3}$~points du \Ord{1}{er} espace
+correspondent $\infty^{3}$~plans distincts. \DPchg{Ecrivons}{Écrivons}
+\[
+\Omega = Ax' + By' + Cz' + D
+\]
+%% -----File: 260.png---Folio 252-------
+ou
+\[
+A = ux + vy + wz + h, \quad
+B = u'  x + \ldots, \quad
+C = u'' x + \ldots, \quad
+D = u'''x + \ldots;
+\]
+pour avoir la transformée d'une surface
+\[
+f(x',y',z') = 0
+\]
+il faut prendre l'enveloppe des plans $\Omega = 0$, $x'\Add{,} y'\Add{,} z'$~étant liés
+par la relation précédente, ce qui donne
+\[
+\frac{A}{\ \dfrac{\dd f}{\dd x'}\ } =
+\frac{B}{\ \dfrac{\dd f}{\dd y'}\ } =
+\frac{C}{\ \dfrac{\dd f}{\dd z'}\ } =
+\frac{D}{\ \dfrac{\dd f}{\dd t'}\ }\Add{.}
+\]
+Telles sont les équations de la transformation. Il faudra que
+l'on en puisse tirer $x, y, z$: donc que les formes $A, B, C, D$, soient
+indépendantes, et alors l'ensemble des plans $\Omega = 0$ constitue
+bien l'ensemble de tous les plans de l'espace. La transformation
+précédente est une \emph{transformation dualistique}. L'ensemble
+des transformations de contact forme évidemment un groupe;
+une transformation de contact peut se décomposer en transformations
+de contact plus simples.
+
+Prenons pour nouvelles variables
+\[
+X = A, \qquad
+Y = B, \qquad
+Z = C, \qquad
+T = D;
+\]
+alors
+\[
+\Omega = Xx' + Yy' + Zz' + 1 = 0\Add{,}
+\]
+la transformation est une transformation par polaires réciproques
+par rapport à la sphère
+\[
+x^2 + y^2 + z^2 + 1 = 0\Add{,}
+\]
+et toute transformation dualistique se ramène à la transformation
+précédente suivie d'une transformation projective.
+
+Considérons celles de ces transformations qui sont \emph{symétriques}
+ou \emph{involutives}, telles que le plan homologue d'un
+%% -----File: 261.png---Folio 253-------
+point soit le même, qu'on considère le point comme appartenant
+à l'un ou à l'autre espace. Les équations
+\[
+\Omega(x,y,z,x',y',z') = 0, \qquad
+\Omega(x',y',z',x,y,z) = 0,
+\]
+doivent être équivalentes; on doit donc avoir, $k$~étant un
+facteur constant
+\[
+\Omega(x,y,z,x',y',z') = k \Omega(x',y',z',x,y,z);
+\]
+faisons $x' = x$, $y' = y$, $z' = z$,
+\[
+\Omega(x,y,z,x,y,z) = k \Omega(x,y,z,x,y,z);
+\]
+alors ou bien $\Omega(x,y,z,x,y,z) = 0$, ou bien $k = 1$.
+
+Si $\Omega = 0$, le plan correspondant à un point passe par ce
+point. On a
+\begin{multline*}%[** TN: Filled in missing terms, added break]
+x(ux + vy + wz + h) + y(u'x + v'y + w'z + h') \\
+  + z (u''x + v''y + w''z + h'')
+  + u'''x + v'''y + w'''z + h''' = 0,
+\end{multline*}
+ce qui revient à écrire que le déterminant
+\[
+\begin{vmatrix}
+u  &  v  &  w  &   h \\
+u'  & v'  &  w' &   h' \\
+u'' & v''  & w'' &   h'' \\
+u''' & v''' & w''' &  h'''
+\end{vmatrix}
+\]
+est un déterminant, symétrique gauche, donc de la forme
+\[
+\left\lvert
+\begin{array}{@{}rrrr@{}}
+  0   &    C  &    -B  &   \phantom{-}P \\
+  -C   &    0   &    A  &  Q \\
+  B   &  -A   &     0 &    R  \\
+  -P    & -Q   &   -R &    0
+\end{array}
+\right\rvert
+\]
+et l'équation devient
+\[
+\Omega = x'(\DPtypo{c}{C}y - Bz + P)
+       + y'(-Cx + Az + Q)
+       + z'(Bx - Ay + R)
+       - Px - \DPtypo{A}{Q}y - Rz = 0\Add{,}
+\]
+ou
+\begin{multline*}%[** TN: Added break]
+A(yz' - zy') + B(zx' - xz') + C(xy' - yx') \\
+  + P(x' - x) + Q (y' - y) + R(z' - z) = 0\Add{,}
+\end{multline*}
+%% -----File: 262.png---Folio 254-------
+équation d'un complexe linéaire. Le lieu des points $(x',y',z')$
+associés au point $(x,y,z)$ est le plan polaire du point $(x,y,z)$
+par rapport au complexe. Le plan polaire d'un point est la
+multiplicité transformée de ce point et réciproquement. La
+transformée d'une droite est sa conjuguée, une droite du complexe
+est à elle-même sa transformée. \Card{2} multiplicités transformées
+sont les \Card{2} multiplicités focales d'une congruence de
+droites du complexe et réciproquement: à une courbe correspond
+en général une développable; à une courbe du complexe
+correspond la développable de ses tangentes.
+
+Si nous prenons maintenant la solution $k = 1$, nous avons
+\[
+x'(ux + vy + wz + h) + \ldots = x(ux' + vy' + wz' + h) + \ldots\Add{,}
+\]
+la forme $\Omega$ est symétrique en $x\Add{,}y\Add{,}z$, $x'\Add{,}y'\Add{,}z'$, et on a
+\begin{multline*}%[** TN: Added break]
+\Omega = axx' +  byy' + czz' \\
+  + m(yz' + zy') + n(zx' + xz') + p(xy' + yx') \\
+  + r(x + x') + s(y + y') + t(z + z') + u\Add{.}
+\end{multline*}
+\DPtypo{les}{Les} \Card{2} points $(x,y,z)$\Add{,} $(x',y,' z')$ sont conjugués par rapport à la
+quadrique
+\[
+ax^2 + by^2 + cz^2 + 2myz + 2nzx + 2pxy + 2rx + 2sy + 2tz + u = 0\Add{.}
+\]
+Nous avons la transformation par polaires réciproques.
+
+D'une façon générale, pour avoir les équations d'une
+transformation de contact définie par une seule relation
+$\Omega = 0$, on écrira que la relation
+\[
+dz' - p'\, dx' - q'\, dy' = 0\Add{,}
+\]
+est conséquence des relations
+\[
+dz - p\, dx - q\, dy = 0, \qquad
+d\Omega = 0;
+\]
+ce qui donne
+\[
+dz' - p'\, dx' - q'\, dy'
+  = \lambda (dz - p\, dx - q\, dy) + \mu\, d\Omega\Add{.}
+\]
+%% -----File: 263.png---Folio 255-------
+En identifiant, on a \Card{6} équations; si entre elles on élimine~$\lambda\Add{,} \mu$,
+on a \Card{4} équations qui jointes à $\Omega = 0$ donnent $x',y',z',p',q'$
+en fonction de~$x,y,z,p,q$.
+
+Passons enfin au cas où on a \Card{2} relations d'élimination
+\[
+\Tag{(4)}
+\Omega(x,y,z,x',y',z') = 0, \qquad
+\Theta(x,y,z,x',y',z') = 0.
+\]
+A un point~$M$ du \Ord{1}{er} espace correspond dans le \Ord{2}{e} espace une
+courbe~$(c')$. A une courbe lieu de $\infty^{1}$~points correspond une
+surface engendrée par $\infty^{1}$~courbes; à une surface~$(S)$ lieu de
+$\infty^{2}$~points correspond une congruence de courbes; une telle
+congruence a en général une surface focale, tangente à toutes
+ces courbes, et qui sera la transformée de la surface~$(S)$.
+
+Pour avoir les équations d'une telle transformation, on
+écrira que la relation
+\[
+dz'- p'\, dx' - q'\, dy' = 0
+\]
+est conséquence des relations
+\[
+dz - p\, dx - q\, dy = 0, \qquad
+d\Omega = 0, \qquad
+d\Theta = 0;
+\]
+ce qui donne
+\[
+dz' - p'\, dx' - q'\, dy'
+  = \lambda (dz - p\, dx - q\, dy) + \mu d\Omega +  \nu d\Theta.
+\]
+En identifiant on a \Card{6} équations; si entre elles on \DPtypo{elimine}{élimine}
+$\lambda, \mu, \nu$ on a \Card{3} équations qui jointes à $\Omega = 0$, $\Theta = 0$, donnent
+les formules de transformation.
+
+\Section{Transformation de Sophus Lie.}
+{3.}{} Supposons les équations~\Eq{(4)} bilinéaires. A un point~$M(x\Add{,}y\Add{,}z)$
+correspond une droite~$D'$. Aux $\infty^{3}$~points~$M$ correspond un
+complexe de droites~$D'$, soit~$(K')$. De même à tous les points
+du \Ord{2}{e} espace correspond dans le \Ord{1}{er} espace un complexe~$(K)$.
+Considérons une seule des équations~\Eq{(4)}; à chaque point~$M$
+%% -----File: 264.png---Folio 256-------
+correspond un plan~$P'$; l'autre équation au même point~$M$ fait
+correspondre un plan~$Q'$ et la droite~$D'$ est l'intersection des
+plans $P'\Add{,} Q'$ qui correspondent au point~$M$ dans les \Card{2} homographies
+les plans~$P'$ correspondent homographiquement aux plans~$Q'$; le
+complexe~$(K')$ est le complexe des droites intersections des
+plans qui se correspondent dans \Card{2} homographies. (C'est le complexe
+de Reye, ou \emph{complexe tétraédral}; les droites sont coupées
+par un tétraèdre en \Card{4} points dont le rapport anharmonique
+est constant. Le rapport anharmonique des \Card{4} plans menés par
+une droite du complexe et par les \Card{4} sommets du tétraèdre est
+constant (Von~Staudt). Le complexe~$(K')$ est du \Ord{2}{e} degré,
+la surface des singularités est constituée par les \Card{4} faces
+du tétraèdre). A une courbe~$(c)$ correspond une surface
+réglée du complexe~$(K')$. A une surface~$(S)$ correspond une
+congruence de droites appartenant au complexe~$(K')$; cette
+congruence admet \Card{2} multiplicités focales. A un élément de
+contact du \Ord{1}{er} espace correspondent \Card{2} éléments de contact de
+l'autre.
+
+Cherchons les équations des \Card{2} complexes. Soient
+\[
+\Omega = Ax' + By' + Cz' + D, \qquad
+\Theta = Lx' + My' + Nz' + P\Add{.}
+\]
+Soit $M'(x',y'z')$ un point du \Ord{2}{e} espace. Soit $D$ la droite correspondante;
+si $(x,y,z)$ et $(x_0,y_0,z_0)$ sont \Card{2} points de cette
+droite, on a
+\begin{alignat*}{2}
+&\Omega (x,y,z,x',y',z') = 0, &
+&\Theta (x,y,z,x'\DPtypo{.}{,}y',z') = 0\Add{,} \\
+&\Omega (x_0,y_0,z_0, x',y',z') = 0, \qquad &
+&\Theta (x_0,y_0,z_0,x',y',z') = 0.
+\end{alignat*}
+\DPchg{Eliminons}{Éliminons} $x',y,'z'$ entre ces \Card{4} équations, nous avons
+%% -----File: 265.png---Folio 257-------
+\[
+\begin{vmatrix}
+A    &  B    &  C    &  D   \\
+A_0  &  B_0  &  C_0  &  D_0 \\
+L    &  M    &  N    &  P   \\
+L_0  &  M_0  &  N_0  &  P_0
+\end{vmatrix} = 0;
+\]
+c'est l'équation du complexe. En développant par la règle de
+Laplace, on trouvera une équation du \Ord{2}{e} degré par rapport aux
+coordonnées de la droite. Le complexe~$(K)$, et de même le complexe~$(K')$,
+est en général du \Ord{2}{e} degré.
+
+A une courbe~$(c)$ correspond une surface réglée engendrée
+par la droite~$D'$. Cherchons si cette surface réglée peut être
+développable. Les droites~$D'$ ont pour équations
+\[
+Ax' + By'+ Cz' + D = 0, \qquad
+Lx' + My'+ Nz' + P = 0;
+\]
+$x,y,z$, et par suite $A,B,C,D$, étant fonctions d'un paramètre~$t$.
+Exprimons que cette droite rencontre la droite infiniment
+voisine: nous adjoignons a ses équations les équations:
+\[
+x'\, dA + y'\, dB + z'\, dC + dD = 0, \qquad
+x'\, dL + y'\, dM + z'\, dN + dP = 0;
+\]
+d'où la condition
+\[
+\begin{vmatrix}
+A   &  B   &  C   &  D  \\
+L   &  M   &  N   &  P  \\
+dA  &  dB  &  dC  &  dD \\
+dL  &  dM  &  dN  &  dP
+\end{vmatrix} = 0.
+\]
+Or\Add{,} l'équation du complexe~$(K)$ peut s'écrire
+\[
+\begin{vmatrix}
+A         &  B         &  C         &  D        \\
+L         &  M         &  N         &  P        \\
+\Delta A  &  \Delta B  &  \Delta C  &  \Delta D \\
+\Delta L  &  \Delta M  &  \Delta N  &  \Delta P
+\end{vmatrix} = 0,
+\]
+$A,B,C,D$ étant des fonctions linéaires, les accroissements~$\Delta$
+sont proportionnels aux différentielles~$d$. La courbe~$(c)$ est
+%% -----File: 266.png---Folio 258-------
+donc telle que sa tangente appartient au complexe~$(K)$. Aux
+courbes du \Ord{1}{er} complexe correspondent des développables engendrées
+par les tangentes aux courbes du \Ord{2}{e} complexe. Au
+point~$M$ d'une courbe~$(c)$ du \Ord{1}{er} complexe correspond une génératrice~$T'$
+d'une développable, soit $M'$ son point de contact avec
+l'arête de rebroussement; si on considère un élément linéaire
+formé d'un point~$M$ et d'une droite~$D$ du \Ord{1}{er} complexe passant
+par ce point, il lui correspondra un élément linéaire déterminé
+du \Ord{2}{e} complexe. Les courbes des \Card{2} complexes se correspondent
+ainsi par points et par tangentes.
+
+\Illustration{266a}
+Soit une surface~$(S)$, et supposons le complexe~$(K)$ effectivement
+du \Ord{2}{e} degré. Considérons un point~$M$ de la surface et
+le plan tangent~$P$. Le cône du complexe~$(K)$ de sommet~$M$ est
+coupé par le plan~$P$ suivant \Card{2} droites
+$D, D_1$ qui appartiennent au complexe~$(K)$\Add{.}
+Par chaque point de~$(S)$ passent ainsi
+\Card{2} droites du complexe~$(K)$ tangentes à
+la surface. Par tout point de la
+surface~$(S)$ passent donc \Card{2} courbes
+$(\gamma)\Add{,} (\gamma_1)$ du complexe~$(K)$ situées sur cette surface. Au point~$M$
+correspond une droite~$D'$ du complexe~$(K')$. A la droite~$D$ du complexe~$(K)$
+correspond un point~$M'$ de~$D'$; et de même à la droite~$D$,
+correspond un point~$M_1'$ de~$D'$. Aux courbes $(\gamma)\Add{,} (\gamma_1)$ du complexe~$(K)$
+correspondent \Card{2} courbes $(\gamma')\Add{,} (\gamma_1')$ du complexe~$(K')$
+tangentes en $M'\Add{,} M_1'$ à la droite~$D'$. Si le point~$M$ décrit la
+courbe~$(\gamma)$, les droites~$D'$ ont pour enveloppe la courbe~$(\gamma_1')$,
+%% -----File: 267.png---Folio 259-------
+et si $M$~décrit~$(\gamma_1)$, $D'$~enveloppe~$(\gamma_1')$. Si on considère la
+congruence des droites~$D'$ correspondant aux points~$M$ de la surface~$(S)$,
+les courbes~$(\gamma_t')$ sont les arêtes de rebroussement
+d'une des familles de développables de cette congruence et les
+courbes~$(\gamma_1')$ sont les arêtes de rebroussement de l'autre famille.
+Les courbes~$(\gamma')$ engendrent une des nappes de la surface
+focale, les courbes~$(\gamma_1')$ engendrent l'autre nappe. Le
+plan tangent en~$M'$ à la multiplicité focale est le plan osculateur
+à~$(\gamma_1')$, et par suite le plan tangent au cône du complexe~$(K')$
+de sommet~$M_1'$. Un élément de contact correspondant
+à l'élément~$(M,P)$ est formé du point~$M'$ et du plan tangent au
+cône du complexe~$(K')$ qui a pour sommet~$M_1'$. L'autre élément
+correspondant à~$(M,P)$ est formé du point~$M_1'$ et du plan tangent
+au cône du complexe~$(K')$ qui a pour sommet~$M'$.
+
+Si la surface~$(S)$ est une surface du complexe~$(K)$, tangente
+en chacun de ses points au cône du complexe, les droites
+$D,~D_1$ sont confondues; alors les \Card{2} éléments de contact correspondant
+à l'élément~$(M\Add{,}P)$ sont confondus, et la surface~$(S')$
+définie par ces éléments est une surface du complexe~$(K')$.
+
+\Paragraph{Remarques.} Les seuls cas possibles sont les suivants:
+
+\ParItem{\Primo.} Les complexes~$(K)\Add{,} (K')$ sont effectivement du \Ord{2}{e} degré.
+On démontre alors, comme nous l'avons dit précédemment,
+qu'ils sont tous \Card{2} tétraédraux.
+
+\ParItem{\Secundo.} Un des complexes est linéaire. On démontre que
+l'autre est constitué par les droites qui rencontrent une
+conique.
+
+\ParItem{\Tertio.} Les \Card{2} complexes sont linéaires. On démontre qu'ils
+%% -----File: 268.png---Folio 260-------
+sont tous \Card{2} spéciaux. Ce cas donne la \emph{transformation d'Ampère},
+définie par les équations
+\[
+x' = p, \qquad
+y' = -y, \qquad
+z' = -z - px\DPtypo{.}{,}\qquad
+p' = x, \qquad
+q' = -q\Add{.}
+\]
+
+\Section{Transformation des droites en sphères.}
+{4.}{} Prenons en particulier:
+\[
+\Omega = x + iy + x'z + z' = 0, \qquad
+\Theta = x' (x - iy) - z - y' = 0
+\]
+L'équation du \Ord{1}{er} complexe est:
+\[
+\left\lvert
+\begin{array}{@{}crrc@{}}
+z - z_0 &  0 &  \phantom{-}0 & x + iy - (x_0 + iy_0) \\
+z_0     &  0 &             1 & x_0 + iy_0 \\
+x - iy  & -1 &             0 & -z \\
+x_0 -iy_0 - (x-iy) & 0 &   0 & z - z_0
+\end{array}
+\right\rvert
+= 0,
+\]
+ce qui devient:
+\[
+(x - x_0)^2 + (y - y_0)^2 + (z - z_0)^ 2 = 0,
+\]
+c'est-à-dire:
+\[
+\Tag{(K)}
+a^2 + b^2 + c^ 2 = 0.
