texlive[60380] Master/texmf-dist: profcollege (31aug21)
commits+karl at tug.org
commits+karl at tug.org
Tue Aug 31 22:51:46 CEST 2021
Revision: 60380
http://tug.org/svn/texlive?view=revision&revision=60380
Author: karl
Date: 2021-08-31 22:51:45 +0200 (Tue, 31 Aug 2021)
Log Message:
-----------
profcollege (31aug21)
Modified Paths:
--------------
trunk/Master/texmf-dist/doc/latex/profcollege/ProfCollege-doc.pdf
trunk/Master/texmf-dist/doc/latex/profcollege/ProfCollege-doc.zip
trunk/Master/texmf-dist/tex/latex/profcollege/ProfCollege.sty
Modified: trunk/Master/texmf-dist/doc/latex/profcollege/ProfCollege-doc.pdf
===================================================================
(Binary files differ)
Modified: trunk/Master/texmf-dist/doc/latex/profcollege/ProfCollege-doc.zip
===================================================================
(Binary files differ)
Modified: trunk/Master/texmf-dist/tex/latex/profcollege/ProfCollege.sty
===================================================================
--- trunk/Master/texmf-dist/tex/latex/profcollege/ProfCollege.sty 2021-08-31 20:51:18 UTC (rev 60379)
+++ trunk/Master/texmf-dist/tex/latex/profcollege/ProfCollege.sty 2021-08-31 20:51:45 UTC (rev 60380)
@@ -3,7 +3,7 @@
% or later, see http://www.latex-project.org/lppl.txtf
\NeedsTeXFormat{LaTeX2e}
-\ProvidesPackage{ProfCollege}[2021/08/22 v0.99-f Aide pour l'utilisation de LaTeX au collège]
+\ProvidesPackage{ProfCollege}[2021/09/01 v0.99-g Aide pour l'utilisation de LaTeX au collège]
\RequirePackage{verbatim}
@@ -99,6 +99,8 @@
\RequirePackage{stackengine}
\RequirePackage[thicklines]{cancel}
+\RequirePackage{fontawesome5}%Pour l'environnement Twitter
+
\RequirePackage{nicematrix}%pour le tableur
\let\myoldmulticolumn\multicolumn
@@ -155,6 +157,17 @@
\end{tikzpicture}%
}
+\newcommand\LogoTW[2]{%
+\setbox1=\hbox{\includegraphics[scale=#2]{#1}}
+\begin{tikzpicture}%
+ \clip (0,0) circle (4mm);
+ \draw (0,0) circle (4mm);
+ \node[xshift=0mm, yshift=0mm, inner xsep=0pt, inner ysep=0pt] (0,0) {%
+ \includegraphics[scale=#2]{#1}%
+ };%
+\end{tikzpicture}%
+}%
+
\makeatletter
\def\Dotfill{%
\leavevmode
@@ -506,6 +519,166 @@
}
%%%
+% Twitter
+%%%
+\setKVdefault[Twitter]{Largeur=0.95\linewidth,Auteur=Christophe,Date=\today,Url=ViveLaTeX,EchelleLogo=0.035,Logo=DrStrange,Publie=false}
+
+\NewEnviron{Twitter}[1][]{%
+ \useKVdefault[Twitter]%
+ \setKV[Twitter]{#1}%
+ \xdef\EchelleLogo{\useKV[Twitter]{EchelleLogo}}%
+ \begin{tcolorbox}[%
+ enhanced,%
+ overlay unbroken and first={%
+ \node[anchor=west,xshift=3em,yshift=-2em] at (frame.north west) {\textbf{\useKV[Twitter]{Auteur}}~{\color{gray}@\ttfamily \useKV[Twitter]{Url} - \useKV[Twitter]{Date}}};