+\]
+Le complexe~$(K)$ est le complexe des droites minima. Cherchons
+le \Ord{2}{e} complexe. Il suffit de \DPtypo{considerer}{considérer} \Card{2} points~$(x',y',z')$
+correspondant au même point~$(x,y,z)$. Nous avons:
+\[
+\begin{vmatrix}
+0      &         0         &  x' - x'_0  &  z' - z'_0 \\
+1      &          i        &       x'_0  &       z'_0 \\
+x'- x'_0  &  -i(x' - x'_0) &   0          & -(y' - y'_0) \\
+x'_0     &      -ix'_0     &  \llap{$-$} 1 &     y'_0
+\end{vmatrix} = 0,
+\]
+ce qui devient:
+\[
+(x' - x'_0) (x'y'_0 - y'x'_0) - (z' - z'_0) (x' - x'_0) = 0,
+\]
+c'est-à-dire:
+\[
+a(r + c) = 0\DPtypo{.}{;}
+\]
+la solution $a = 0$ est singulière, et on a pour le complexe~$(K')$
+\[
+\Tag{(K')}
+r + c =0.
+\]
+Nous avons ainsi une \emph{correspondance entre un complexe spécial
+%% -----File: 269.png---Folio 261-------
+du \Ord{2}{e} degré et un complexe linéaire}. Le cône du complexe~$(K)$
+sera le cône isotrope. A chaque élément de contact du \Ord{1}{er} espace
+correspondent \Card{2} éléments de contact du \Ord{2}{e} espace conjugués
+par rapport au complexe~$(K')$, car d'une façon générale
+les points~$M'\Add{,} M_1'$, sont sur une droite~$D'$ de~$(K')$ et le plan associé
+à~$M'$ est ici le plan polaire de~$M_1'$ et inversement.
+
+Partons d'une sphère; prenons \Card{2} génératrices d'un système;
+ce sont des droites minima~$D, D_1$. Le \Ord{2}{e} système de génératrices
+est entièrement défini, car chacune d'elles doit
+rencontrer $D, D_1$, et le cercle imaginaire à l'infini. Aux \Card{2} droites~$D, D_1$
+correspondent \Card{2} points~$M', M_1'$. Considérons une génératrice
+isotrope~$\Delta$ rencontrant~$D, D_1$, il lui correspond un
+point~$\mu'$; $\Delta$~rencontrant la droite~$D$, la droite~$M' \mu'$ est une
+droite du complexe linéaire, et de même~$M_1' \mu'$; donc $\mu_1$~est le
+pôle d'un plan passant par~$M_1'M'$. Lorsque $\Delta$~décrit la sphère,
+le plan~$\mu' M' M_1'$ tourne autour de~$M'M_1'$, et le lieu de~$\mu'$ est la
+droite conjuguée de~$M_1' M'$. A la sphère correspond une droite.
+En partant du \Ord{2}{e} système de génératrices, au système~$D$,~$D_1$, correspondrait
+la droite~$M_1'M'$. \emph{A une sphère correspondent en
+réalité \Card{2} droites conjuguées par rapport au complexe linéaire.}
+
+Ceci peut se voir par le calcul. Prenons la droite~$(\Delta')$
+\[
+\Tag{(\Delta')}
+x' = az' - q, \qquad
+y' = bz' + p;
+\]
+$c = 1$ et~$r = - ap - bq$. La surface réglée correspondante est
+engendrée par les droites
+\[
+x + iy + z(az' - q) + z' = 0, \qquad
+(az'- q)(x - iy) - z - bz' - p = 0\Add{,}
+\]
+ou
+\[
+x + iy - qz + z'(az + 1) = 0, \qquad
+-q(x - iy) - z - p + z' \bigl[a(x - iy) - b\bigr] = 0\Add{.}
+\]
+%% -----File: 270.png---Folio 262-------
+\DPchg{Eliminant}{Éliminant} $z'$ on a la surface
+\[
+(x + iy - qz) \bigl[a(x - iy) - b\bigr]
+   + (az + 1) \bigl[q(x - iy) + z + p\bigr] = 0\Add{,}
+\]
+ou
+\[
+a(x^2 + y^2+ z^2) - b(x + iy) + q(x - iy) + (c - r)z + p = 0\Add{,}
+\]
+c'est l'équation d'une sphère et il est facile de voir que ce
+peut être une sphère quelconque, en choisissant~$(\Delta')$ convenablement.
+
+Cherchons la conjuguée de~$(\Delta')$ par rapport à~$(K')$. Nous
+avons à exprimer que le complexe~$(K')$ et les complexes spéciaux
+$(\Delta), (\Delta')$ appartiennent à un même faisceau. Ce qui donne
+\begin{alignat*}{3}%[** TN: Added break, rearranged terms]
+\lambda a + \lambda' a' &= 0, &
+\lambda b + \lambda' b' &= 0, &
+\lambda c + \lambda' c' + 1 &= 0, \\
+\lambda p + \lambda' p' &= 0,\qquad &
+\lambda q + \lambda' q' &= 0,\qquad &
+\lambda r + \lambda' r' + 1 &= 0;
+\end{alignat*}
+prenons $\lambda' = -\lambda$, nous avons:
+\begin{alignat*}{3}%[** TN: Added break, rearranged terms]
+a' &= a, & b' &= b, & c' &= c + \frac{1}{\lambda}, \\
+p' &= p, \qquad & q' &= q, \qquad & r' &= r + \frac{1}{\lambda};
+\end{alignat*}
+prenons $\lambda = -\dfrac{1}{c+r}$, alors $c' = -r$ et $r'= -c$, et l'on voit
+que l'on retrouve la même sphère.
+
+Les formules de la transformation se calculent par la
+méthode générale. On trouve:
+\begin{gather*}
+x = -\frac{z'}{2}
+  + \frac{1}{2}\, \frac{x' (x' p + y' q) + y' - p'}{q' + x'}, \qquad
+y =  \frac{i z'}{2}
+  - \frac{i}{2}\, \frac{x' (x' p + y' q) - y' + p'}{q' + x'}, \\
+z =  -\frac{x'p' + y'q'}{q' + x'}, \qquad
+p =   \frac{x'q' + 1}{q' - x'}, \qquad
+q = -i\frac{x'q' - 1}{q' - x'}.
+\end{gather*}
+Cette transformation de Sophus Lie, changeant des droites qui
+se rencontrent en sphères tangentes, c'est-à-dire des droites
+qui ont un élément de contact commun en sphères ayant un élément
+de contact commun, réalise par suite la correspondance
+%% -----File: 271.png---Folio 263-------
+signalée dans les chapitres précédents entre les droites et
+les sphères.
+
+Par exemple elle transforme une surface réglée en surface
+canal; une quadrique en cyclide de Dupin; une surface
+développable en surface canal isotrope; une bande asymptotique
+d'une surface en une bande de courbure de la transformée;
+de sorte qu'on peut dire qu'elle transforme les lignes asymptotiques
+en lignes de courbure.
+
+On vérifiera facilement qu'elle transforme un complexe
+linéaire de droites en une famille de $\infty^{2}$~sphères coupant une
+sphère fixe sous un angle constant; et que cet angle constant
+est droit, lorsque le complexe linéaire est en involution avec
+le complexe~$(K')$.
+
+\Section{Transformation des lignes asymptotiques.}
+{5.}{} Proposons-nous de trouver toutes les transformations
+de contact qui changent les lignes asymptotiques d'une
+surface quelconque en les lignes asymptotiques de la transformée
+de cette surface; c'est-à-dire qui changent toute
+bande asymptotique en une bande asymptotique. Remarquons à
+cet effet qu'une telle transformation changera toute multiplicité~$M_2$
+sur laquelle les bandes asymptotiques ne dépendent
+pas seulement de constantes arbitraires, mais dépendent de
+fonctions arbitraires, en une multiplicité~$M_2$ de même nature.
+Or\Add{,} les bandes asymptotiques (ou de rebroussement) étant définies
+par les équations
+\[
+dz - p\, dx - q\, dy = 0, \qquad
+dp\, dx + dq\, dy = 0,
+\]
+on devra considérer comme bande asymptotique, dans la question
+%% -----File: 272.png---Folio 264-------
+actuelle, $\infty^{1}$~éléments de contact ayant le même point\Add{,}
+\[%[** TN: Inline parenthetical remark in original]
+dx = dy = dz = 0,\qquad  p = f(q)\Add{,}
+\]
+c'est-à-dire un cône élémentaire.
+
+Et, dès lors, les $M_2$~particulières en question sont les plans,
+les droites et les points. Les transformations cherchées
+échangent donc entre elles les figures qui sont des droites,
+des points ou des plans. De là plusieurs cas à examiner:
+
+\ParItem{\Primo.} Si la transformation est ponctuelle, elle échange les
+points en points, les plans en plans, et les droites en droites.
+C'est par suite une \emph{transformation homographique}.
+
+\ParItem{\Secundo.} Si la transformation est une transformation de contact
+de la première espèce, c'est-à-dire fait correspondre à chaque
+point du premier espace~$(E)$ une surface du second espace~$(E')$,
+elle change les points de~$(E)$ en les plans de~$(E')$; et comme
+elle fait alors correspondre aussi à chaque point de~$(E')$ une
+surface de~$(E)$, elle change les points de~$(E')$ en les plans de~$(E)$;
+et dès lors elle change les points en plans, les plans
+en points, et les droites en droites. Si donc on la compose
+avec une transformation par polaires réciproques, on obtient
+une transformation homographique; et, par suite, elle s'obtient
+en composant une transformation homographique avec une
+transformation par polaires réciproques. C'est donc une
+\emph{transformation dualistique}.
+
+\ParItem{\Tertio.} Si la transformation est une transformation de contact
+de la deuxième espèce, c'est-à-dire si à tout point de l'un
+des espaces correspond dans l'autre une courbe, à tout point
+de l'un des espaces correspondra dans l'autre une droite. Or,
+prenons dans l'espace~$(E)$ quatre points $P_1,P_2,P_3,P_4$ non situés
+%% -----File: 273.png---Folio 265-------
+dans un même plan, et soient $D_1,D_2,D_3,D_4$ les droites qui leur
+correspondent dans l'espace~$(E')$. Il existe au moins une
+droite~$\Delta$ ayant avec chacune des quatre droites $D_1,D_2,D_3,D_4$
+un élément de contact commun; et à~$\Delta$ devrait correspondre dans~$(E)$
+un point, un plan ou une droite ayant un élément de contact
+commun avec chacun des quatre points $P_1,P_2,P_3,P_4$. Or\Add{,} il
+n'en existe pas. Donc \emph{ce troisième cas est impossible}.
+
+Les seules transformations pouvant répondre à la question
+sont donc homographiques ou dualistiques. Mais toute
+transformation de contact changeant les droites en droites
+répond à la question, car elle changera une famille de génératrices
+d'une développable, dont chacune a un élément de
+contact commun avec la génératrice infiniment voisine, en les
+génératrices d'une autre développable; et, par suite, la bande
+de rebroussement de la première développable ou la bande de
+rebroussement de la seconde.
+
+Nous pouvons donc conclure:
+
+\ParItem{\Primo.} \emph{Les transformations homographiques et les transformations
+dualistiques changent les lignes asymptotiques en lignes
+asymptotiques; et ce sont les seules transformations de
+contact possédant cette propriété.}
+
+\ParItem{\Secundo.} \emph{Ces transformations sont aussi les seules transformations
+de contact changeant toute droite en une droite.}
+
+\Section{Transformations des lignes de courbure.}
+{6.}{} La transformation de contact des droites en sphères,
+de Lie, permet de déduire immédiatement des résultats
+précédents toutes les transformations de contact qui
+%% -----File: 274.png---Folio 266-------
+changent les lignes de courbure d'une surface quelconque en
+les lignes de courbure de sa transformée.
+
+On voit de plus que ce sont aussi celles qui changent
+toute sphère en une sphère. On aurait pu du reste refaire un
+raisonnement direct analogue à celui du \Numero~précédent, en partant
+des multiplicités~$M_2$ pour lesquelles les bandes de courbure
+dépendent de fonctions arbitraires.
+
+Cherchons, plus spécialement, celles des transformations
+considérées qui sont des transformations ponctuelles. Dans
+la transformation de Lie, les points de l'espace~$(E)$ correspondent
+aux droites d'un complexe linéaire~$(K')$. Les transformations
+cherchées proviennent donc des transformations
+projectives ou dualistiques qui laissent invariant ce complexe\Add{.}
+On les obtient en composant avec la transformation par polaires
+réciproques définie par ce complexe~$(K')$ l'une quelconque
+des transformations projectives qui laissent le complexe
+invariant.
+
+Ainsi se trouve établie une correspondance entre le
+groupe projectif d'un complexe linéaire et le groupe des
+transformations ponctuelles qui changent toute sphère en
+sphère. Ce dernier est ce qu'on appelle le groupe conforme;
+on sait que ses transformations s'obtiennent en combinant des
+inversions, des homothéties et des déplacements.
+
+Parmi les transformations de contact qui changent les
+lignes de courbure en lignes de courbure figurent les \emph{dilatations},
+dans lesquelles chaque élément de contact subit une
+translation perpendiculaire à son plan et d'amplitude donnée,
+c'est-à-dire dans lesquelles chaque surface est remplacée par
+%% -----File: 275.png---Folio 267-------
+une surface parallèle.
+
+Parmi ces transformations figurent aussi les transformations
+de \DPtypo{Ribeaucour}{Ribaucour} qui seront définies au chapitre~XIII\@.
+
+
+\ExSection{XI}
+
+\begin{Exercises}
+\item[50.] \DPchg{Etudier}{Étudier} la congruence des droites définies par les équations
+\[
+A\lambda + B\mu + C = 0, \qquad
+A_{1}\lambda + B_{1}\mu + C_{1} = 0,
+\]
+où $A,B,C$, $A_{1},B_{1},C_{1}$ sont des fonctions linéaires des coordonnées
+et $\lambda, \mu$ des paramètres arbitraires. Discuter en particulier la
+question des droites passant par un point, des droites rencontrant
+une droite fixe, des droites situées dans un plan, des
+multiplicités focales.
+
+\item[51.] Démontrer les résultats énoncés à la fin du~\No3 de ce chapitre.
+
+\item[52.] Démontrer par le calcul les propriétés de la transformation
+de Lie énoncées à la fin du~\No4 de ce chapitre.
+\end{Exercises}
+%% -----File: 276.png---Folio 268-------
+
+
+\Chapitre{XII}{\DPchg{Systemes}{Systèmes} Triples Orthogonaux.}
+
+
+\Section{Théorème de Dupin.}
+{1.}{} L'emploi des coordonnées rectangulaires revient à
+celui d'un système de \Card{3} plans orthogonaux. On peut généraliser
+et employer comme surfaces coordonnées un système triple
+quelconque:
+\[
+\phi(x,y,z) = u, \qquad
+\psi(x,y,z) = v, \qquad
+\chi(x,y,z) = w;
+\]
+ces formules transforment les coordonnées $u,v,w$ en coordonnées
+$x,y,z$\DPtypo{,}{.} Si nous résolvons les équations précédentes en $x,y,z$, ce
+que nous supposons possible, nous aurons
+\[
+\Tag{(1)}
+x = f(u,v,w), \qquad
+y = g(u,v,w), \qquad
+z = h(u,v,w).
+\]
+On emploie en général un \emph{système triple orthogonal}. Cherchons
+donc à exprimer que les équations~\Eq{(1)} définissent un système
+triple orthogonal. Les intersections des surfaces \Card{2} à \Card{2} doivent
+être orthogonales. Les surfaces des \Card{3} familles s'obtiendront
+en faisant successivement $u = \cte$, $v = \cte$, $w = \cte[]$.
+Les intersections des surfaces \Card{2} à \Card{2} sont respectivement
+$(v = \cte, w = \cte)$\Add{,} $(w = \cte, u = \cte)$\Add{,} $(u = \cte, v = \cte)$, et les
+directions des tangentes sont respectivement $\dfrac{\dd f}{\dd u}, \dfrac{\dd g}{\dd u}, \dfrac{\dd h}{\dd u}$;
+$\dfrac{\dd f}{\dd v}, \dfrac{\dd g}{\dd v}, \dfrac{\dd h}{\dd v}$; et
+$\dfrac{\dd f}{\dd w}, \dfrac{\dd g}{\dd w}, \dfrac{\dd h}{\dd w}$. La condition d'orthogonalité est
+\[
+\Tag{(2)}
+\sum \frac{\dd f}{\dd v} \Add{·} \frac{\dd f}{\dd w} = 0, \qquad
+\sum \frac{\dd f}{\dd w} · \frac{\dd f}{\dd u} = 0, \qquad
+\sum \frac{\dd f}{\dd u} \Add{·} \frac{\dd f}{\dd v} = 0.
+\]
+Interprétons ces conditions. Prenons la surface $w = \cte[]$. La
+\Ord{3}{e} condition exprime que sur cette surface les lignes $u = \cte$,
+%% -----File: 277.png---Folio 269-------
+$v = \cte$ sont orthogonales, et les \Card{2} premières expriment que
+$\dfrac{\dd f}{\dd w}, \dfrac{\dd g}{\dd w}, \dfrac{\dd h}{\dd w}$ est une direction perpendiculaire aux tangentes à
+ces \Card{2} courbes, et par suite, que c'est la direction de la
+normale; soient $l,m,n$ ces \Card{3} coefficients de direction. \DPtypo{Differentions}{Différentions}
+la \Ord{3}{e} relation par rapport à~$w$ nous avons
+\[
+\sum \frac{\dd f}{\dd u}\, \frac{\dd^2 f}{\dd v\, \dd w} +
+\sum \frac{\dd f}{\dd v}\, \frac{\dd^2 f}{\dd u\, \dd w} = 0,
+\]
+ou
+\[
+\sum \frac{\dd f}{\dd u}\, \frac{\dd l}{\dd v} +
+\sum \frac{\dd f}{\dd v}\, \frac{\dd l}{\dd u} = 0.
+\]
+Or\Add{,} on a
+\[
+\sum l\, \frac{\dd f}{\dd u} = 0, \qquad
+\sum l\, \frac{\dd f}{\dd v} = 0,
+\]
+d'où
+\[
+\sum l\, \frac{\dd^2 f}{\dd u\, \dd v}
+  = -\sum \frac{\dd l}{\dd v}\, \frac{\dd f}{\dd u}, \qquad
+\sum l\, \frac{\dd^2 f}{\dd u\, \dd v}
+  = -\sum \frac{\dd l}{\dd u}\, \frac{\dd f}{\dd v},
+\]
+et la condition précédente s'écrit
+\[
+\sum l\, \frac{\dd^2 f}{\dd u\, \dd v}  = 0,
+\]
+ou $F' = 0$, ce qui exprime que les lignes $u = \cte$, $v = \cte$,
+c'est-à-dire les intersections de la surface $w = \cte$ avec les
+surfaces $u = \cte$ et $v = \cte$ sont conjuguées sur cette surface,
+et comme elles sont orthogonales, ce sont des lignes de courbure.
+D'où le \emph{Théorème de Dupin: sur chaque surface d'un
+système triple orthogonal, les intersections avec les autres
+surfaces de ce système sont des lignes de courbure}.
+
+\Section{\DPchg{Equation}{Équation} aux dérivées partielles de Darboux.}
+{2.}{} Proposons-nous de rechercher les systèmes triples
+orthogonaux. Prenons une famille de surfaces:
+\[
+\Tag{(1)}
+\phi(x,y,z) = u
+\]
+et cherchons à déterminer \Card{2} autres familles constituant avec
+celle-ci un système triple orthogonal. Prenons dans l'espace
+%% -----File: 278.png---Folio 270-------
+un point~$M$; par ce point~$M$ passe une surface~$u$; prenons les
+tangentes~$MT, MT'$ en $M$~à ses lignes de courbure. Ces droites
+sont parfaitement déterminées; si $p, q, -1$ sont les coefficients
+de direction de~$MT$, ce sont des fonctions connues de~$x, y, z$.
+De même pour~$MT'$. Il faudra alors qu'en chaque point~$M$,
+une surface d'une autre famille, soit
+\[
+%[** TN: Looks like \Psi in original, but using \psi for consistency]
+%[** TN: No equation number (2) in original]
+\Tag{(2)}
+\psi (x,y,z) = v,
+\]
+soit normale à~$MT$; il faudra donc que $p, q$ soient les dérivées
+partielles de~$z$ par rapport à $x, y$, ($z$~étant défini par l'équation
+précédente). On aura donc
+\[
+\frac{\dd \psi}{\dd x} + p \frac{\dd \psi}{\dd z} = 0, \qquad
+\frac{\dd \psi}{\dd y} + \Err{}{q} \frac{\dd \psi}{\dd z} = 0.
+\]
+Ces équations ne sont pas compatibles en général: pour
+qu'elles le soient, il faut et il suffit, d'après la théorie
+des systèmes complets, que l'on ait:
+\[
+\Tag{(3)}
+\frac{\dd q}{\dd x} + p \frac{\dd q}{\dd z} =
+\frac{\dd p}{\dd y} + q \frac{\dd p}{\dd z}\Add{,}
+\]
+équation aux dérivées partielles du \Ord{3}{e} ordre, puisque $p, q$
+s'expriment en fonction des dérivées \Ord{1}{ères} et \Ord{2}{mes} de~$\phi$ par
+rapport à $x, y, z$. Ainsi donc \emph{une famille de surfaces données ne
+peut en général faire partie d'un système triple orthogonal}.
+Si la condition~\Eq{(3)} est réalisée, la fonction~$\psi(x,y,z)$ est déterminée
+à une fonction arbitraire près, et nous avons une
+\Ord{2}{e} famille de surfaces dont chacune coupe à angle droit chacune
+des surfaces~$(S)$ de la famille~$\phi(x,y,z) = \const.$\ suivant
+une ligne de courbure de cette surface~$(S)$. Et, d'après le
+théorème de Joachimsthal, l'intersection de chaque surface~$(S_1)$
+de cette seconde famille avec chaque surface~$(S)$ de la première
+est aussi ligne de courbure sur~$(S_1)$.
+%% -----File: 279.png---Folio 271-------
+
+En résumé nous avons deux familles de surfaces
+\begin{align*}
+\Tag{(S)}
+\phi(x,y,z) &= \const. \\
+\Tag{(S_1)}
+\psi(x,y,z) &= \const.
+\end{align*}
+qui se coupent orthogonalement suivant des courbes qui sont
+lignes de courbure à la fois pour les deux surfaces qui s'y
+croisent. Et il reste à étudier si l'on peut déterminer une
+troisième famille de surfaces
+\[
+\Tag{(S_2)}
+\chi(x,y,z) = \const.
+\
+\]
+qui constitue avec les deux premières un système triple orthogonal,
+c'est-à-dire à étudier le système d'équations linéaires
+aux dérivées partielles dont dépend la fonction inconnue~$\chi$:
+\[
+\Tag{(4)}
+\left\{
+\begin{aligned}
+\frac{\dd \phi}{\dd x} · \frac{\dd \chi}{\dd x} +
+\frac{\dd \phi}{\dd y} · \frac{\dd \chi}{\dd y} +
+\frac{\dd \phi}{\dd z} · \frac{\dd \chi}{\dd z} &= 0, \\
+%
+\frac{\dd \psi}{\dd x} · \frac{\dd \chi}{\dd x} +
+\frac{\dd \psi}{\dd y} · \frac{\dd \chi}{\dd y} +
+\frac{\dd \psi}{\dd z} · \frac{\dd \chi}{\dd z} &= 0.
+\end{aligned}
+\right.
+\]
+Posons, pour abréger,
+\begin{align*}
+Af &= \frac{\dd \phi}{\dd x} · \frac{\dd f}{\dd x}
+    + \frac{\dd \phi}{\dd y} · \frac{\dd f}{\dd y}
+    + \frac{\dd \phi}{\dd z} · \frac{\dd f}{\dd z}, \\
+Bf &= \frac{\dd \phi}{\dd x} · \frac{\dd f}{\dd x}
+    + \frac{\dd \phi}{\dd y} · \frac{\dd f}{\dd y}
+    + \frac{\dd \phi}{\dd z} · \frac{\dd f}{\dd z};
+\end{align*}
+et, d'après la théorie des systèmes complets d'équations linéaires,
+la condition nécessaire et suffisante pour l'\Err{intégralité}{intégrabilité}
+du système~\Eq{(4)} est que l'équation:
+\[
+\left(A \frac{\dd\psi}{\dd x} - B \frac{\dd\phi}{\dd x}\right)
+  \Del{·} \frac{\dd\chi}{\dd x} +
+\left(A \frac{\dd\psi}{\dd y} - B \frac{\dd\phi}{\dd y}\right)
+  \Del{·} \frac{\dd\chi}{\dd y} +
+\left(A \frac{\dd\psi}{\dd z} - B \frac{\dd\phi}{\dd z}\right)
+  \Del{·} \frac{\dd\chi}{\dd z} = 0
+\]
+soit une conséquence algébrique des équations~\Eq{(4)}, c'est-à-dire
+que l'on ait:
+\[
+\Tag{(5)}
+\begin{vmatrix}
+A \mfrac{\dd\psi}{\dd x} - B \mfrac{\dd\phi}{\dd x} &
+  \mfrac{\dd\phi}{\dd x} & \mfrac{\dd\psi}{\dd x} \\
+A \mfrac{\dd\psi}{\dd y} - B \mfrac{\dd\phi}{\dd y} &
+  \mfrac{\dd\phi}{\dd y} & \mfrac{\dd\psi}{\dd y} \\
+A \mfrac{\dd\psi}{\dd z} - B \mfrac{\dd\phi}{\dd z} &
+  \mfrac{\dd\phi}{\dd z} & \mfrac{\dd\psi}{\dd z}
+\end{vmatrix}
+= 0.
+\]
+%% -----File: 280.png---Folio 272-------
+Cette condition se simplifie. Remarquons en effet que
+\begin{align*}
+A \frac{\dd\psi}{\dd x} + B \frac{\dd\phi}{\dd x}
+  &= \frac{\dd\phi}{\dd x}\, \frac{\dd^2\psi}{\dd x^2}
+   + \frac{\dd\phi}{\dd y}\, \frac{\dd^2\psi}{\dd y\, \dd x}
+   + \frac{\dd\phi}{\dd z}\, \frac{\dd^2\psi}{\dd z\, \dd x} \\
+%
+  &+ \frac{\dd\psi}{\dd x}\, \frac{\dd^2\phi}{\dd x^2}\,
+   + \frac{\dd\psi}{\dd y}\, \frac{\dd^2\phi}{\dd y\, \dd x}
+   + \frac{\dd\psi}{\dd z} · \frac{\dd^2\phi}{\dd z\, \dd x} \\
+%
+  &= \frac{\dd}{\dd x} \left\{
+       \frac{\dd\phi}{\dd x}\, \frac{\dd\psi}{\dd x}
+     + \frac{\dd\phi}{\dd y}\, \frac{\dd\psi}{\dd y}
+     + \frac{\dd\phi}{\dd z}\, \frac{\dd\psi}{\dd z}
+  \right\},
+\end{align*}
+c'est-à-dire que l'on a, à cause de l'\DPtypo{orthogonalite}{orthogonalité} des surfaces $(S)$~et~$(S_1)$,
+\[
+A \frac{\dd\psi}{\dd x} + B \frac{\dd\phi}{\dd x} = 0.
+\]
+On voit de même que l'on a aussi
+\[
+A \frac{\dd\psi}{\dd y} + B\frac{\dd\phi}{\dd y} = 0, \qquad
+A \frac{\dd\psi}{\dd z} + B\frac{\dd\phi}{\dd z} = 0.
+\]
+Par suite, la condition~\Eq{(4)} peut s'écrire simplement
+\[
+\Tag{(6)}
+\begin{vmatrix}
+A \mfrac{\dd\psi}{\dd x} & \mfrac{\dd\phi}{\dd x} & \mfrac{\dd\psi}{\dd x} \\
+A \mfrac{\dd\psi}{\dd y} & \mfrac{\dd\phi}{\dd y} & \mfrac{\dd\psi}{\dd y} \\
+A \mfrac{\dd\psi}{\dd z} & \mfrac{\dd\phi}{\dd z} & \mfrac{\dd\psi}{\dd z}
+\end{vmatrix}
+= 0.
+\]
+Or, pour une valeur quelconque de $x,y,z$, les dérivées~$\dfrac{\dd \psi}{\dd x},\dfrac{\dd \psi}{\dd y},\dfrac{\dd \psi}{\dd z}$ sont les coefficients de direction~$l,m,n$ de la normale
+à la surface~$(S_1)$ qui passe par le point de coordonnées~$x,y,z$;
+et~$\dfrac{\dd\phi}{\dd x},\dfrac{\dd\phi}{\dd y},\dfrac{\dd\phi}{\dd z}$ sont les coefficients de direction de la
+normale à la surface~$(S)$ qui passe par ce même point, c'est-à-dire
+de la tangente à une ligne de courbure de~$(S_1)$; en désignant
+par~$dx,dy,dz$ un déplacement effectué suivant la direction
+de cette tangente, on aura donc
+\[
+\frac{\ \dfrac{\dd\phi}{\dd x}\ }{dz} =
+\frac{\ \dfrac{\dd\phi}{\dd y}\ }{dy} =
+\frac{\ \dfrac{\dd\phi}{\dd z}\ }{dx} =
+\frac{A \dfrac{\dd\psi}{\dd x}}{dl} =
+\frac{A \dfrac{\dd\psi}{\dd y}}{dm} =
+\frac{A \dfrac{\dd\psi}{\dd z}}{dn},
+\]
+%% -----File: 281.png---Folio 273-------
+et la condition~\Eq{(6)} deviendra:
+\[
+\begin{vmatrix}
+dl & dx & l \\
+dm & dy & m \\
+dn & dz & n
+\end{vmatrix}
+= 0\DPtypo{.}{,}
+\]
+ce qui est une identité puisque la différentiation~$d$ considérée
+a lieu suivait une ligne de courbure.
+
+La condition d'intégrabilité du système~\Eq{(4)} est donc remplie,
+et la troisième famille~$(S_2)$ existe toujours et est
+entièrement déterminée. On peut donc énoncer les résultats
+suivants:
+
+\ParItem{\Primo.} \emph{Il existe une équation aux dérivées partielles du troisième
+ordre $\bigl($l'équation \Eq{(3)}$\bigr)$ qui exprime la condition nécessaire
+et suffisante pour qu'une fonction $\phi(x,y,z)$ fournisse
+une famille de surfaces~$(S)$ faisant partie d'un système triple
+orthogonal. Si la famille~$(S)$ est donnée, les deux autres familles $(S_1)$~et~$(S_2)$ sont entièrement déterminées}.
+
+\ParItem{\Secundo.} \emph{Pour que deux familles de surfaces, $(S)$~et~$(S_1)$, fassent
+partie d'un système triple orthogonal, il faut et il suffit
+qu'elles se coupent à angle droit, et que les intersections
+soient lignes de courbure sur les surfaces~$(S)$, ou sur les
+surfaces~$(S_1)$.}
+
+On peut remarquer enfin, que si l'on \DPchg{connait}{connaît} les lignes
+de courbure~$(C_1)$ des surfaces~$(S_1)$ par exemple, qui ne sont
+pas les intersections des surfaces~$(S_1)$ et des surfaces~$(S)$,
+et les lignes de courbure~$(C)$ d'une seule surface~$(S)$, chaque
+%% -----File: 282.png---Folio 274-------
+surface~$(S_2)$ sera engendrée par les courbes~$(C_1)$ qui s'appuient
+sur une même courbe~$(C)$.
+
+\Section{Systèmes triples orthogonaux contenant une surface.}
+{3.}{} Cherchons maintenant si une surface donnée peut
+faire partie d'un système triple orthogonal. Traçons
+sur cette surface les lignes de courbure, et menons les
+normales à la surface en tous les points de ces lignes, elles
+engendrent deux familles de développables orthogonales à la
+surface donnée. En adjoignant à cette surface les surfaces
+parallèles, on a un système triple orthogonal.
+
+\Paragraph{Remarque \1.} Les surfaces parallèles à une surface~$(S)$
+en dérivent par une transformation de contact définie par
+l'équation
+\[
+(X - x)^2 + (Y - y)^2 + (Z - z)^2 - r^2 = 0\Add{,}
+\]
+où $r$~est une constante arbitraire; en effet la surface parallèle
+est l'enveloppe d'une famille de sphères de rayon
+constant ayant leurs centres sur la surface~$(S)$. Cette
+transformation de contact s'appelle dilatation.
+
+%[** TN: Roman numeral in original]
+\Paragraph{Remarque \2.} Lorsqu'on sait qu'une famille de surfaces
+appartient à un système triple orthogonal, on peut sur
+chacune de ces surfaces déterminer les lignes de courbure,
+et les autres familles du système sont engendrées par les
+trajectoires orthogonales des surfaces qui s'appuient sur
+les lignes de courbure. Dans le cas particulier d'une famille
+de surfaces parallèles, les trajectoires orthogonales
+sont les normales à ces surfaces.
+%% -----File: 283.png---Folio 275-------
+
+\Section{Systèmes triples orthogonaux contenant une famille de plans.}
+{4.}{} Considérons une famille de plans~$P$; les trajectoires
+orthogonales s'obtiennent, comme on l'a vu à propos
+des surfaces moulures, en faisant rouler un plan mobile
+sur la développable enveloppe des plans~$P$. Prenons
+dans le plan deux systèmes de courbes orthogonales, ce qui
+est toujours possible, car si nous nous donnons l'un des
+systèmes
+\[
+\phi(x,y) = a,
+\]
+l'autre se détermine par l'intégration de l'équation
+\[
+\dfrac{dx}{\ \dfrac{\dd\phi}{\dd x}\ } =
+\dfrac{dy}{\ \dfrac{\dd\phi}{\dd y}\ }.
+\]
+On peut engendrer les autres familles du système triple
+orthogonal au moyen de ces courbes des plans~$P$, assujetties
+à rencontrer les trajectoires orthogonales; ces familles
+sont ainsi constituées de surfaces moulures. On peut ainsi,
+au moyen du Théorème de Dupin, retrouver leurs lignes de
+courbure.
+
+\Section{Systèmes triples orthogonaux contenant une famille de sphères.}
+{5.}{} Le fait que toute famille de plans fait partie
+d'un système triple orthogonal tient au fond à ce que
+toute courbe d'un plan est ligne de courbure du plan; de
+sorte qu'une famille de surfaces orthogonales aux plans
+donnés satisfera à la condition \DPtypo{necessaire}{nécessaire} et suffisante
+pour qu'il existe une troisième famille complétant le système
+triple orthogonal.
+
+Le même fait sera donc vrai aussi pour une famille de
+sphères. Et pour construire un système triple orthogonal
+%% -----File: 284.png---Folio 276-------
+quelconque contenant la famille de sphères~$(S)$ donnée, il
+suffira: \Primo.~de prendre sur une des sphères deux familles
+de courbes~$(C), (C_1)$ orthogonales; \Secundo.~de déterminer les trajectoires
+orthogonales~$(T)$ des sphères~$(S)$. Car alors les
+courbes~$(T)$ qui s'appuient sur les courbes~$(C)$, et les courbes~$(T)$
+qui s'appuient sur les courbes~$(C_1)$ engendreront les
+surfaces des deux familles $(S_1)$~et~$(S_2)$ formant avec les sphères~$(S)$
+le système triple cherché.
+
+Tout revient donc à résoudre les deux problèmes suivants:
+\Primo.~déterminer sur une sphère un système orthogonal quelconque;
+\Secundo.~déterminer les trajectoires orthogonales d'une famille de sphères.
+
+Le premier problème se ramène immédiatement au problème
+analogue dans le plan au moyen d'une projection stéréographique.
+
+\DPchg{Etudions}{Étudions} donc le second:
+
+Si nous considérons \Card{2} sphères de la famille, les trajectoires
+orthogonales établissent entre elles une correspondance
+point par point, et cette correspondance d'après ce qui précède,
+sera telle qu'à un système orthogonal sur l'une des
+sphères corresponde un système orthogonal sur l'autre. Or,
+deux directions rectangulaires sont conjuguées harmoniques
+par rapport aux directions isotropes, et dans une transformation
+ponctuelle quelconque, le rapport anharmonique des tangentes
+est un invariant, donc les directions isotropes se
+transforment en directions isotropes, les génératrices rectilignes
+de l'une des sphères se transforment en génératrices
+%% -----File: 285.png---Folio 277-------
+rectilignes de l'autre; et le rapport anharmonique de \Card{2} directions
+quelconques avec les directions isotropes restant
+constant, les angles se conservent; la transformation est
+conforme.
+
+Soit alors
+\[
+\sum (x - \Err{x}{x_0})^2 - R^2 = 0\Add{,}
+\]
+l'équation générale des sphères considérées, dépendant d'un
+paramètre~$t$\DPtypo{,}{.}
+
+Les considérations précédentes nous conduisent à introduire
+les génératrices rectilignes. On posera donc:
+\begin{align*}
+x - x_0 + i(y - y_0) &= \lambda \bigl[(z - z_0) + R\bigr], \\
+x - x_0 - i(y - y_0) &= \frac{-1}{\lambda} \bigl[(z - z_0) - R\bigr]
+  = \mu \bigl[z - z_0 + R\bigr];
+\end{align*}
+d'où:
+\begin{gather*}%[** TN: Re-arranged]
+(z - z_0)\left(\mu + \frac{1}{\lambda}\right)
+  = R \left(\frac{1}{\lambda} - \mu\right),\qquad
+z - z_0 = R\, \frac{1 - \lambda \mu}{1 + \lambda \mu}, \\
+\begin{aligned}
+x - x_0 + i(y - y_0) &= \frac{2R \lambda}{1 + \lambda \mu}, \\
+x - x_0 - i(y - y_0) &= \frac{2R \mu}{1 + \lambda \mu}.
+\end{aligned}
+\end{gather*}
+Les équations \DPtypo{différentièllés}{différentielles} à des trajectoires orthogonales
+sont:
+\[
+\frac{dx}{x - x_0} =
+\frac{dy}{y - y_0} =
+\frac{dz}{z - z_0} =
+\frac{d(x + iy)}{x - x_0 + i(y - y_0)} =
+\frac{d(x - iy)}{x - x_0 - i(y - y_0)}\Add{.}
+\]
+\DPtypo{en}{En} égalant le \Ord{3}{e} rapport successivement aux \Card{2} derniers, et
+posant:
+\[
+dA = \frac{d(x_0 + i y_0)}{2R}, \qquad
+dB = \frac{d(x_0 - i y_0)}{2R}, \qquad
+dC = \frac{dz_0}{2\Err{R_0}{R}},
+\]
+\DPtypo{On}{on} obtient \Card{2} équations de Riccati
+\[
+\frac{d\lambda}{dt}
+  = \lambda^2 \frac{dB}{dt} - 2\lambda \frac{dC}{dt} - \frac{dA}{dt},
+\qquad
+\frac{d \mu}{dt}
+  = \mu^2 \frac{dA}{dt} - 2\mu \frac{dC}{dt} - \frac{dB}{dt}.