+ \node[anchor=center,xshift=1em+2mm,yshift=-2em] at (frame.north west) {\LogoTW{\useKV[Twitter]{Logo}}{\EchelleLogo}};
+ \node[xshift=-1em,yshift=-2em] at (frame.north east) {\color{gray}...};
+ \coordinate[yshift=1em] (A) at (frame.south west);
+ \coordinate[yshift=1em] (B) at (frame.south east);
+ \node[] (C1) at ($(A)!0.1!(B)$) {\faComment[regular]\ifboolKV[Twitter]{Publie}{~\fpeval{randint(1,10)}}{}};
+ \node[] (C2) at ($(A)!0.325!(B)$) {\faRetweet\ifboolKV[Twitter]{Publie}{~\fpeval{randint(1,10)}}{}};
+ \node[] (C3) at ($(A)!0.55!(B)$) {\faHeart[regular]\ifboolKV[Twitter]{Publie}{~\fpeval{randint(1,10)}}{}};
+ \node[] (C4) at ($(A)!0.775!(B)$) {\faShareSquare};
+ },
+ colback=white,
+ colframe=gray!15,
+ top=2em,
+ left=3em,
+ bottom=2em]
+ \vspace*{0.5em}\par
+ \BODY%
+ \end{tcolorbox}
+}
+
+%%%
+% Facebook
+%%%
+\setKVdefault[Facebook]{Largeur=0.95\linewidth,Auteur=Christophe,Date=\today,Heure=3:14,EchelleLogo=0.035,Logo=DrStrange,Publie=false}
+
+\NewEnviron{Facebook}[1][]{%
+ \useKVdefault[Facebook]%
+ \setKV[Facebook]{#1}%
+ \xdef\EchelleLogo{\useKV[Facebook]{EchelleLogo}}%
+ \begin{tcolorbox}[%
+ enhanced,%
+ overlay unbroken and first={%
+ \node[anchor=west,xshift=3em,yshift=-1em] at (frame.north west) {\textbf{\useKV[Facebook]{Auteur}}};
+ \node[anchor=west,xshift=3em,yshift=-2em] at (frame.north west) {\scriptsize\color{gray}\useKV[Facebook]{Date}, \useKV[Facebook]{Heure}};
+ \node[anchor=center,xshift=1em+2mm,yshift=-1.5em] at (frame.north west) {\LogoTW{\useKV[Facebook]{Logo}}{\EchelleLogo}};
+ \node[xshift=-1em,yshift=-1.5em] at (frame.north east) {\bfseries\color{gray}...};
+ \coordinate[yshift=1.15em] (A) at (frame.south west);
+ \coordinate[yshift=1.15em] (B) at (frame.south east);
+ \coordinate[xshift=0.5em,yshift=1.8em] (A1) at (frame.south west);
+ \coordinate[xshift=-0.5em,yshift=1.8em] (B1) at (frame.south east);
+ \coordinate[xshift=0.5em,yshift=0.5em] (A2) at (frame.south west);
+ \coordinate[xshift=-0.5em,yshift=0.5em] (B2) at (frame.south east);
+ \ifboolKV[Facebook]{Publie}{%
+ \coordinate[xshift=1em,yshift=1em] (A3) at (A1);
+ \draw[blue,fill=blue] (A3) circle (1.5mm);
+ \node[] at (A3) {\tiny\color{white}\faThumbsUp};
+ \node[anchor=west] at (A3) {~\scriptsize\fpeval{randint(1,150)}};
+ \node[anchor=east,xshift=-1em,yshift=1em] at (B1) {\scriptsize\fpeval{randint(2,20)} commentaires~\fpeval{randint(2,10)} partages};
+ }{}
+ \draw[gray] (A1)--(B1);
+ \draw[gray] (A2)--(B2);
+ \node[] (C1) at ($(A)!0.15!(B)$) {\footnotesize\faThumbsUp{}~\bfseries J'aime};
+ \node[] (C2) at ($(A)!0.5!(B)$) {\footnotesize\faComment*[regular]~\bfseries Commenter}; \node[] (C3) at ($(A)!0.85!(B)$) {\footnotesize\faShareSquare~\bfseries Partager};
+ },
+ colback=white,
+ colframe=gray!15,
+ top=2em,
+ left=3em,