+\]
+Si on \DPchg{connait}{connaît} une trajectoire orthogonale, on a une solution
+%% -----File: 286.png---Folio 278-------
+de chaque équation, et la résolution du problème est ramenée
+à \Card{2} quadratures. Pour \Card{2} trajectoires orthogonales, on n'a
+plus qu'une seule quadrature, et pour \Card{3} trajectoires orthogonales,
+le problème s'achève sans quadrature. On a alors,
+comme intégrale générale de la première équation:
+\[
+\frac{\lambda - \lambda_1}{\lambda - \lambda_2} :
+\frac{\lambda_3 - \lambda_1}{\lambda_3 - \lambda_2} =
+\frac{\lambda^0 - \lambda^0_1}{\lambda^0 - \lambda^0_2} :
+\frac{\lambda^0_3 - \lambda^0_1}{\lambda^0_3 - \lambda^0_\DPtypo{1}{2}},
+\]
+en désignant par l'indice zéro les valeurs qui correspondent
+à~$t = t_0$. C'est donc une relation de la forme:
+\[
+\lambda = \frac{M \lambda^0 + N}{P \lambda^0 + Q}.
+\]
+On aura de même, pour la seconde équation de Riccati, une
+intégrale de la forme
+\[
+\mu = \frac{R \mu^0 + S}{T \mu^0 + U};
+\]
+et ces deux formules définissent la correspondance établie par
+les trajectoires orthogonales entre la sphère qui correspond
+à la valeur~$t_0$ du paramètre, et la sphère qui correspond à la
+valeur~$t$ du paramètre.
+
+On \DPchg{reconnait}{reconnaît} alors que cette transformation change les
+cercles d'une des sphères en cercles de l'autre; et par projection
+stéréographique elle deviendrait une des transformations
+planes du groupe des \Err{}{des transformations par }rayons vecteurs réciproques.
+
+\Section{Systèmes triples orthogonaux particuliers.}
+{6.}{} Rappelons comme systèmes triples orthogonaux
+particuliers, le système des quadriques homofocales
+\[
+\frac{x^2}{a - \lambda} +
+\frac{y^2}{b - \lambda} +
+\frac{z^2}{c - \lambda} - 1 = 0;
+\]
+et le système des cyclides du quatrième degré homofocales
+\[
+\frac{x^2}{a - \lambda} +
+\frac{y^2}{b - \lambda} +
+\frac{z^2}{c - \lambda} +
+\frac{(x^2 + y^2 + z^2 - R^2)^2}{4R^2 (d - \lambda)} -
+\frac{(x^2 + y^2 + z^2 + R^2)\Err{}{^2}}{4R^2 (e - \lambda)} = 0.
+\]
+%% -----File: 287.png---Folio 279-------
+On vérifie qu'on obtient un autre système, formé de cyclides
+de Dupin du troisième degré, en considérant les surfaces
+lieux des points de contact des plans tangents menés, par un
+point d'un des axes, à une famille de quadriques homofocales.
+
+
+\ExSection{XII}
+
+\begin{Exercises}
+\item[53.] On considère une famille de $\infty^{1}$~\DPchg{paraboloides}{paraboloïdes}~$P$ ayant mêmes
+plans principaux. Comment faut-il choisir ces \DPchg{paraboloides}{paraboloïdes}
+pour que la congruence des génératrices rectilignes d'un même
+système de tous ces \DPchg{paraboloides}{paraboloïdes} soit une congruence de normales?
+Montrer qu'alors les \DPchg{paraboloides}{paraboloïdes}~$P$ constituent l'une
+des trois familles d'un système triple orthogonal et trouver
+les deux autres familles. Montrer qu'on peut choisir les \DPchg{paraboloides}{paraboloïdes}~$P$
+plus particulièrement de manière que l'une de
+ces autres familles soit encore formée de \DPchg{paraboloides}{paraboloïdes}; et
+donner, dans ce cas, la signification géométrique des deux
+familles de \DPchg{paraboloides}{paraboloïdes}.
+\end{Exercises}
+%% -----File: 288.png---Folio 280-------
+
+
+\Chapitre{XIII}{Congruences de \DPchg{Spheres}{Sphères} et \DPchg{Systemes}{Systèmes} Cycliques.}
+
+\Section{Généralités.}
+{1.}{} Nous appellerons \emph{congruence de sphères} une famille
+de $\infty^{2}$~sphères~$(\Sigma)$
+\[
+\Tag{(1)}
+\sum (x - f)^2 - r^2 = 0,
+\]
+$f,g,h,r$ étant fonctions de \Card{2} paramètres~$u,v$. Le lieu des centres
+de ces sphères est une surface~$(S)$
+\[
+x = f(u,v), \qquad
+y = g(u,v), \qquad
+r = h(u,v).
+\]
+Cherchons l'enveloppe de ces sphères. A l'équation~\Eq{(1)} nous
+devrons adjoindre les \Card{2} équations
+\[
+\Tag{(2)}
+\sum (x - f) \frac{\dd f}{\dd u} + r \frac{\dd r}{\dd u} = 0, \qquad
+\sum (x - f) \frac{\dd f}{\dd v} + r \frac{\dd r}{\dd v} = 0.
+\]
+Ces équations~\Eq{(2)} représentent une droite, donc l'enveloppe
+des sphères~$(\Sigma)$ touche chacune de ces sphères en \Card{2} points,
+que l'on appelle \emph{points focaux}; l'enveloppe, que l'on appellera
+\emph{surface focale}, se \DPtypo{decompose}{décompose} donc en \Card{2} nappes~$(F_1), (F_2)$.
+
+Considérons dans la congruence~\Eq{(1)} une famille de $\infty^{1}$~sphères~$(\Sigma)$;
+il suffit de définir~$u, v$ en fonction d'un paramètre~$t$;
+ces sphères admettent une enveloppe, qui touche chacune
+d'elles le long d'un cercle \DPtypo{caracteristique}{caractéristique} ayant pour plan
+\[
+\Tag{(3)}
+\sum (x - f)\, df + r\, dr = 0;
+\]
+lorsque les expressions de~$u, v$ en fonction de~$t$ varient, tous
+ces cercles caractéristiques passent par \Card{2} points fixes, qui
+sont les points focaux de la sphère considérée. Les enveloppes
+%% -----File: 289.png---Folio 281-------
+ainsi obtenues correspondent aux surfaces réglées des congruences
+de droites; on peut les appeler \emph{surfaces canaux} de
+la congruence~\Eq{(1)}.
+
+Cherchons parmi ces surfaces canaux celles pour lesquelles
+chaque sphère est tangente à la sphère infiniment
+voisine. Ce sont en réalité des surfaces réglées à génératrices
+isotropes. Le cercle défini par les équations \Eq{(1)},~\Eq{(3)}
+doit se réduire à un couple de droites isotropes; le plan~\Eq{(3)}
+est tangent à la sphère~\Eq{(1)}. Ce qui donne la condition
+\[
+\Tag{(4)}
+\sum df^2 - dr^2 = 0,
+\]
+équation différentielle du \Ord{1}{re} ordre et du \Ord{2}{e} degré; il y a
+donc \Card{2} familles de sphères spéciales, le point de contact de
+l'une d'elles avec la sphère infiniment voisine étant défini
+par
+\[
+\Tag{(5)}
+\frac{x - f}{df} = \frac{y - g}{dg} = \frac{z - h}{dh}
+  = \frac{-r}{dr}\Add{,}
+\]
+$df, dg, dh$~sont les coefficients de direction du rayon du point
+de contact.
+
+\Illustration[2.75in]{289a}
+L'équation~\Eq{(4)} définit sur la surface~$(S)$ \Card{2} directions $\omega l, \omega l'$; soient $M, M'$~les
+points de contact de la
+sphère~$(\Sigma)$ correspondante
+avec la surface focale~$(F)$;
+la droite~$MM'$ est représentée
+par les \Card{2} équations~\Eq{(2)},
+ou encore, puisque les
+points~$M, M'$ sont sur
+tous les cercles caractéristiques, par les \Card{2} équations~\Eq{(3)}.
+%% -----File: 290.png---Folio 282-------
+qui correspondent aux enveloppes spéciales; or\Add{,} dans ce cas
+l'équation~\Eq{(3)} représente le plan tangent à la sphère en l'un
+des points~$I, I'$; donc les droites~$II', MM'$ sont polaires réciproques
+par rapport à la sphère~$(\Sigma)$. Si nous supposons cette
+sphère réelle, et si les points~$M, M'$ sont réels, $I,I'$~sont
+imaginaires; et inversement. Nous désignerons par~$D$ la droite~$MM'$,
+et par~$\Delta$ la droite~$ll'$; $\omega l, \omega l'$~sont dans le plan tangent
+en~$\omega$ à la surface~$(S)$; $MM'$~est perpendiculaire à ce plan tangent;
+les points~$M, M'$, et par suite les droites~$\omega M, \omega M'$ sont
+symétriques par rapport à ce plan tangent.
+
+Si nous remarquons maintenant que $\omega M$~est normale à la
+\Ord{1}{ère} nappe de la surface focale, et $\omega M'$~normale à la \Ord{2}{e}, nous
+voyons que l'on peut considérer~$\omega M$ comme rayon incident, $\omega M'$~comme
+rayon réfléchi sur la surface~$(S)$, et par suite, nous
+avons une congruence de normales qui se réfléchit sur la surface~$(S)$
+suivant une congruence de normales. La surface~$(S)$
+étant quelconque, donnons-nous une surface~$(F_1)$; nous pourrons
+toujours considérer les sphères ayant leurs centres sur~$(S)$
+et tangentes à~$(F_1)$; $(F_1)$~sera l'une des nappes focales de
+la congruence de sphères ainsi obtenues, et la congruence des
+normales à~$(F_1)$ se réfléchira sur~$(S)$ suivant la congruence
+des normales à~$(F_2)$ \Ord{2}{e} nappe focale. D'où le \emph{Théorème de
+Malus: Les rayons normaux à une surface quelconque se réfléchissent
+sur une surface quelconque suivant les normales à
+une nouvelle surface}.
+
+\Illustration[2.25in]{291a}
+Ce Théorème peut s'étendre aux rayons réfractés. Reprenons
+%% -----File: 291.png---Folio 283-------
+la construction d'Huygens. Considérons une sphère de
+centre~$\omega$; soit $\omega M$~le rayon incident,
+et la normale~$\omega N$ à la surface \DPtypo{réflechissante}{réfléchissante}.
+Construisons une \Ord{2}{e} sphère de centre~$\omega$ et dont le rayon
+soit dans le rapport~$n$ avec le rayon
+de la \Ord{1}{ère}. Considérons le plan tangent~$\omega T$
+à la surface réfléchissante.
+Au point~$M$ où le rayon incident rencontre la \Ord{1}{ère} sphère menons
+le plan tangent à cette sphère, qui rencontre le plan~$\omega l$
+suivant la droite~$T$, et par la droite~$T$ menons le plan~$TP$
+tangent à la \Ord{2}{e} sphère. En appelant~$i\Add{,} i'$ les angles de~$\omega M$
+et~$\omega P$ avec~$\omega N$, on a immédiatement
+\[
+\omega T = \frac{\omega M}{\sin i} = \frac{\omega P}{\sin i'}\Add{,}
+\]
+d'où
+\[
+\frac{\sin i'}{\sin i} = \frac{\omega P}{\omega M} = n\Add{,}
+\]
+$\omega P$~est le rayon réfracté. Partons alors d'une congruence de
+normales, soit $(F_1)$~la surface normale; pour construire les
+rayons réfractés, il faut considérer les sphères~$(\Sigma')$ concentriques
+à~$(\Sigma)$ et de rayon~$nr$; les points~$I, I'$ sont définis
+par les équations~\Eq{(5)}; or\Add{,} ces équations ne changent pas lorsqu'on
+remplace~$r$ par~$nr$, la droite~$II'$ est la même que précédemment,
+et le Théorème s'étend à la réfraction.
+
+\Section{Congruences spéciales.}
+{2.}{} A la congruence de sphères considérée nous avons
+associé \Card{4} congruences de droites: celle des droites~$\omega M$ normales
+à~$(F_1)$, celles des droites~$\omega M'$ normales à~$(F_2)$~celle des
+%% -----File: 292.png---Folio 284-------
+droites~$\Delta$, et celle des droites~$D$.
+
+Supposons~$MM'$~confondus; ils sont confondus aussi avec~$II'$;
+les \Card{2} nappes focales sont confondues; alors le lieu des
+points~$I\Add{,} I'$ confondus est une ligne de courbure de la surface~$(F)$
+et la sphère~$(\Sigma)$ est la sphère de courbure correspondante.
+\emph{La congruence de sphères est alors constituée par des sphères
+de courbure d'une même surface~$(F)$.}
+
+\emph{Réciproquement}. Soit une surface~$(F)$ et ses sphères de
+courbure d'une même famille, la surface~$(F)$ est surface focale
+double de la congruence de ces sphères de courbure.
+
+Toutes les congruences de droites considérées se réduisent
+ici à~$2$, celle des droites~$D$ tangentes à une famille de
+lignes de courbure, et celle des droites~$\Delta$~tangentes à l'autre
+famille. La surface~$(S)$ est alors l'une des nappes de
+la développée de la surface focale double. Aux lignes de
+courbure intégrales~de~\Err{\Eq{(H)}}{\Eq{(4)}} correspond sur la surface~$(S)$ une
+famille de géodésiques. On est ainsi conduit à la détermination
+des géodésiques de~$(S)$ en écrivant que l'équation~\Eq{(4)} a
+une racine double en~$du, dv$. Cette équation s'écrit
+\begin{gather*}
+E\, du^2 + 2F\, du\, dv + G\, dv^2
+  - \left(\frac{\dd r}{\dd u}\, du
+        + \frac{\dd r}{\dd v}\, dv \right)^2 = 0\Add{,} \\
+\left[ E - \left(\frac{\dd r}{\dd u}\right)^2 \right] du^2
+  + 2 \Biggl[ F - \frac{\dd r}{\dd u}\, \frac{\dd r}{\dd v} \Biggr] du\, dv
+  + \left[ G - \left(\frac{\dd r}{\dd v}\right)^2 \right] dv^2 = 0\Add{,}
+\end{gather*}
+pour qu'il y ait une racine double, il faut que
+\[
+\left[ E - \left(\frac{\dd r}{\dd u}\right)^2 \right]
+\left[ G - \left(\frac{\dd r}{\dd v}\right)^2 \right]
+  - \Biggl[ F - \frac{\dd r}{\dd u}\, \frac{\dd r}{\dd v} \Biggr]^2 = 0\Add{,}
+\]
+ou
+\[
+H^2 - \left[ E \left(\frac{\dd r}{\dd v}\right)^2
+  - 2F \frac{\dd r}{\dd u}\, \frac{\dd r}{\dd v}
+  + G \left(\frac{\dd r}{\dd u}\right)^2 \right] = 0\Add{,}
+\]
+équation aux dérivées partielles qui détermine~$r$. Ayant~$r$ on
+obtient la famille de géodésiques correspondante par l'intégration
+%% -----File: 293.png---Folio 285-------
+d'une équation différentielle ordinaire.
+
+\Section{Théorème de Dupin.}
+{3.}{} Supposons que la surface focale ait ses \Card{2} nappes
+distinctes, et cherchons ses lignes de courbure. Soient~$\lambda\Add{,} \mu\Add{,} \nu$
+les cosinus directeurs de~$\omega M$.
+\[
+\lambda = \frac{x - f}{r}, \qquad
+\mu = \frac{y - g}{r}, \qquad
+\nu = \frac{z - h}{r};
+\]
+d'où
+\[
+x = f + \lambda r, \qquad
+y = g + \mu r, \qquad
+z = h + \nu r.
+\]
+Portons ces valeurs de~$x,y,z$ dans les équations~\Eq{(2)}, elles
+deviennent
+\[
+\Tag{(6)}
+\sum \lambda \frac{\dd f}{\dd u} + \frac{\dd r}{\dd u} = 0, \qquad
+\sum \lambda \frac{\dd f}{\dd v} + \frac{\dd r}{\dd v} = 0.
+\]
+Soient~$i, i'$ les angles de $\omega M$~et~$\omega M'$ avec~$\omega N$, ces angles sont
+supplémentaires, $\cos i' = -\cos i$; si $l,m,n$~sont les cosinus
+directeurs de~$\omega N$\DPtypo{.}{,} \DPtypo{On}{on} a
+\[
+\Tag{(7)}
+\sum \lambda l - \cos i = 0.
+\]
+Calculons~$\cos i$. Dans le plan tangent à~$(S)$ soient $\omega U, \omega V$ les
+tangentes aux courbes $v = \cte$, $u = \cte[]$.
+
+\Illustration{293a}
+Les cosinus directeurs de~$\omega U$ sont
+\[
+\frac{1}{\sqrt{E}}\, \frac{\dd f}{\dd u}, \qquad
+\frac{1}{\sqrt{E}}\, \frac{\dd g}{\dd u}, \qquad
+\frac{1}{\sqrt{E}}\, \frac{\dd h}{\dd u};
+\]
+\DPtypo{Ceux}{ceux} de~$\omega V$ sont
+\[
+\frac{1}{\sqrt{G}}\, \frac{\dd f}{\dd v}, \qquad
+\frac{1}{\sqrt{G}}\, \frac{\dd g}{\dd v}, \qquad
+\frac{1}{\sqrt{G}}\, \frac{\dd h}{\dd v}.
+\]
+Soit $\omega \delta$~le segment directeur de~$\omega M$,
+ses projections orthogonales sur~$\omega U, \omega V$ sont:
+\[
+A = -\frac{1}{\sqrt{E}}\, \frac{\dd r}{\dd u}, \qquad
+B = -\frac{1}{\sqrt{G}}\, \frac{\dd r}{\dd v},
+\]
+sa projection sur~$\omega W$ est~$\cos i$. Soit $\theta$~l'angle de~$\omega U$~et~$\omega V$:
+\[
+\Cos \theta = \frac{F}{\sqrt{EG}}.
+\]
+%% -----File: 294.png---Folio 286-------
+\DPtypo{soient}{Soient} $U, V$~les coordonnées par rapport à~$\omega U, \omega V$ de la projection~$\delta'$
+de~$\delta$ sur le plan tangent; elles sont données par les
+équations
+\[
+U \cos\theta + V = B, \qquad
+U + V \cos\theta = A;
+\]
+d'où
+\[
+U \sin^2\theta = A - B \cos\theta, \qquad
+V \sin^2\theta = B - A \cos\theta;
+\]
+d'où encore
+\begin{multline*}
+\sin^{\Err{2}{4}}\theta \bigl[U^2 + V^2 + 2UV \cos\theta\bigr] \\
+  = (A - B \cos\theta)^2 + (B - A \cos\theta)^2
+  + 2 \cos\theta (A - B \cos\theta) (B - A \cos\theta)\DPtypo{.}{,}
+\end{multline*}
+ou
+\[
+\sin^2\theta \bigl[U^2 + V^2 + 2UV \cos\theta\bigr]
+  = A^2 - 2AB \cos\theta + B^2.
+\]
+Donc
+\[
+\omega \delta'{}^2 = U^2 + V^2 + 2UV \cos\theta
+  = \frac{1}{\sin^2\theta} \bigl[A^2 - 2AB \cos\theta + B^2\bigr],
+\]
+et
+\[
+1 = \overline{\omega\delta}^2
+  = \overline{\omega\delta'}^2 + \cos^2 i
+  = \cos^2 i + \frac{1}{\sin^2\theta} \bigl[A^2 - 2AB \cos\theta + B^2\bigr].
+\]
+Or,
+\[
+A^2 - 2AB \cos\theta + B^2
+  = \frac{1}{H^2}\, \Phi\left( \frac{\dd r}{\dd v}, -\frac{\dd r}{\dd u}\right),
+\]
+d'où la formule
+\[
+\sin^2 i = \frac{1}{H^2}\,
+  \Phi\left(\frac{\dd r}{\dd v}, -\frac{\dd r}{\dd u}\right).
+\]
+
+Ceci posé, les lignes de courbure de la surface focale
+sont définies par l'équation
+\[
+\begin{vmatrix}
+  dx & \lambda & d\lambda
+\end{vmatrix} = 0,
+\]
+ou
+\[
+\begin{vmatrix}
+  df + \lambda · dr + r · d\lambda & \lambda & d\lambda
+\end{vmatrix} = 0;
+\]
+qui se réduit à
+\[
+\begin{vmatrix}
+  df & \lambda & d\lambda
+\end{vmatrix} = 0.
+\]
+\DPtypo{multiplions}{Multiplions} par le déterminant
+$\begin{vmatrix}
+  \lambda & \dfrac{\dd f}{\dd u} & \dfrac{\dd f}{\dd v}
+\end{vmatrix}$
+qui n'est
+pas nul, la normale n'étant pas dans le plan tangent. L'équation
+devient
+\[
+\left\lvert
+\begin{array}{@{}lll@{}}
+\sum \lambda\,df & \sum \lambda^2 & \sum \lambda\, d\lambda \\
+\sum \mfrac{\dd f}{\dd u}\, df    & \sum \lambda \mfrac{\dd f}{\dd u} & \sum \mfrac{\dd f}{\dd u}\, d\lambda \\
+\sum \mfrac{\dd f}{\dd v}\, df    & \sum \lambda \mfrac{\dd f}{\dd v} & \sum \mfrac{\dd f}{\dd v}\, d\lambda
+\end{array}
+\right\rvert
+= 0,
+\]
+%% -----File: 295.png---Folio 287-------
+ou, en tenant compte de~\Eq{(6)}
+\[
+\begin{vmatrix}
+  -dr & 1 & 0  \\
+  \sum \mfrac{\dd f}{\dd u}\, df & -\mfrac{\dd r}{\dd u} & \sum \mfrac{\dd f}{\dd u}\, d\lambda \\
+  \sum \mfrac{\dd f}{\dd v}\, df & -\mfrac{\dd r}{\dd v} & \sum \mfrac{\dd f}{\dd v}\, d\lambda
+\end{vmatrix}
+= 0.
+\]
+Multiplions la \Ord{1}{ère} ligne par~$\dfrac{\dd r}{\dd u}$ et ajoutons à la \Ord{2}{e}, puis
+par~$\dfrac{\dd r}{\dd v}$ et ajoutons à la \Ord{3}{e}. Nous obtenons
+\[
+\begin{vmatrix}
+\sum \mfrac{\dd f}{\dd u}\, df - \mfrac{\dd r}{\dd u}\, dr & \sum \mfrac{\dd f}{\dd u}\, d\lambda \\
+\sum \mfrac{\dd f}{\dd v}\, df - \mfrac{\dd r}{\dd v}\, dr & \sum \mfrac{\dd f}{\dd v}\, d\lambda
+\end{vmatrix}
+=0.
+\]
+\DPtypo{les}{Les} éléments de la \Ord{1}{ère} colonne sont les demi-dérivées partielles
+par rapport à~$du\Add{,} dv$ de la forme quadratique
+\[
+\sum df^2 - dr^2 = \Phi_1 (du, dv),
+\]
+qui définit le couple des droites~$\omega I, \omega I'$. Voyons si les éléments
+de la \Ord{2}{e} colonne sont susceptibles d'une interprétation
+analogue. Si nous \DPtypo{differentions}{différentions}~\Eq{(6)} nous obtenons
+\[
+\sum \frac{\dd f}{\dd u}\, d\lambda
+  = -\sum \lambda\, d\left(\frac{\dd f}{\dd u}\right)
+                  - d\left(\frac{\dd r}{\dd u}\right);
+\]
+or,
+\[
+d\left(\frac{\dd r}{\dd u}\right)
+  = \Err{-}{}\frac{1}{2} · \frac{\dd(d^2r)}{\dd(du)},
+\]
+et
+\[
+\sum \lambda · d\left(\frac{\dd f}{\dd u}\right)
+  = \frac{1}{2}\, \frac{\dd(\sum \lambda · d^2f)}{\dd (du)}.
+\]
+Posons
+\[
+\scrA = \sum \lambda\, d^2 f, \qquad
+\scrB = \scrA + d^2r
+\]
+les différentielles secondes étant prises par rapport
+à~$u$~et~$v$;  et l'équation s'écrit,
+\[
+\begin{vmatrix}
+\mfrac{\dd \Phi_1}{\dd\, du} & \mfrac{\dd \scrB}{\dd\, du} \\
+\mfrac{\dd \Phi_1}{\dd\, dv} & \mfrac{\dd \scrB}{\dd\, dv}
+\end{vmatrix}
+= 0:
+\]
+les directions principales de~$(F)$ sont conjuguées harmoniques
+par rapport aux \Card{2} couples $\Phi_1 = 0$~et~$\scrB = 0$. Calculons~$\scrA$. Pour
+%% -----File: 296.png---Folio 288-------
+cela, éliminons $\lambda\Add{,} \mu\Add{,} \nu$ entre les équations \Eq{(6)}\Add{,}~\Eq{(7)} et
+\[
+\sum \lambda\, d^2 f - \scrA = 0;
+\]
+nous obtenons
+\[
+\begin{vmatrix}
+\mfrac{\dd f}{\dd u} & \mfrac{\dd f}{\dd v} & 1 & d^2 f \\
+\dots & \dots & \dots & \dots \\
+\dots & \dots & \dots & \dots \\
+\mfrac{\dd r}{\dd u} & \mfrac{\dd r}{\dd v} & -\cos i & -\scrA
+\end{vmatrix}
+= 0,
+\]
+ce qui donne en développant
+\[
+\scrA H + H \cos i \Psi(du, dv) + H \chi(du, dv) = 0;
+\]
+et:
+\[
+\scrB = d^2 r - \cos i \Psi(du, dv) - \chi(du, dv)
+  = -\Psi_{1}(du, dv) - \cos i \Psi(du, dv).
+\]
+
+Les lignes de courbure de la \Ord{2}{e} nappe sont tangentes aux
+directions conjuguées par rapport à $\Phi_{1} = 0$ et
+\[
+\scrB_{1} = -\Psi_{1}(du, dv) + \Psi(du, dv) = 0;
+\]
+pour que les lignes de courbure se correspondent sur les \Card{2} nappes,
+il faut et il suffit que le couple des directions
+principales soit conjugué par rapport aux \Card{3} couples
+\[
+\Phi_{1} = 0,\qquad
+\Psi_{1} - \cos i \Psi = 0,\qquad
+\Psi_{1} + \cos i \Psi = 0,
+\]
+ou par rapport aux couples
+\[
+\Phi_{1} = 0,\qquad
+\Psi_1 \Err{+ \cos i \Psi}{} = 0,\qquad
+\Psi = 0.
+\]
+\DPchg{Etant}{Étant} conjuguées par rapport à $\Psi = 0$, elles correspondent à
+des lignes conjuguées sur~$(S)$, d'où le \emph{Théorème de Dupin}: si
+les lignes de courbure se correspondent sur les \Card{2} nappes focales,
+les développables des normales correspondantes coupent
+la surface~$(S)$ suivant le même réseau conjugué; et \emph{réciproquement}.
+Donc \emph{la condition nécessaire et suffisante pour que
+les développables d'une congruence de normales se réfléchissent
+sur une surface suivant des développables est qu'elles
+%% -----File: 297.png---Folio 289-------
+déterminent sur la surface un réseau conjugué}.
+
+\Section{Congruence des droites \texorpdfstring{$D$}{D}.}
+{4.}{} Cherchons les développables de la congruence des
+droites~$D$; elles sont définies par l'équation
+\[
+\begin{vmatrix}
+dx & dy & dz \\
+ l &  m &  n \\
+dl & dm & dn
+\end{vmatrix}
+= 0;
+\]
+or
+\[
+x = f + r \lambda,\qquad
+y = g + r \mu,\qquad
+z = h + r \nu,
+\]
+et l'équation devient
+\[
+\begin{vmatrix}
+df + r\, d\lambda + \lambda\, dr & l & dl
+\end{vmatrix} = 0.
+\]
+Multiplions le \Ord{1}{er} membre par le déterminant non nul
+\[
+\begin{vmatrix}
+l & \mfrac{\dd f}{\dd u} & \mfrac{\dd f}{\dd v}
+\end{vmatrix}
+\]
+nous avons:
+\[
+\begin{vmatrix}
+r \sum l\, d \lambda + dr \sum \lambda l & 1 & 0 \\
+\sum \mfrac{\dd f}{\dd u}\, df + r \sum \mfrac{\dd f}{\dd u}\, d\lambda
+  + dr \sum \lambda \mfrac{\dd f}{\dd u} & 0 & \sum \mfrac{\dd f}{\dd u}\, dl \\
+\sum \mfrac{\dd f}{\dd v}\, df + r \sum \mfrac{\dd f}{\dd v}\, d\lambda
+  + dr \sum \lambda \mfrac{\dd f}{\dd v} & 0 & \sum \mfrac{\dd f}{\dd v}\, dl
+\end{vmatrix}
+= 0,
+\]
+ou
+\[
+\begin{vmatrix}
+\sum \mfrac{\dd f}{\dd u}\, df + r \sum \mfrac{\dd f}{\dd u}\, d\lambda
+  + dr \sum \lambda \mfrac{\dd f}{\dd u} & \sum \mfrac{\dd f}{\dd u}\, dl \\
+\sum \mfrac{\dd f}{\dd v}\, df + r \sum \mfrac{\dd f}{\dd v}\, d\lambda
+  + dr \sum \lambda \mfrac{\dd f}{\dd v} & \sum \mfrac{\dd f}{\dd v}\, dl
+\end{vmatrix}
+= 0.
+\]
+
+Les éléments de la dernière colonne sont les demi-dérivées
+partielles par rapport à~$du, dv$ de la forme $\Psi(du, dv) = \Err{}{-}\sum df · dl$.
+Quant aux éléments de la \Ord{1}{ère} ligne, remarquons que
+\[
+\sum \frac{\dd f}{\dd u}\, d\lambda = \frac{1}{2} · \frac{\dd \scrB}{\dd u},
+\qquad
+\sum \frac{\dd f}{\dd v}\, d\lambda = \frac{1}{2} · \frac{\dd \scrB}{\dd v},
+\]
+où
+\[
+\scrB = -\Psi_{1} - \cos i · \Psi.
+\]
+Enfin, si nous remarquons que les points $M, M'$ sont définis par
+les relations:
+\[
+\sum \lambda\, \frac{\dd f}{\dd u} + \frac{\dd r}{\dd u} = 0,\qquad
+\sum \lambda\, \frac{\dd f}{\dd v} + \frac{\dd r}{\dd v} = 0,
+\]
+%% -----File: 298.png---Folio 290-------
+nous avons:
+\[
+\sum \frac{\dd f}{\dd u}\, df
+  + dr \sum \lambda\, \frac{\dd f}{\dd u}
+  = \sum \frac{\dd f}{\dd u}\, df - \frac{\dd r}{\dd u}\, dr
+  = \frac{1}{2}\, \frac{d\Phi_{1}}{\dd\, du},\quad
+\text{et l'analogue};
+\]
+de sorte que les éléments de la \Ord{1}{ère} colonne sont les demi-dérivées
+partielles par rapport à~$du, dv$ de la forme $\Phi_{1} - r(\Psi_{1} + \Psi \cos i)$.
+
+\emph{Les développables de la congruence correspondent
+sur la surface~$(S)$ aux directions conjuguées par rapport
+à}
+\begin{alignat*}{2}
+\Psi &= 0,\qquad && \Phi_{1} - r [\Psi_{1} + \Psi \cos i] = 0, \\
+\intertext{ou par rapport à}
+\Psi &= 0, && \Phi_{1} - r \Psi_{1} = 0;
+\end{alignat*}
+le résultat ne change pas si on change $i$~en~$\pi - i$, et les
+développables de la congruence des droites~$D$ correspondent
+sur la surface~$(S)$ à un réseau conjugué.
+
+Considérons les plans focaux; un plan focal est parallèle
+à la direction $l, m, n$, et à la direction $dl, dm, dn$. Mais
+\[
+l^2 + m^2 + n^2 = 1,
+\]
+d'où
+\[
+l\, dl + m\, dm + n\, dn = 0;
+\]
+$dl, dm, dn$ correspondent à la direction du plan focal parallèle
+au plan tangent à la surface. Or\Add{,} les \Card{2} directions correspondant
+aux \Card{2} plans focaux, donc aux \Card{2} développables, étant
+conjuguées, on a
+\[
+\sum dl · \delta f = 0;
+\]
+le plan focal est perpendiculaire à la direction $\delta f, \delta g, \delta h$ qui
+correspond à l'autre plan focal. \emph{Chaque plan focal est perpendiculaire
+à la direction de la surface~$(S)$ correspondant à
+l'une des développables.}
+%% -----File: 299.png---Folio 291-------
+
+\Section{Congruence des droites \texorpdfstring{$\Delta$}{Delta}.}
+{5.}{} La droite~$\Delta$ est l'intersection des plans tangents à la sphère en~$M$ et à la surface~$(S)$ en~$\omega$
+\[
+\sum \lambda\,(X - f) - r = 0,  \qquad \sum l\,(X - f) = 0.
+\]
+Cherchons les développables. Exprimons que la droite précédente
+rencontre la droite infiniment voisine. Cela donne
+\[
+\sum d\lambda\, (X - f) - \sum \lambda\, df - dr = 0, \qquad
+\sum dl\, (X - f) - \sum l\, df = 0;
+\]
+conditions qui se simplifient en remarquant que $\sum l\, df = 0$, et
+$\sum \lambda\, df + dr = 0$. Il reste:
+\[
+\Tag{(1)}
+\sum d\lambda (X - f) = 0, \qquad
+\sum dl (X - f) = 0.
+\]
+Exprimons que les équations obtenues sont compatibles, nous
+avons l'équation qui définit les développables
+\[
+\Tag{(2)}
+\begin{vmatrix}
+  l & d\lambda & dl
+\end{vmatrix} = 0.
+\]
+Multiplions encore par le déterminant non nul
+\[
+\begin{vmatrix}
+  l & \dfrac{\dd f}{\dd u} & \dfrac{\dd f}{\dd v}
+\end{vmatrix},
+\]
+nous obtenons
+\[
+\begin{vmatrix}
+1 & \sum l\, d\lambda   &   0 \\
+0 & \sum d\lambda\, \mfrac{\dd f}{\dd u} & \sum dl\, \mfrac{\dd f}{\dd u} \\
+0 & \sum d\lambda\, \mfrac{\dd f}{\dd v} & \sum dl\, \mfrac{\dd f}{\dd v}
+\end{vmatrix}
+= 0,
+\]
+ou
+\[
+\begin{vmatrix}
+\sum \mfrac{\dd f}{\dd u}\, d\lambda & \sum \mfrac{\dd f}{\dd u}\, dl \\
+\sum \mfrac{\dd f}{\dd v}\, d\lambda & \sum \mfrac{\dd f}{\dd v}\, dl
+\end{vmatrix}
+= 0;
+\]
+les éléments de la \Ord{1}{ère} colonne sont les demi-dérivées partielles
+par rapport à~$du, dv$ de la forme~$\scrB = -\Psi_1 - \Psi \cos i$.
+Ceux de la \Ord{2}{e} colonne sont les demi-dérivées partielles de~$\Psi$.
+\emph{Les développables de la congruence des droites~$\Delta$ correspondent
+sur la surface~$(S)$ au réseau conjugué par rapport aux
+couples}
+\[
+\Psi = 0, \qquad \Psi_1 = 0.
+\]
+%% -----File: 300.png---Folio 292-------
+
+Quant aux points focaux, ils sont définis par les équations
+de~$\Delta$ et les équations~\Eq{(1)}, compatibles en vertu de la
+relation~\Eq{(2)}. On en déduit que les directions joignant~$\omega$ aux
+points focaux sont définies par les relations
+\[
+\sum  l · \delta f = 0,\qquad
+\sum dl · \delta f = 0,\qquad
+\sum d\lambda · \delta f = 0;
+\]
+la \Ord{1}{ère} exprime que ces droites sont dans le plan tangent à~$(S)$,
+la \Ord{2}{e} que ce sont les tangentes conjuguées de~$(S)$ qui
+correspondent aux développables.
+
+\Illustration{300a}
+Supposons que les \Card{2} congruences précédentes se correspondent
+par développables. Les \Card{2} réseaux conjugués déterminés
+sur la surface~$(S)$ sont confondus; il faut alors que les \Card{3} couples
+$\Psi = 0$, $\Psi_{1} = 0$, $\Phi_{1} - r \Psi_{1} = 0$, ou $\Psi = 0$, $\Psi_{1} = 0$, $\Phi_{1} = 0$ appartiennent
+à une même involution, et alors les lignes de
+courbure se correspondent sur les \Card{2} nappes de la surface~$(F)$.
+Nous avons alors sur la surface~$(S)$ un réseau conjugué qui
+correspond aux développables des \Card{4} congruences
+$\omega M, \omega M', D, \Delta$. Les points
+focaux~$F\Add{,} F'$ de~$\Delta$ sont sur les tangentes
+aux \Card{2} courbes conjuguées qui passent
+par~$\omega$, les droites $MF, MF'$ sont les
+tangentes aux lignes de courbure en~$M$
+de la surface~$(F)$, la droite~$D$ est perpendiculaire au plan~$F \omega F'$,
+et ses plans focaux sont perpendiculaires à $\omega F$~et~$\omega F'$.
+Les développables de la congruence des droites~$D$ coupent les
+\Card{2} nappes de l'enveloppe~$(F)$ suivant leurs lignes de courbure.