+ bottom=4em]
+ %\vspace*{0.5em}\par
+ \BODY%
+ \end{tcolorbox}
+}
+
+%%%
+% Instagram
+%%%
+\setKVdefault[Instagram]{Largeur=0.95\linewidth,Auteur=Christophe,Expediteur=Pierre,Date=\today,Temps=34,Publie=false,Logo=DrStrange,LogoEx=tiger,EchelleLogo=0.035,Texte={}}
+
+\NewEnviron{Instagram}[1][]{%
+ \useKVdefault[Instagram]%
+ \setKV[Instagram]{#1}%
+ \xdef\EchelleLogo{\useKV[Instagram]{EchelleLogo}}%
+ \begin{tcolorbox}[%
+ enhanced,%
+ underlay unbroken and first={%
+ \node[anchor=west,xshift=3em,yshift=-1.5em] at (frame.north west) {\textbf{\useKV[Instagram]{Expediteur}}};
+ \node[anchor=center,xshift=1em+2mm,yshift=-1.5em] at (frame.north west) {\LogoTW{\useKV[Instagram]{LogoEx}}{\EchelleLogo}};
+ \node[xshift=-1em,yshift=-1.5em,rotate=90] at (frame.north east) {\bfseries\color{gray}...};
+ \coordinate[yshift=-3em] (HA) at (frame.north west);
+ \coordinate[yshift=-3em] (HB) at (frame.north east);
+ \draw[gray!15] (HA)--(HB);
+ \coordinate[yshift=7em] (BA) at (frame.south west);
+ \coordinate[yshift=7em] (BB) at (frame.south east);
+ \draw[gray!15] (BA)--(BB);
+ \coordinate[xshift=1em,yshift=6em] (A) at (frame.south west);
+ \node[anchor=west] at (A) {\bfseries\faHeart[regular]\quad\faComment[regular]\quad\faPaperPlane};
+ \coordinate[xshift=-1em,yshift=6em] (A1) at (frame.south east);
+ \node[anchor=east] at (A1) {\bfseries\faBookmark[regular]};
+ \coordinate[xshift=1em,yshift=5em] (B) at (frame.south west);
+ \node[anchor=west] at (B) {\footnotesize\bfseries\fpeval{randint(10,30)} J'aime};
+ \coordinate[xshift=1em,yshift=4em] (C) at (frame.south west);
+ \node[anchor=west] at (C) {\textbf{\useKV[Instagram]{Expediteur}}~\useKV[Instagram]{Texte}};
+ \node[anchor=center,xshift=2em,yshift=2.25em] at (frame.south west) {\LogoTW{\useKV[Instagram]{Logo}}{\EchelleLogo}};
+ \node[anchor=west,xshift=4em,yshift=2.25em] at (frame.south west) {\textcolor{gray!50}{Ajouter un commentaire\dots}};
+ \node[anchor=east,xshift=-1em,yshift=2.25em] at (frame.south east) {\textcolor{red}{\faHeart}\quad\textcolor{Gold}{\faHandSpock}\quad\textcolor{gray!50}{\faPlusCircle}};
+ \node[anchor=west,xshift=1em,yshift=0.5em] at (frame.south west) {\scriptsize\color{gray} Il y a \useKV[Instagram]{Temps} secondes};
+ },
+ colback=white,
+ colframe=gray!15,
+ top=3em,
+ left=3em,
+ bottom=7em]
+ \BODY%
+ \end{tcolorbox}
+}
+
+%%%
+% Snapchat
+%%%
+\setKVdefault[Snapchat]{Largeur=0.95\linewidth,Auteur=Christophe,Date=\today,Temps=34,Logo=DrStrange,EchelleLogo=0.035,Texte=Envoyer un Chat}
+
+\NewEnviron{Snapchat}[1][]{%
+ \useKVdefault[Snapchat]%
+ \setKV[Snapchat]{#1}%
+ \xdef\EchelleLogo{\useKV[Snapchat]{EchelleLogo}}%
+ \begin{tcolorbox}[%
+ enhanced,%
+ underlay unbroken and first={%