+%% -----File: 301.png---Folio 293-------
+
+\Section{Le système triple de \DPtypo{Ribeaucour}{Ribaucour}.}
+{6.}{} Plaçons-nous dans ce dernier cas; soit $(\gamma)$ une des
+courbes conjuguées de la surface~$(S)$; quand $\omega$~décrit~$(\gamma)$, le
+point~$M$ décrit une ligne de courbure~$(K)$ de la surface~$(F)$
+qui sera tangente à~$MF$, et la droite~$\Delta$ enveloppe une courbe~$(c)$
+lieu de~$F'$. Considérons la sphère~$(\sigma)$ de centre~$F'$ et passant
+par~$M$; cette sphère a pour enveloppe une surface canal~$(E)$;
+la sphère~$(\sigma)$ ayant son rayon~$MF'$ perpendiculaire à~$MF$
+est constamment tangente à la courbe~$(K)$, donc le point~$M$ est
+un point du cercle caractéristique~$(H)$; le plan de ce cercle
+est perpendiculaire à la droite~$\Delta$ tangente à~$(c)$, son centre~$H$
+sera le pied de la perpendiculaire abaissée de~$M$ sur~$\Delta$; ce
+cercle sera donc orthogonal à la sphère~$(\Sigma)$ au point~$M$ et au
+point~$M'$ symétrique par rapport au plan~$F \omega F'$; et la surface~$(E)$
+est engendrée par le cercle orthogonal à la sphère~$(\Sigma)$
+aux points $M, M'$; ce cercle tangent en~$M$ à~$\omega M$ reste orthogonal
+à la ligne de courbure~$(K)$, or\Add{,} il est ligne de courbure sur la surface~$(E)$,
+donc $(K)$~est aussi ligne de courbure sur la surface~$(E)$.
+Si nous faisons varier~$(K)$ nous obtenons une famille
+de surfaces~$(E)$ qui seront toutes orthogonales à $(F_{1}),(F_{2})$, et
+qui les couperont suivant leurs lignes de courbure. Si maintenant
+nous prenons sur~$(F)$ le \Ord{2}{e} système de lignes de courbure,
+nous devrons considérer les sphères de centres~$F$ et passant
+par~$M$; le cercle caractéristique sera encore le cercle~$(H)$;
+de plus, $FM$~et~$F'M$ étant perpendiculaires, les sphères $(\sigma)\Add{,} (\sigma')$ %[** TN: Erratum corrected]
+sont orthogonales, donc aussi leurs enveloppes $(E)\Add{,} (E')$. Nous
+avons donc \Card{2} familles de surfaces canaux qui se coupent
+orthogonalement suivant des lignes de courbure, les cercles~$(H)$;
+%% -----File: 302.png---Folio 294-------
+donc elles appartiennent à un système triple orthogonal;
+autrement dit les cercles~$(H)$ sont orthogonaux à une famille
+de surfaces, à laquelle appartiennent $(F), (F')$; et ils établissent
+une correspondance entre les points~$M, M'$, donc entre
+les lignes de courbure de ces surfaces. Donc \emph{lorsque les cercles~$(H)$
+d'une congruence sont orthogonaux aux \Card{2} nappes focales~$(F_1)\Add{,} (F_2)$
+d'une congruence de sphères, et s'ils établissent
+une correspondance entre les lignes de courbure de ces \Card{2} nappes,
+ils sont orthogonaux à une infinité de surfaces sur lesquelles
+les lignes de courbure se correspondent; ces surfaces
+appartiennent à un système triple orthogonal dont les deux
+autres familles sont constituées par les surfaces canaux engendrées
+par ceux des cercles~$(H)$ qui s'appuient sur une des
+lignes de courbure de~$(F_1)$~ou~$(F_2)$}. De telles congruences de
+cercles s'appellent \emph{systèmes cycliques}.
+
+\Section{Congruences de cercles et systèmes cycliques.}
+{7.}{} Considérons une famille de $\infty^{2}$~cercles, et cherchons
+s'il existe des surfaces normales à tous ces cercles.
+Soit~$K$ l'un d'eux, $C(x_0, y_0, z_0)$~son centre, $\rho$~son rayon,
+$x_0, y_0, z_0, \rho$~étant fonctions de \Card{2} paramètres~$u, v$.
+Pour définir le plan
+de ce cercle nous définirons \Card{2} directions
+rectangulaires $CA(a, b, c)$~et~$CB(a', b', c')$ passant par le centre du
+cercle, et nous fixerons la position
+%% -----File: 303.png---Folio 295-------
+d'un point~$M$ sur le cercle par l'angle~$(CA, CM) = t$. Les coordonnées
+de~$M$ par rapport au système~$CAB$ sont $\rho \cos t$,~$\rho \sin t$, et
+ses coordonnées~$x, y, z$ sont
+\begin{alignat*}{4}%[** TN: Rebroken]
+x &= x_0 &&+ \rho (a \cos t &&+ a' \sin t) &&= x_0 + \rho \alpha', \\
+y &= y_0 &&+ \rho (b \cos t &&+ b' \sin t) &&= y_0 + \rho \beta', \\
+z &= z_0 &&+ \rho (c \cos t &&+ c' \sin t) &&= z_0 + \rho \gamma'.
+\end{alignat*}
+
+\Illustration[1.5in]{302a}
+\noindent Cherchons à déterminer~$t$ de façon que la surface lieu du
+point correspondant admette pour normale la tangente au cercle
+au point~$M$, dont nous désignerons par $\alpha, \beta, \gamma$ les cosinus directeurs.
+Nous obtenons la condition
+\[
+\sum \alpha\, dx = 0,
+\]
+équation aux différentielles totales des surfaces cherchées.
+Développons cette équation; $\alpha, \beta, \gamma$ sont les projections du segment
+directeur de la direction~$CM'$ correspondant à~$t + \frac{\pi}{2}$
+\[
+\alpha = -a \sin t + a' \cos t, \qquad
+\beta  = -b \sin t + b' \cos t, \qquad
+\gamma = -c \sin t + c' \cos t\DPtypo{;}{.}
+\]
+Maintenant
+\begin{alignat*}{5}%[** TN: Rebroken, filled in last two equations]
+dx &= dx_0 &&+ \alpha' · d\rho &&+ \rho \alpha · dt &&+ \rho (\cos t · da &&+ \sin t · da'), \\
+dy &= dy_0 &&+ \beta'  · d\rho &&+ \rho \beta  · dt &&+ \rho (\cos t · db &&+ \sin t · db'), \\
+dz &= dz_0 &&+ \gamma' · d\rho &&+ \rho \gamma · dt &&+ \rho (\cos t · dc &&+ \sin t · dc');
+\end{alignat*}
+d'où
+\begin{gather*}
+\sum \alpha\, dx
+  = \sum \alpha · dx_0 + \rho · dt
+  + \rho \left[\cos t\sum \alpha\, da + \sin t\sum \alpha\, da'\right] = 0, \\
+%
+\sin t \sum a\, dx + \cos t\sum a'\, dx_0 + \rho\,dt
+  + \rho \left[\cos^2 t \sum a'\, da - \sin^2 t\sum a\, da'\right] = 0\Add{.}
+\end{gather*}
+Mais
+\[
+\sum aa'= 0,
+\]
+d'où en différentiant
+\[
+\sum a\, da' + \sum a'\, da = 0;
+\]
+et l'équation devient
+\begin{gather*}
+-\sin t\sum a\, dx_0 + \cos t\sum a'\, dx_0
+  + \rho\, dt - \rho\sum a\, da' = 0, \\
+\Tag{(3)}
+dt = \sum a\, da' + \frac{1}{\rho}\sum a\, dx_0 \sin t
+  - \frac{1}{\rho} \sum a'\, dx_0 \cos t.
+\end{gather*}
+%% -----File: 304.png---Folio 296-------
+Posons
+\begin{gather*}
+\Tag{(4)}
+\tg \frac{t}{2} = w, \\
+t = 2 \arctg w,
+\end{gather*}
+Nous obtenons
+\[
+\frac{2\, dw}{1 + w^2}
+  = \sum a\, da' + \frac{1}{\rho} \sum a\, dx_0 · \dfrac{2w}{1 + w^2}
+  - \frac{1}{\rho} \sum a'\, dx_0 \frac{1 - w^2}{1 + w^2},
+\]
+ou
+\[
+\Tag{(5)}
+2\, dw = (1 + w^2) \sum a\, da' + \frac{2w}{\rho} \sum a\, dx_0
+  + \frac{w^2 - 1}{\rho} \sum a'\, dx_0,
+\]
+équation qui jouit de propriétés analogues à celles de l'équation
+de Riccati. Elle peut se mettre sous la forme
+\[
+dw = A\, du + A'\, dv + w (B\, du + B'\, dv) + w^2 (C\, du + C'\, dv),
+\]
+qui se décompose en \Card{2} équations aux dérivées partielles:
+\[
+\Tag{(6)}
+\frac{\dd w}{\dd u} = A + B w + C w^2, \qquad
+\frac{\dd w}{\dd v} = A' + B' w + C' w^2;
+\]
+d'où la condition nécessaire et suffisante d'intégrabilité
+\begin{multline*}%[** TN: Added break]
+\frac{\dd A}{\dd v} + w \frac{\dd B}{\dd v} + w^2 \frac{\dd C}{\dd v}
+  + (B + 2Cw) (A' + B' w + C' w^2) \\
+  - \left[ \frac{\dd A'}{\dd u}
+    + w\frac{\dd B'}{\dd u} + w^2\frac{\dd \Err{c}{C}'}{\dd u}\right]
+  - (B' + 2C' w) (A + Bw + Cw^2) = 0.
+\end{multline*}
+
+Toute intégrale du système~\Eq{(6)} satisfait à cette condition,
+qui est de la forme
+\[
+L + Mw + Nw^2 = 0.
+\]
+Si cette condition n'est pas identiquement satisfaite, il ne
+peut y avoir d'autres solutions que celles de l'équation
+précédente, qui en admet \Card{2}. Si l'on veut qu'il y en ait une
+infinité, cette condition doit être identiquement satisfaite,
+et comme elle est du \Ord{2}{e} degré, il suffit qu'elle soit satisfaite
+par \Card{3} fonctions. Les conditions pour qu'il en soit
+ainsi sont
+\begin{alignat*}{5}%[** TN: Rebroken]
+L &= \frac{\dd A}{\dd v} &&- \frac{\dd A'}{\dd u} &&+ &&BA' - AB' &&= 0, \\
+M &= \frac{\dd B}{\dd v} &&- \frac{\dd B'}{\dd u} &&+ 2(&&CA' - AC') &&= 0\DPtypo{.}{,} \\
+N &= \frac{\dd C}{\dd v} &&- \frac{\dd C'}{\dd u} &&+ &&CB'- BC' &&= 0\Add{.}
+\end{alignat*}
+%% -----File: 305.png---Folio 297-------
+\emph{Si les cercles sont normaux à \Card{3} surfaces, ils sont normaux à
+une infinité de surfaces.}
+
+Il est facile de construire des cercles normaux à \Card{2} surfaces,
+car il existe des sphères tangentes aux \Card{2} surfaces, et
+les cercles orthogonaux à ces sphères aux points de contact
+sont normaux aux \Card{2} surfaces. Si les lignes de courbure se
+correspondent sur les \Card{2} surfaces, on a un système cyclique,
+système de cercles normaux à $\infty^{1}$~surfaces. Réciproquement,
+supposons \Card{2} surfaces normales aux cercles, les conditions
+d'intégrabilité se réduiront à une seule; d'autre part, si on
+a une enveloppe de sphères, pour exprimer que les lignes de
+courbure se correspondent sur les \Card{2} nappes, on obtient aussi
+une seule condition. Cherchons à montrer que ces conditions
+sont identiques.
+
+Supposons donc qu'il existe une surface normale à tous
+ces cercles, supposons qu'elle corresponde à $t = 0$, ou $w = 0$,
+l'équation~\Eq{(5)} admet la solution $w = 0$, d'où la condition
+\[
+\sum a\, da' - \frac{1}{\rho} \sum a'\, dx_0 = 0;
+\]
+et l'équation devient
+\[
+\Tag{(7)}
+dw = w^2 \sum a\, da' + \frac{w}{\rho} \sum a\, dx_0.
+\]
+Soit $M_0(x,y,z)$ le point correspondant à $t = 0$
+\begin{alignat*}{3}%[** TN: Filled in last two columns]
+x    &= x_0 + \rho a, & y   &= y_0 + \rho b, & z    &= z_0 + \rho c, \\
+x_0  &= x   - \rho a, & y_0 &= y   - \rho b, & z_0  &= z   - \rho c, \\
+dx_0 &= dx - \rho\, da - a\, d\rho,\quad &
+dy_0 &= dy - \rho\, db - b\, d\rho,\quad &
+dz_0 &= dz - \rho\, dc - c\, d\rho;
+\end{alignat*}
+d'où
+\[
+\sum a\, dx_0 = \sum a\, dx - d\rho.
+\]
+Si maintenant nous considérons la normale~$(l,m,n)$ en~$M_0$ à~$(\Sigma)$,
+c'est la tangente au cercle, et \Eq{(7)}~devient
+%% -----File: 306.png---Folio 298-------
+\[
+dw = w^2 \sum a\, dl + \frac{w}{\rho} \left(\sum a\, dx - d\rho\right),
+\]
+ou
+\[
+\frac{dw}{w} + \frac{d\rho}{\rho} = w \sum a\, dl + \frac{1}{\rho} \sum a\, dx.
+\]
+
+\Illustration[2.25in]{306a}
+Nous introduisons ainsi la quantité
+\[
+\Tag{(8)}
+\rho w = r,
+\]
+et nous obtenons
+\begin{align*}
+\frac{dr}{r} &= \frac{r}{\rho} \sum a\, dl + \frac{1}{\rho} \sum a\, dx, \\
+dr &= \frac{r^2}{\rho} \sum a\, dl + \frac{r}{\rho} \sum a\, dx.
+\end{align*}
+Or
+\[
+r = \rho \tg \frac{t}{2},
+\]
+ce qui montre que $r$~est le rayon de la sphère~$(\Sigma)$ tangente
+aux surfaces lieux de $M$~et~$M_0$; son centre est le point~$\omega$ intersection
+des tangentes au cercle en $M$~et~$M_0$.
+
+Supposons maintenant qu'il existe une \Ord{2}{e} surface normale
+aux cercles. Posons
+\begin{gather*}
+\Tag{(9)}
+\frac{1}{r} = S, \\
+dr = -r^2 · dS;
+\end{gather*}
+et l'équation devient
+\begin{gather*}
+-r^2 · dS = \frac{r^2}{\rho} \sum a\, dl + \frac{r}{\rho} \sum a\, dx, \\
+\Tag{(10)}
+dS + \frac{S}{\rho} \sum a\, dx + \frac{1}{\rho} \sum a\, dl = 0.
+\end{gather*}
+Soit $S_1$\DPtypo{,}{} la solution connue
+\[
+dS_1 + \frac{S_1}{\rho} \sum a\, dx + \frac{1}{\rho} \sum a\, dl = 0,
+\]
+d'où en retranchant
+\begin{gather*}
+d(S - S_1) + \frac{S - S_1}{\rho} \sum a\, dx = 0, \\
+\Tag{(11)}
+d\log (S - S_1) = -\frac{1}{\rho} \sum a\, dx.
+\end{gather*}
+Pour que l'équation ait d'autres intégrales, il faut que $\dfrac{1}{\rho} \sum a\, dx$
+%% -----File: 307.png---Folio 299-------
+soit différentielle exacte. Or\Add{,} nous avons
+\[
+\frac{\dd S_1}{\dd u} + \frac{S_1}{\rho} \sum a \frac{\dd x}{\dd u}
+  + \frac{1}{\rho} \sum a \frac{\dd l}{\dd u} = 0, \quad
+\frac{\dd S_1}{\dd v} + \frac{S_1}{\rho} \sum a \frac{\dd x}{\dd v}
+  + \frac{1}{\rho} \sum a \frac{\dd l}{\dd v} = 0.
+\]
+Supposons que les lignes coordonnées soient lignes de courbure.
+Les formules d'Olinde Rodrigues donnent
+\begin{alignat*}{3}%[** TN: Filled in missing columns in two systems below]
+\frac{\dd l}{\dd u} &= -\frac{1}{R}\, \frac{\dd x}{\dd u}, \qquad &
+\frac{\dd m}{\dd u} &= -\frac{1}{R}\, \frac{\dd y}{\dd u}, \qquad &
+\frac{\dd n}{\dd u} &= -\frac{1}{R}\, \frac{\dd z}{\dd u}, \\
+%
+\frac{\dd l}{\dd v} &= -\frac{1}{R'}\, \frac{\dd x}{\dd v}, \qquad &
+\frac{\dd m}{\dd v} &= -\frac{1}{R'}\, \frac{\dd y}{\dd v}, \qquad &
+\frac{\dd n}{\dd v} &= -\frac{1}{R'}\, \frac{\dd z}{\dd v}.
+\end{alignat*}
+Posons
+\[
+- \frac{1}{R} = T,  \qquad  - \frac{1}{R'} = T',
+\]
+\begin{alignat*}{3}
+\frac{\dd l}{\dd u} &= T \frac{\dd x}{\dd u}, \qquad &
+\frac{\dd m}{\dd u} &= T \frac{\dd y}{\dd u}, \qquad &
+\frac{\dd n}{\dd u} &= T \frac{\dd z}{\dd u}, \\
+%
+\frac{\dd l}{\dd v} &= T' \frac{\dd x}{\dd v}, \qquad &
+\frac{\dd m}{\dd v} &= T' \frac{\dd y}{\dd v}, \qquad &
+\frac{\dd n}{\dd v} &= T' \frac{\dd z}{\dd v};
+\end{alignat*}
+Alors
+\[
+\sum a \frac{\dd l}{\dd u} = T \sum a \frac{\dd x}{\dd u}, \qquad
+\sum a \frac{\dd l}{\dd v} = T'\sum a \frac{\dd x}{\dd v},
+\]
+et les conditions pour que $S_1$\DPtypo{,}{} soit intégrale deviennent
+\[
+\frac{\dd S_1}{\dd u} + (S_1 + T)  \frac{\displaystyle\sum a \dfrac{\dd x}{\dd u}}{\rho} = 0,
+\qquad
+\frac{\dd S_1}{\dd v} + (S_1 + T') \frac{\displaystyle\sum a \dfrac{\dd x}{\dd v}}{\rho} = 0;
+\]
+d'où
+\[
+-\frac{1}{\rho} \sum a\, dx
+  = \frac{1}{S_1 + T}\,  \frac{\dd S_1}{\dd u}\, du
+  + \frac{1}{S_1 + T'}\, \frac{\dd S_1}{\dd v}\, dv\Add{.}
+\]
+Exprimons que le \Ord{2}{e} membre est une différentielle exacte,
+nous aurons l'équation aux dérivées partielles des systèmes
+cycliques
+\[
+\Tag{(12)}
+\Omega
+  = \frac{\dd}{\dd v} \left(\frac{1}{S_1 + T}\,  \frac{\dd S_1}{\dd u}\right)
+  - \frac{\dd}{\dd u} \left(\frac{1}{S_1 + T'}\, \frac{\dd S_1}{\dd v}\right)
+  = 0.
+\]
+Montrons que \emph{cette condition exprime que les lignes de courbure
+se correspondent sur les surfaces $M_0, M_1$}. D'après le
+Théorème de Dupin, pour qu'il en soit ainsi il faut et il
+suffit que ces lignes de courbure correspondent à un réseau
+conjugué sur la surface lieu de~$\omega$. Soient $X,Y,Z$ les coordonnées
+de~$\omega$:
+\[
+X = x + \frac{1}{S} l, \qquad
+Y = y + \frac{1}{S} m, \qquad
+Z = z + \frac{1}{S} n;
+\]
+pour que sur cette surface les courbes $u = \cte$, $v = \cte$ forment
+%% -----File: 308.png---Folio 300-------
+un réseau conjugué, il faut que
+\[
+\Tag{(13)}
+\begin{vmatrix}
+  \mfrac{\dd^2 X}{\dd u\, \dd v} &
+  \mfrac{\dd X}{\dd u} &
+  \mfrac{\dd X}{\dd v}
+\end{vmatrix} = 0.
+\]
+Mais
+\begin{align*}
+\frac{\dd X}{\dd u}
+  &= \frac{\dd x}{\dd u} + \frac{T}{S}\, \frac{\dd x}{\dd u}
+  + l\frac{\dd \left(\dfrac{1}{s}\right)}{\dd u}
+  = \left(1 + \frac{T}{S}\right) \frac{\dd x}{\dd u}
+  + l \frac{\dd \left(\dfrac{1}{s}\right)}{\dd u},
+  &&\text{\dots, \qquad \dots,} \\
+\frac{\dd X}{\dd v}
+  &= \left(1 + \frac{T'}{S}\right) \frac{\dd x}{\dd v}
+  + l\frac{\dd \left(\dfrac{1}{s}\right)}{\dd v},
+  &&\text{\dots, \qquad \dots;}
+\end{align*}
+relations qu'on peut encore écrire
+\begin{alignat*}{3}
+\frac{\dd X}{\dd u} &= \frac{S + T}{S^2}
+  &&\left[S \frac{\dd x}{\dd u} - \frac{1}{S + T}\, \frac{\dd s}{\dd u} l\right],
+  &&\text{\dots, \qquad \dots,} \\
+\frac{\dd X}{\dd v} &= \frac{S + T'}{S^2}
+  &&\left[S \frac{\dd x}{\dd v} - \frac{1}{S + T'}\, \frac{\dd s}{\dd v} l\right],
+  \qquad&&\text{\dots, \qquad \dots,}
+\end{alignat*}
+dans le déterminant~\Eq{(13)} nous pouvons remplacer $\dfrac{\dd^2 X}{\dd u\, \dd v}$ par
+\[%[** TN: Not displayed in original]
+\frac{\dd}{\dd v} \left(M \frac{\dd X}{\dd u}\right)
+  - \frac{\dd}{\dd u} \left(N \frac{\dd X}{\dd v}\right)
+\]
+à condition que $M - N \neq 0$; nous prendrons
+$M = \dfrac{S^2}{S + T}$ et $N = \dfrac{S^2}{S + T'}$; nous avons alors à vérifier la relation
+\[
+\begin{vmatrix}
+\mfrac{\dd S}{\dd v} · \mfrac{\dd x}{\dd u}
+  - \mfrac{\dd S}{\dd u}\, \mfrac{\dd x}{\dd v}
+  - \mfrac{T'}{S + T}\, \mfrac{\dd S}{\dd u} · \mfrac{\dd x}{\dd v}
+  + \mfrac{T}{S + T'}\, \mfrac{\dd S}{\dd v} · \mfrac{\dd X}{\dd u}
+  - \Omega l \\
+S \mfrac{\dd x}{\dd u} - \mfrac{1}{S + T}\,  \mfrac{\dd S}{\dd u}\, l \\
+S \mfrac{\dd X}{\dd v} - \mfrac{1}{S + T'}\, \mfrac{\dd S}{\dd v}\, l
+\end{vmatrix} = 0\DPtypo{;}{.}
+\]
+Multiplions la \Ord{2}{e} ligne par $-\dfrac{S + T + T'}{S(S + T')}\, \dfrac{\dd S}{\dd v}$, la \Ord{3}{e} par
+$\dfrac{S + T + T'}{S(S + T)}\, \dfrac{\dd S}{\dd u}$ et ajoutons à la \DPtypo{1ère}{\Ord{1}{ère}}, nous obtenons
+\[
+\Omega S^2 \begin{vmatrix}
+  1 & \mfrac{\dd x}{\dd u} & \mfrac{\dd x}{\dd v}
+\end{vmatrix} = 0,
+\]
+or\Add{,} le déterminant n'est pas nul, $S$~non plus, donc $\Omega = 0$ et
+réciproquement. Les conditions sont identiques.
+
+\Section{Transformation de contact de \DPtypo{Ribeaucour}{Ribaucour}.}
+{}{Remarque.} Considérons une sphère fixe de centre~$\omega$, et
+les cercles~$(K)$ orthogonaux à cette sphère; considérons une
+surface~$(S)$, un de ses points~$M$, et l'élément de contact en ce
+point; il y a un cercle~$(K)$ et un seul passant par~$M$ et normal
+%% -----File: 309.png---Folio 301-------
+en ce point à la surface~$(S)$. Donc à la surface~$(S)$ correspond
+une congruence de cercles qui lui sont orthogonaux; de
+plus ces cercles étant orthogonaux à la sphère~$(\omega)$ en \Card{2} points
+sont orthogonaux à \Card{3} surfaces\Add{;} ils constituent donc un système
+cyclique. Soient $P\Add{,} P'$ les points où le cercle~$(K)$ rencontre la
+sphère; déterminons sur ce cercle le point~$M'$ tel que $\Ratio{M}{M'}{P}{P'} = \cte[]$.
+Le lieu du point~$M'$ est une surface normale à~$(K)$, puisque
+l'équation~\Eq{(5)} a mêmes propriétés que l'équation de Riccati\Add{.}
+A l'élément de contact de la surface~$(S)$ au point~$M$ correspond
+ainsi un élément de contact d'une autre surface; les lignes
+de courbure se correspondent sur les \Card{2} surfaces, et nous avons
+ainsi un groupe de transformations de contact conservant les
+lignes de courbure.
+
+Ces résultats subsistent si on prend les cercles~$(K)$
+normaux à un plan fixe.
+
+\Section{Surfaces de Weingarten.}
+{8.}{} Nous avons considéré des congruences de sphères telles
+que les lignes de courbure se correspondent sur les \Card{2} nappes
+focales. Aux sphères, la transformation de S.~Lie fait
+correspondre des droites, et aux lignes de courbure correspondent
+les lignes asymptotiques. Nous aurons donc à considérer
+des congruences de droites telles que les asymptotiques
+se correspondent sur les \Card{2} nappes focales. Nous nous
+bornerons au cas où la congruence est une congruence de normales,
+et le problème revient ainsi à chercher les surfaces
+telles que les asymptotiques se correspondent sur les \Card{2} nappes
+de la développée.
+
+Soit donc une surface~$(\Sigma)$ sur laquelle nous prendrons
+les lignes de courbure pour lignes coordonnées; soient $l,m,n$
+%% -----File: 310.png---Folio 302-------
+les cosinus directeurs de la normale, $R, R'$ les rayons de
+courbure principaux. Les \Card{2} nappes de la développée ont pour
+équations
+\begin{alignat*}{3}
+\Tag{(S)}
+X &= x + Rl, & Y &= y + Rm, & Z &= z + Rn; \\
+\Tag{(S')}
+X' &= x + R'l, \qquad & Y' &= y + R'm, \qquad & Z' &= z + R'n.
+\end{alignat*}
+Cherchons les asymptotiques de~$(S), (S')$; et exprimons que les
+équations différentielles en $u, v$ sont les mêmes. Ici les lignes
+coordonnées formant un réseau orthogonal et conjugué, on
+a
+\begin{gather*}
+ds^{2} = E\, du^{2} + G\, dv^{2}, \\
+\sum l\, d^{2}x = L\, du^{2} + N\, dv^{2};
+\end{gather*}
+et
+\[
+\frac{1}{R} = \frac{L}{E}, \qquad
+\frac{1}{R'} = \frac{N}{G},
+\]
+d'où
+\[
+\sum l\, d^{2} x = \frac{E}{R}\, du^{2} + \frac{G}{R'}\, dv^{2}.
+\]
+Les formules d'O.~Rodrigues donnent
+\begin{alignat*}{3}%[** TN: Filled in last two columns]
+\frac{\dd x}{\dd u} &= -R\, \frac{\dd l}{\dd u}, &
+\frac{\dd y}{\dd u} &= -R\, \frac{\dd m}{\dd u}, &
+\frac{\dd z}{\dd u} &= -R\, \frac{\dd n}{\dd u}, \\
+\intertext{d'où}
+\frac{\dd l}{\dd u} &= -\frac{1}{R}\, \frac{\dd x}{\dd u}, &
+\frac{\dd m}{\dd u} &= -\frac{1}{R}\, \frac{\dd y}{\dd u}, &
+\frac{\dd n}{\dd u} &= -\frac{1}{R}\, \frac{\dd z}{\dd u}, \\
+\intertext{et}
+\frac{\dd l}{\dd v} &= -\frac{1}{R'}\, \frac{\dd x}{\dd v}, \qquad &
+\frac{\dd m}{\dd v} &= -\frac{1}{R'}\, \frac{\dd y}{\dd v}, \qquad &
+\frac{\dd n}{\dd v} &= -\frac{1}{R'}\, \frac{\dd z}{\dd v};
+\end{alignat*}
+et par conséquent
+\begin{align*}%[** TN: Rebroken]
+dX &= dx + R\, dl + l\, dR \\
+   &= \frac{\dd x}{\dd u}\, du + \frac{\dd x}{\dd v}\, dv
+  - R \left(\frac{1}{R} · \frac{\dd x}{\dd u} du
+    + \frac{1}{\DPtypo{R}{R'}}\, \frac{\dd x}{\dd v} dv\right) + l\, dR \\
+   &= \left(1 - \frac{R}{R'}\right) \frac{\dd x}{\dd v} dv + l\, dR,
+\end{align*}
+formules qui montrent que la normale à~$(S)$ a pour coefficients
+de direction $\dfrac{\dd x}{\dd u}, \dfrac{\dd y}{\dd u}, \dfrac{\dd z}{\dd u}$.
+
+On a donc sur cette surface~$(S)$
+\[
+ds^{2} = \left(1 - \frac{R}{R'}\right)^{2} G\, dv^{2} + dR^{2},
+\]
+ce qui met en évidence sur la surface~$(S)$ une famille de géodésiques
+$v = \cte$, et leurs trajectoires orthogonales $R = \cte[]$.
+%% -----File: 311.png---Folio 303-------
+L'équation \DPtypo{differentielle}{différentielle} des asymptotiques est
+\begin{align*}
+&\sum dl · dX = 0, \\
+\intertext{ou}
+&\sum d\left(\frac{\dd x}{\dd u}\right) · dX = 0.
+\end{align*}
+Développons cette équation. Le coefficient de $\left(1 - \dfrac{R}{R'}\right) · dv$ est
+\[
+\sum \frac{\dd x }{ \dd v}\, d\left(\frac{\dd x}{\dd u}\right)
+  = du \sum \frac{\dd x}{\dd v} · \frac{\dd^{2} x}{\dd u^{2}}
+  + dv \sum \frac{\dd x}{\dd v} · \frac{\dd^{2} x}{\dd u\, \dd v};
+\]
+\DPtypo{Or}{or} on a
+\[
+\sum \frac{\dd x}{\dd u} · \frac{\dd x}{\dd v} = 0;
+\]
+d'où
+\[
+\sum \frac{\dd x}{\dd v} · \frac{\dd^{2} x}{\dd u^{2}}
+  = -\sum \frac{\dd x}{\dd u} · \frac{\dd^{2} x}{\dd u\, \dd v}
+  = -\frac{1}{2} · \frac{\dd E}{\dd v},
+\]
+et
+\[
+\sum \frac{\dd x}{\dd v} · \frac{\dd^{2} x}{\dd u\, \dd v}
+  = \frac{1}{2} · \frac{\dd G}{\dd u}.
+\]
+Le coefficient de~$dR$ est
+\[
+\sum l\, d\left(\frac{\dd x}{\dd u}\right)
+  = \sum l\, \frac{\dd^{2} x}{\dd u^{2}} · du
+  + \sum l\, \frac{\dd^{2} x}{\dd u\, \dd v}\, dv
+  = \frac{E}{R}\, du,
+\]
+d'où l'équation aux asymptotiques
+\[
+\frac{1}{2}\left(1 - \frac{R}{R'}\right)
+  \left[- \frac{\dd E}{\dd v}\, du\, dv
+        + \frac{\dd G}{\dd u}\, dv^{2}\right]
+      + \frac{E}{R}\, dR\, du = 0.
+\]
+Les courbes $u = \cte$, $v = \cte$ correspondent à des courbes conjuguées
+sur la surface~$(S)$, donc le coefficient de~$du\, dv$ dans
+l'équation précédente est nul:
+\[
+\Tag{(1)}
+-\frac{1}{2} \left(1 - \frac{R}{R'}\right) \frac{\dd E}{\dd v}
+  + \frac{E}{R}\, \frac{\dd R}{\dd v} = 0;
+\]
+et l'équation devient
+\[
+\frac{1}{2} \left(1 - \frac{R}{R'}\right) \frac{\dd G}{\dd u}\, dv^{2}
+  + \frac{E}{R}\, \frac{\dd R}{\dd u}\, du^{2} = 0.
+\]
+De même sur la surface~$(S')$ on obtiendra la condition
+\[
+\Tag{(2)}
+-\frac{1}{2} \left(1 - \frac{R'}{R}\right) · \frac{\dd G}{\dd u}
+  + \frac{G}{R'}\, \frac{\dd R'}{\dd u} = 0,
+\]
+de sorte que l'équation aux asymptotiques de~$(S)$ peut s'écrire
+\[
+-\frac{G}{R'{}^2}\, \frac{\dd R'}{\dd u}\, dv^{2}
+  + \frac{E}{R^{2}}\, \frac{\dd R}{\dd u}\, du^{2} = 0,
+\]
+ou
+\[
+G\, \frac{\dd\left(\dfrac{1}{R'}\right)}{\dd u}\, dv^{2}
+  - E\, \frac{\dd\left(\dfrac{1}{R}\right)}{\dd u}\, du^{2} = 0;
+\]
+%% -----File: 312.png---Folio 304-------
+et de même pour~$(S')$
+\[
+E\, \frac{\dd \left(\dfrac{1}{R}\right)}{\dd v}\, du^{2}
+  - G\, \frac{\dd \left(\dfrac{1}{R'}\right)}{\dd v}\, dv^{2} = 0.
+\]
+Pour que ces équations soient identiques, il faut et il suffit
+que l'on ait
+\[
+\begin{vmatrix}
+\mfrac{\dd \left(\dfrac{1}{R}\right)}{\dd u} &
+\mfrac{\dd \left(\dfrac{1}{R'}\right)}{\dd u} \\
+\mfrac{\dd \left(\dfrac{1}{R}\right)}{\dd v} &
+\mfrac{\dd \left(\dfrac{1}{R'}\right)}{\dd v}
+\end{vmatrix}
+= 0,
+\]
+c'est-à-dire que $\dfrac{1}{R}$~soit fonction de~$\dfrac{1}{R'}$. \emph{Les rayons de courbure
+sont fonctions l'un de l'autre \(\DPtypo{Ribeaucour}{Ribaucour}\).} Ces surfaces
+s'appellent \emph{surfaces de Weingarten, ou surfaces~$W$}. Les surfaces
+minima en sont un cas particulier $(R + R' = 0)$.
+
+Supposons que nous partions d'une surface~$(W)$: $R'$~est
+fonction de~$R$, et la condition~\Eq{(2)} montre que
+\[
+\frac{\dd \log G}{\dd u} = \Psi(R)\, \frac{\dd R}{\dd u},
+\]
+d'où
+\begin{gather*}
+\log G = \chi(R) + \theta(v), \\
+G = e^{\chi(R)}\, e^{\theta(v)} = F(R)\, K(v);
+\end{gather*}
+et sur la développée
+\[
+ds^{2} = \Theta^{2}(R)\, K(v)\, dv^{2} + dR^{2}.
+\]
+Posons
+\[
+\sqrt{K(v)}\, dv = dV,
+\]
+nous avons
+\[
+ds^{2} = dR^{2} + \Theta^{2}(R)\, dV^{2},
+\]
+forme caractéristique de l'élément d'arc des surfaces de révolution
+rapportées aux méridiens et aux parallèles. Si nous
+rapportons la méridienne à l'arc, ses équations sont
+\begin{alignat*}{3}
+x &= \Theta (s), & y &= 0,   & z &= \Theta_{1}(s);\\
+\intertext{et celles de la surface de révolution sont}
+x &= \Theta(s)\cos V, \qquad & y &= \Theta(s)\sin V, \qquad &z &= \Theta_{1}(s).
+\end{alignat*}
+%% -----File: 313.png---Folio 305-------
+\emph{On voit ainsi que les développées de toute surface~$W$ sont applicables
+sur des surfaces de révolution, les méridiens correspondant
+à une famille de géodésiques et les parallèles à
+leurs trajectoires orthogonales.}
+
+\Paragraph{Application.} Supposons la surface~$W$ à courbure totale
+constante. En changeant d'unité on peut toujours écrire
+\begin{align*}
+&RR' = - 1, \\
+&R' = - \frac{1}{R};
+\end{align*}
+la condition~\Eq{(2)} s'écrit
+\begin{gather*}
+\left(1 + \frac{1}{R^{2}}\right) \frac{\dd G}{\dd u}
+  = -\frac{2G}{R}\, \frac{\dd R}{\dd u}, \\
+\frac{\dd \log G}{\dd u}
+  = -\frac{2R}{R^{2} + 1}\, \frac{\dd R}{\dd u}
+  = -\frac{\dd \log (R^{2} + 1)}{\dd u}, \\
+G = \frac{1}{R^{2} + 1}\, K(v), \\
+dS^{2} = (R^{2} + 1) · dV^{2} + dR^{2}.
+\end{gather*}
+Posons
+\[
+\Theta(R) = \sqrt{R^{2} + 1}\Add{,}
+\]
+la méridienne de la surface de révolution est donc telle que l'on ait
+\[
+x = \sqrt{s^{2} + 1},
+\]
+d'où
+\[
+s = \sqrt{x^{2} - 1}.
+\]
+Cherchons~$z$.
+\begin{align*}
+dx^{2} + dz^{2} &= ds^{2} = dx^{2} · \frac{x^{2}}{x^{2} - 1}, \\
+dz^{2} &= \frac{dx^{2}}{x^{2} - 1}, \\
+dz &= \frac{dx}{\sqrt{x^{2} - 1}},
+\end{align*}
+\vspace*{-\belowdisplayskip}%
+\vspace*{-\abovedisplayskip}%
+\begin{gather*}
+z = \DPtypo{L}{\log}(x + \sqrt{x^{2} - 1}),\vphantom{\bigg|} \\
+x + \sqrt{x^{2} - 1} = e^{z},
+\end{gather*}
+d'où
+\[
+x - \sqrt{x^{2} - 1} = e^{-z};
+\]
+d'où
+\[
+x = \frac{1}{2}\, (e^{z} + e^{-z})\Add{,}
+\]
+%% -----File: 314.png---Folio 306-------
+\emph{les \Card{2} nappes de la développée d'une surface à courbure totale
+constante sont applicables sur l'alysséide}. c\Add{.}à\Add{.}d.\ sur la surface
+engendrée par une chaînette qui tourne autour de sa base.
+
+
+\ExSection{XIII}
+
+\begin{Exercises}
+\item[54.] Soit $S$ une surface quelconque et $\Pi$~un plan quelconque. On
+considère toutes les sphères~$U$ ayant leurs centres sur~$S$ et
+coupant le plan~$\Pi$ sous un angle constant~$\phi$ tel que l'on ait
+$\cos\phi = \dfrac{1}{\DPtypo{K}{k}}$. Soit $S'$ la surface déduite de~$S$ en réduisant les
+%[** TN: Fraction has imaginary value, but also appears in typeset edition]
+ordonnées de~$S$ perpendiculaires à~$\Pi$ dans le rapport $\dfrac{\sqrt{1-k^2}}{1}$.
+
+Les sphères~$U$ enveloppent une surface à deux nappes. Montrer
+que leurs lignes de courbure correspondent point par point à
+celles de~$S'$. Examiner le cas où $S$~est du second degré.
+
+\item[55.] De chaque point~$M$ d'une surface~$S$ comme centre, on décrit un
+cercle~$K$ situé dans le plan tangent à~$S$, et dont le rayon soit
+égal à une constante donnée.
+
+%[** Regularized formatting of parts]
+\Primo. Déterminer les familles de
+$\infty^{1}$~cercles~$K$ qui engendrent une surface sur laquelle ces cercles
+soient lignes de courbure. Lieux des centres des sphères
+dont une telle surface est l'enveloppe.
+
+\Secundo. Trouver la condition
+nécessaire et suffisante pour que les cercles~$K$ forment
+%% -----File: 328.png---Folio 320-------
+un système cyclique. Cette condition étant supposée
+remplie, soit~$S_{1}$, l'une des surfaces normales aux cercles~$K$;
+montrer que les lignes de courbure de~$S_{1}$, correspondent à celles
+de~$S$, quand on fait correspondre à chaque point~$M$ de~$S$
+le point~$M_{1}$ du cercle~$K$ correspondant où $S_{1}$~est normal à~$K$.
+
+\Tertio. Montrer que $S_{1}$~a une courbure totale constante, et que
+la congruence de droites qui a~$S,S_{1}$ pour surfaces focales est
+une congruence de normales.
+
+\Quarto. Soit $C$~l'un des centres de
+courbure principaux de~$S$ en~$M$, et $C_{1}$~le centre de courbure
+principal de~$S_{1}$ en~$M_{1}$, qui correspond à~$C$. \DPchg{Etudier}{Étudier} la congruence
+des droites~$CC_{1}$.
+
+\item[56.] \DPchg{Etant}{Étant} donnée une surface~$S$, on désigne par~$C$ l'une quelconque
+des lignes de courbure de l'une des familles, par~$C'$ l'une
+quelconque des lignes de courbure de l'autre famille, de
+sorte qu'en un point~$M$ de~$S$ se croisent une courbe~$C$ et une
+courbe~$C'$. Soient $\omega, \omega'$ les centres de courbure principaux
+correspondant à ces deux courbes; et soient $G,G'$ les centres
+de courbure géodésique de ces deux courbes.
+
+% [** TN: Regularized formatting of parts]
+\Primo. Que peut-on
+dire des congruences définies respectivement par les quatre
+droites $MG, MG', G\omega, G'\omega'$?
+
+\Secundo. Soit $(\gamma)$~le cercle osculateur
+à~$C$ en~$M$. Démontrer que $(\gamma)$~engendre une surface canal
+quand $M$~décrit une courbe~$C'$. Trouver les sphères dont cette
+surface canal est l'enveloppe.
+
+\Tertio. Montrer que si $S$~fait
+partie de l'une des familles d'un système triple orthogonal,
+les cercles osculateurs aux trajectoires orthogonales des
+surfaces de cette famille, construits aux divers points de~$S$
+forment un système cyclique.
+
+\item[57.] Soit $O$ un point fixe, et $S$~une surface quelconque; en un point
+quelconque~$M$ de~$S$ on mène le plan tangent~$P$ et de~$O$ on abaisse
+la perpendiculaire sur~$P$; soit $H$ non pied.
+
+%[** TN: Regularized formatting of parts]
+\Primo. Trouver les
+courbes de~$S$ qui, en chacun de leurs points~$M$, admettent $MH$~pour
+normale.
+
+\Secundo. Soit $HI$ la médiane du triangle~$OHM$; la
+congruence des droites~$HI$ est une congruence de normales.
+Trouver les surfaces normales à toutes ces droites. Montrer
+que leurs lignes de courbure correspondent à un réseau de
+courbes conjuguées décrites par~$M$ sur~$S$.
+
+\Tertio. Soit $K$ le
+point où le plan perpendiculaire à~$MO$ rencontre~$MH$; et soit~$(\gamma)$
+le cercle de centre~$K$, passant en~$O$, et \DPtypo{situe}{situé} dans le
+plan~$MOK$. Les cercles~$(\gamma)$ forment un système cyclique.
+
+\item[58.] De chaque point~$M$ du \DPchg{paraboloide}{paraboloïde}
+\[
+\Tag{(P)}
+xy - az = 0,
+\]
+comme centre, on décrit une sphère~$\Sigma$ tangente au plan~$\DPtypo{xoy}{xOy}$.
+Soit $A$ le point de contact de~$\Sigma$ avec ce plan et $B$~le second
+point de contact de~$\Sigma$ avec son enveloppe.
+
+%[** TN: Regularized formatting of parts]
+\Primo. Quelles courbes
+doit décrire~$M$ sur~$(P)$ pour que $AS$~engendre une développable?
+Ces courbes forment sur~$P$ un réseau conjugué, et leurs
+tangentes en chaque point~$M$ sont perpendiculaires aux plans
+focaux de la congruence engendrée par~$AS$.
+
+\Secundo. Déterminer
+les lignes de courbure de l'enveloppe de~$\Sigma$; les normales
+menées à cette enveloppe le long de chaque ligne de courbure
+découpent sur~$(P)$ un réseau \DPtypo{conjugue}{conjugué}.
+
+\Tertio. On considère le
+cercle~$C$ normal à~$\Sigma$ en $A$~et~$B$. Montrer qu'il y a une infinité
+de surfaces normales à tous les cercles~$C$, et les déterminer.
+
+\Quarto. Montrer que ces surfaces forment l'une des familles d'un
+système triple orthogonal, et achever de déterminer ce
+système.
+\end{Exercises}
+
+%[** TN: Exercises moved to the end of the respective chapters]
+%% -----File: 315.png---Folio 307-------
+% Chapitre II: #8 -- #11
+%% -----File: 316.png---Folio 308-------
+% Chapitre III: #12 -- #14
+%% -----File: 317.png---Folio 309-------
+% #14 -- #17
+%% -----File: 318.png---Folio 310-------
+% Chapitre IV: #18 -- #20; Chapitre V: #21 -- #26
+%% -----File: 319.png---Folio 311-------
+% #26 -- #28
+%% -----File: 320.png---Folio 312-------
+% #28; Chapitre VI: #29 -- #31
+%% -----File: 321.png---Folio 313-------
+% #31; Chapitre VII: #32 -- #36
+%% -----File: 322.png---Folio 314-------
+% #36 -- #39
+%% -----File: 323.png---Folio 315-------
+% #39 -- Chapitre VIII: #40 -- #41
+%% -----File: 324.png---Folio 316-------
+% #41 -- #42
+%% -----File: 325.png---Folio 317-------
+% Chapitre IX: #43 -- #44; Chapitre X: #45 -- #47
+%% -----File: 326.png---Folio 318-------
+% #47 -- #49; Chapitre XI: #50 -- #52; Chapitre XII: #53
+%% -----File: 327.png---Folio 319-------
+% #53; Chapitre XIII: #54 -- #55
+%% -----File: 329.png---Folio 321-------
+% #55 -- #56
+%% -----File: 330.png---Folio 322-------
+% #57 -- #58
+%% -----File: 331.png---Folio 323-------
+% #58
+
+\iffalse%%[** TN: Raw DP formatter code for errata]
+\Section{Errata\Add{.}}
+
+\DPnote{[** d^o means dito, used
+\newcommand{\dito}{\qquad d$^\text{o}$\qquad}]}
+
+\begin{tabular}{ccll}
+Page & Ligne && Lire \\
+[x]7  & 6      & Au lieu de $- \dfrac{1}{6R^2} ds$ & $- \dfrac{1}{6R^2} ds^3$ \\
+[x]7  & Figure & [Illustration] & [Illustration] \\
+[x]7  & 15     & Au lieu de $T\ 0$ &  $T < 0$ \\
+[x]11 & avant dernière & \dito  le lieu des sphères & le lieu des centres des sphères. \\
+[x]16 & 7      & \dito  $- \dfrac{1}{6R} ds^3$ & $- \dfrac{1}{6R^2} ds^3$ \\
+[x]17 & 8      & \dito  secteur  & vecteur \\
+[x]19 & 15     & \dito  $\sum \dfrac{\dd x}{\dd u} \dfrac{\dd x}{\dd v^2}$ & $\sum \dfrac{\dd x}{\dd u} \dfrac{\dd x}{\dd v}$ \\
+[x]20 & 8 en remontant & \dito  $(S') (S')$ & $(S) (S')$ \\
+[x]21 & 3      & \dito            $x = F$   & $x = F_1$ \\
+[x]   &        & \phantom{\dito } $y = G_2$ & $y = G_1$ \\
+[x]   &        & \phantom{\dito } $z = H$   & $z = H_1$ \\
+[x]21 & 7 en remontant & \dito  $\dfrac{E_1}{E} = \dfrac{F_1}{F} = \dfrac{G_1}{G}$ & $\dfrac{E}{E_1} = \dfrac{F}{F_1} = \dfrac{G}{G_1}$ \\
+[x]22 & 13, 14 & \dito  $dd(uv),d\beta(uv)$ & $d\alpha(u,v),d\beta(u,v)$ \\
+[x]29 & 9 en remontant & \dito  $\dfrac{a\sin\theta}{R}$ & $- \dfrac{a\sin\theta}{R}$ \\
+[x]30 & 9      & \dito  $\sum a, \dfrac{d^2x}{ds}$ & $\sum a, \dfrac{d^{2}x}{ds^2}$ \\
+[x]31 & 3 en remontant & \dito  $\Omega^2 (du\,d^{2}v - \dots$ & $- H^2 (du\,d^{2}v$ \\
+[x]33 & Figure & la lettre M &  Manque au sommet de l'angle $\theta$. \\
+[x]35 & dernière & Au lieu de $1$  & $l$ \\
+[x]37 & 15     & \dito  $r\,d^{2}x$ & $r\,dx^2$ \\
+[x]37 & 10 en remontant & \dito   elipse  & ellipse \\
+\end{tabular}
+%% -----File: 332.png---Folio 324-------
+\begin{tabular}{ccll}
+\emph{Page} & \emph{Ligne} &&  \emph{Lire} \\
+[x]41  & dernière        & Au lieu de $= - c$  &  $= - c^2$ \\
+[x]49  & 3 en remontant  & \dito $h()$  & $h (u)$ \\
+[x]53  & 4 en remontant  & \dito pour coordonnées  &  pour lignes coordonnées \\
+[x]57  & 3 en remontant  & \dito $E\, du + F'\, dv$  &  E'\,du + F'\,dv \\
+[x]74  & 3               & \dito $A' < 0$  &   $A A'< 0$ \\
+[x]75  & 5 en remontant  & \dito $\dots dv\,dv - \dots dv\,du \dots$  &  $d\gamma \dots d\gamma$ \\
+[x]87  & 3 en remontant  & \dito $MG$ fait avec $MG$  &  $MG_1$ fait \dots \\
+[x]94  & 1               & \dito par $u,v,w,\dots a,s,$  &  par $u',v',w',\dots$ a~$s$, \\
+[x]114 & 8 en remontant  & \dito $z = (v)\phi$  &   $z = \phi(v)$ \\
+[x]142 & 4 en remontant  & \dito $\dfrac{1}{du} \dfrac{2du}{1+w^2}$  &  $\dfrac{1}{du} \dfrac{2dw}{1+w^2}$ \\
+[x]152 & 10              & \multicolumn{2}{l}{Mettre le chiffre (3) à l'équation précédente} \\
+[x]156 & 12 en remontant & Au lieu de sur $(\phi)$    &   sur $(\phi')$ \\
+[x]163 & 10 en remontant & \dito   $\dfrac{-r\,dr}{dr}$  &  $\dfrac{-r\,dr}{dr^2}$ \\
+[x]165 & Figure          & $(T)$  &  $(\Gamma)$ \\
+[x]169 & 2 en remontant  & \dito   $+\zeta\left(-\dfrac{a}{R} - \dfrac{a''}{T} -\right)ds$   &  $+\eta \left(-\dfrac{a}{R} - \dfrac{a''}{T}\right)ds$ \\
+[x]175 & 10 en remontant & \dito   résolues en $x,y$  &  résolues en $x_1,y_1$ \\
+[x]176 & 2               & \dito   avec $0,x$  &  $0x_1$  \\
+[x]188 & 2               & \dito   $\rho = \cte$  &  $\rho = 0$ \\
+[x]189 &                 & \dito   $M M$  &   $M M_1$  \\
+[x??]189 & 8 en remontant  & \dito  $x = x + \rho x \text{etc}\dots$  &  $= x + \rho x,$ \\
+[x]191 & 12              & \dito   $x$  &  $x_1$
+\end{tabular}
+%% -----File: 333.png---Folio 325-------
+\begin{tabular}{ccll}
+\emph{Page} & \emph{Ligne} &&  \emph{Lire} \\
+[x]191 &  10 en remontant &  Au lieu de $\dfrac{\dd Q}{\dd \lambda}$  &  $\dfrac{\dd Q}{\dd \lambda} \dfrac{\dd x}{\dd \mu$ \\
+[x]191 &  7  en remontant &  \dito  $\dfrac{\dd x}{\dd \mu$  &  $\dfrac{\dd x'}{\dd \mu}$ \\
+[x]191 &  7  en remontant &  \dito   car $\dfrac{\dd^2 x}{\dd \lambda\, \dd \mu}$ etc\dots  &  car $\dfrac{\dd\DPtypo{'}{^2} x}{\dd \lambda\, \dd \mu} = \dfrac{\dd x_1}{\dd \mu}$ et $\dfrac{\dd x}{\dd \lambda} = x_1$ \\
+[x]193 &  4               &  \dito   $(S)$  &  (5) \\
+[x]19[** TN: sic, presumed 193] &  12     &  \dito   $x_1$  &  $x$  \\
+[x]199 &  6 en remontant  &  \dito   $(S_1)$  &  $(S'_1)$  \\
+[x]199 &  2 en remontant  &  \dito   l'homothétique de~$M$  &  l'homothétique de~$M_1$ \\
+[x]200 &  3               &  \dito   au rayon~$OM$  &   au rayon~$OM_1$  \\
+[x]200 &  7               &  \dito   $OM$  &   $OM_1$  \\
+[x]200 &  8               &  \dito   $(S)$ (au commencement de la ligne)  &  $(S_1)$  \\
+[x]204 &  14              &  \dito   $= 0 \lambda$ &  $= 0$ \\
+[x]207 &  6 en remontant  &  \dito   $f(x,y,z,U}{W,V}{W) = 0$  &  $f(x,y,z,-U}{W,-V}{W) = 0$ \\
+[x?]212 &  9               &  \dito   les développables de l'une des familles  &  ces développables \\
+[x]212 &  11              &  \dito   indépendant de la développable     &      indépendant de la congruence \\
+[x]222 &  13              &  \dito   $p = cY-bz$ \dots  &  ajouter: ou encore $p = yz-zY$ et l'équation écrite à la ligne
+ 18 sera $\chi (X-x, Y-y, Z-z, yZ-zY, zX-xZ, xY-yX) = 0$ \\
+[x]223 &  11 en remontant &  \dito   $p = yz' - zy$  &  $p = yz' - zy'$ \\
+[x]    &                  &          $q = zx' - xz$  &  $q = zx'- xz'$ \\
+[x]224 &  9               &  \dito   $p_{ik} = \begin{vmatrix}x & x_k \\ y & y_k \end{matrix}$ & $p_{ik} = \begin{vmatrix}x_i & x_k \\ y_i & y_k \end{matrix}$
+\end{tabular}
+%% -----File: 334.png---Folio 326-------
+\begin{tabular}{ccll}
+\emph{Page} & \emph{Ligne} &&  \emph{Lire} \\
+[x]232 &  8               & Au lieu de $+ c R$  &  $+ C R$ \\
+[x]243 &  4               & \dito    focal &  polaire \\
+[x]270 &  1               & \dito   $\dfrac{\dd \Psi}{dy} + \dfrac{\dd \Psi}{dz} = 0$  &  $\dfrac{\dd \Psi}{\dd y} + q \dfrac{\dd \Psi}{\dd z} = 0$ \\
+[x]271 &  5 en remontant  & \dito   intégralité &  intégrabilité \\
+[x]277 &  6               & \dito   $\sum (x-x)^2$  & $\sum (x-x_0)^2$ \\
+[x]277 &  4 en remontant  & \dito   $dC = \dfrac{dz_0}{2R_0}$  &  $dC = \dfrac{dz_0}{2R}$ \\
+[x]278 &  6 en remontant  & \dito   du groupe des rayons &  du groupe des transformations par rayons\dots \\
+[x]278 &  1 en remontant  & \dito   $-\dfrac{(x^2+y^2+z^2+R^2)}{4R^2 (e-\lambda)}$  &  $-\dfrac{(x^2+y^2+z^2+R^2)^2}{4R^2 (e-\lambda)}$ \\
+[x]284 &  11 en remontant &  \dito   $(H)$  &  (4) \\
+[x]285 &  Figure          & [Illustration] &  [Illustration] \\
+[x]286 &  7               & \dito   $\sin^2 \theta [$  &  $\sin^4 \theta [$ \\
+[x]287 &  8 en remontant  & \dito   $= -\dfrac{1}{2}$.  &  $= \dfrac{1}{2}$. \\
+[x]288 &  9 en remontant  & \dito   $\Psi_1 + \cos i \Psi = 0$  &  $\Psi_1 = 0$ \\
+[x]289 &  6 en remontant  & \dito   $\sum df · dl$.  &  $-\sum df · dl$. \\
+[x]293 &  4 en remontant  & \dito   $('\sigma) (\sigma)$  & $(\sigma) (\sigma')$ \\
+[x]296 &  13              & \dito   $w^2 \dfrac{\dd c'}{\dd u}$  &  $w^2 \dfrac{\dd C' }{\dd u}]$
+\end{tabular}
+\fi
+
+
+%%%% LICENSE %%%%
+\backmatter
+\pagenumbering{Alph}
+\phantomsection
+\pdfbookmark[0]{License.}{License}
+\fancyhead[C]{LICENSE}
+\SetPageNumbers
+
+\begin{PGtext}
+End of Project Gutenberg's Leçons de Géométrie Supérieure, by Ernest Vessiot
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+% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %
+%                                                                         %
+% End of Project Gutenberg's Leçons de Géométrie Supérieure, by Ernest Vessiot
+%                                                                         %
+% *** END OF THIS PROJECT GUTENBERG EBOOK LEÇONS DE GÉOMÉTRIE SUPÉRIEURE ***
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diff --git a/LICENSE.txt b/LICENSE.txt
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+This eBook, including all associated images, markup, improvements,
+metadata, and any other content or labor, has been confirmed to be
+in the PUBLIC DOMAIN IN THE UNITED STATES.
+
+Procedures for determining public domain status are described in
+the "Copyright How-To" at https://www.gutenberg.org.
+
+No investigation has been made concerning possible copyrights in
+jurisdictions other than the United States. Anyone seeking to utilize
+this eBook outside of the United States should confirm copyright
+status under the laws that apply to them.
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+Project Gutenberg (https://www.gutenberg.org) public repository for
+eBook #35052 (https://www.gutenberg.org/ebooks/35052)