+ \node[anchor=west,xshift=3em,yshift=-1em] at (frame.north west) {\textbf{\useKV[Snapchat]{Auteur}}};
+ \node[anchor=west,xshift=3em,yshift=-2em] at (frame.north west) {\scriptsize\color{gray} il y a \useKV[Snapchat]{Temps}~min};
+ \node[anchor=center,xshift=1em+2mm,yshift=-1.5em] at (frame.north west) {\LogoTW{\useKV[Snapchat]{Logo}}{\EchelleLogo}};
+ \node[xshift=-1em,yshift=-1.5em,rotate=90] at (frame.north east) {\bfseries...};
+ \node[xshift=-3em,yshift=-1.5em] at (frame.north east) {\faBell[regular]};
+ \coordinate[xshift=2em,yshift=2em] (P1) at (frame.south west);
+ \coordinate[xshift=4.5em,yshift=2em] (P2) at (frame.south west);
+ \coordinate[xshift=-3em,yshift=2em] (P4) at (frame.south west);
+ \coordinate[xshift=-2em,yshift=2em] (P3) at (frame.south east);
+ \coordinate[xshift=4.5em,yshift=1em] (P5) at (frame.south west);
+ \coordinate[xshift=4.5em,yshift=3em] (P8) at (frame.south west);
+ \coordinate[xshift=-4.5em,yshift=1em] (P6) at (frame.south east);
+ \coordinate[xshift=-4.5em,yshift=3em] (P7) at (frame.south east);
+ \draw (P1) circle (1em);
+ \node at (P1) {\faCamera};
+ \draw (P3) circle (1em);
+ \node[xshift=-0.125em,rotate=-45] at (P3) {\faLocationArrow};
+ \node[anchor=west,inner sep=0pt] at (P2) {\useKV[Snapchat]{Texte}};
+ \draw (P5) -- (P6) arc(270:450:1em) -- (P7) -- (P8) arc(90:270:1em) -- cycle;
+ },
+ colback=white,
+ colframe=gray!15,
+ top=3em,
+ left=3em,
+ bottom=3em]
+ \BODY%
+ \end{tcolorbox}
+}
+
+%%%
% Bon de sortie
%%%
\newtcolorbox{Sortie}{%
@@ -1189,6 +1362,151 @@
}
%%%
+% Triominos
+%%%
+\setKVdefault[ClesTriomino]{Longueur=5cm,Etages=3,AffichagePiece=false}
+\defKV[ClesTriomino]{Piece=\setKV[ClesTriomino]{AffichagePiece=true}}%
+
+\def\TraceTriomino#1{%
+ \ifluatex
+ \mplibforcehmode
+ \begin{mplibcode}
+ u:=\useKV[ClesTriomino]{Longueur};
+ Rayon:=0.75*u*sqrt(3)/6;
+ Etages:=\useKV[ClesTriomino]{Etages};
+ pair A,B,C,D,E,F;
+ A=(0,0);
+ B-A=Etages*u*(1,0);
+ C=rotation(B,A,60);
+ D=(1/Etages)[C,A];
+ E=(1/Etages)[C,B];
+ F=C;
+ trace polygone(A,B,C);
+ for k=1 upto Etages-1:
+ trace (k/Etages)[C,A]--(k/Etages)[C,B];
+ trace (k/Etages)[A,C]--(k/Etages)[A,B];
+ trace (k/Etages)[B,A]--(k/Etages)[B,C];
+ endfor;
+ pair G[];color H[];%Couleur pour garder l'orientation des textes...
+ G[1]=iso(D,E,F);
+ H1=blue;
+ n=1;
+ for k=1 upto Etages-1:
+ for l=0 upto (2*k):
+ n:=n+1;
+ if (l mod 2=0):
+ G[n]=G[1] shifted(k*(D-F)+(l div 2)*(E-D));
+ H[n]=blue;
+ else:
+ G[n]=symetrie(G[1],D,E) shifted((k-1)*(D-F)+(l div 2)*(E-D));
+ H[n]=green;
+ fi;
+ endfor;
+ endfor;
+ % affichage des textes
+ nba=0;
+ for p_=#1:
+ if (nba mod 3)=1:
+ if H[(nba div 3)+1]=blue:
+ label(TEX(p_) rotated 120,pointarc(cercles(G[(nba div 3)+1],Rayon),30));
+ else:
+ label(TEX(p_) rotated 180,pointarc(cercles(G[(nba div 3)+1],Rayon),90));
+ fi;
+ elseif (nba mod 3)=2:
+ if H[(nba div 3)+1]=blue:
+ label(TEX(p_),pointarc(cercles(G[(nba div 3)+1],Rayon),270));
+ else:
+ label(TEX(p_) rotated 60,pointarc(cercles(G[(nba div 3)+1],Rayon),330));
+ fi;
+ else:
+ if H[(nba div 3)+1]=blue:
+ label(TEX(p_) rotated 240,pointarc(cercles(G[(nba div 3)+1],Rayon),150));
+ else:
+ label(TEX(p_) rotated 300,pointarc(cercles(G[(nba div 3)+1],Rayon),210));
+ fi;
+ fi;
+ nba:=nba+1;
+ endfor;
+ \end{mplibcode}
+ \else
+ \begin{mpost}[mpsettings={u:=\useKV[ClesTriomino]{Longueur}; Etages:=\useKV[ClesTriomino]{Etages};}]
+ Rayon:=0.75*u*sqrt(3)/6;
+ pair A,B,C,D,E,F;
+ A=(0,0);
+ B-A=Etages*u*(1,0);
+ C=rotation(B,A,60);
+ D=(1/Etages)[C,A];
+ E=(1/Etages)[C,B];
+ F=C;
+ trace polygone(A,B,C);
+ for k=1 upto Etages-1:
+ trace (k/Etages)[C,A]--(k/Etages)[C,B];
+ trace (k/Etages)[A,C]--(k/Etages)[A,B];
+ trace (k/Etages)[B,A]--(k/Etages)[B,C];
+ endfor;
+ pair G[];color H[];%Couleur pour garder l'orientation des textes...
+ G[1]=iso(D,E,F);
+ H1=blue;
+ n=1;
+ for k=1 upto Etages-1:
+ for l=0 upto (2*k):
+ n:=n+1;
+ if (l mod 2=0):
+ G[n]=G[1] shifted(k*(D-F)+(l div 2)*(E-D));
+ H[n]=blue;
+ else:
+ G[n]=symetrie(G[1],D,E) shifted((k-1)*(D-F)+(l div 2)*(E-D));
+ H[n]=green;
+ fi;
+ endfor;
+ endfor;
+ % affichage des textes
+ nba=0;
+ for p_=#1:
+ if (nba mod 3)=1:
+ if H[(nba div 3)+1]=blue:
+ label(LATEX(p_) rotated 120,pointarc(cercles(G[(nba div 3)+1],Rayon),30));
+ else:
+ label(LATEX(p_) rotated 180,pointarc(cercles(G[(nba div 3)+1],Rayon),90));
+ fi;
+ elseif (nba mod 3)=2:
+ if H[(nba div 3)+1]=blue:
+ label(LATEX(p_),pointarc(cercles(G[(nba div 3)+1],Rayon),270));
+ else:
+ label(LATEX(p_) rotated 60,pointarc(cercles(G[(nba div 3)+1],Rayon),330));
+ fi;
+ else:
+ if H[(nba div 3)+1]=blue:
+ label(LATEX(p_) rotated 240,pointarc(cercles(G[(nba div 3)+1],Rayon),150));
+ else:
+ label(LATEX(p_) rotated 300,pointarc(cercles(G[(nba div 3)+1],Rayon),210));
+ fi;
+ fi;
+ nba:=nba+1;
+ endfor;
+ \end{mpost}
+ \fi
+}
+
+\newtoks\toklisteTriomino%
+\def\UpdatetoksTriomino#1\nil{\addtotok\toklisteTriomino{"#1",}}%
+
+\newcommand\Triomino[2][]{%
+ \useKVdefault[ClesTriomino]%
+ \setKV[ClesTriomino]{#1}%
+ \setsepchar{§}%\ignoreemptyitems%
+ \readlist*\ListeTriominos{#2}%
+ \toklisteTriomino{}
+ \ifboolKV[ClesTriomino]{AffichagePiece}{%
+ \setKV[ClesTriomino]{Etages=1}%
+ \TraceTriomino{"\ListeTriominos[\fpeval{3*\useKV[ClesTriomino]{Piece}-2}]","\ListeTriominos[\fpeval{3*\useKV[ClesTriomino]{Piece}-1}]","\ListeTriominos[\fpeval{3*\useKV[ClesTriomino]{Piece}}]"}%
+ }{%
+ \foreachitem\compteur\in\ListeTriominos{\expandafter\UpdatetoksTriomino\compteur\nil}%
+ \TraceTriomino{\the\toklisteTriomino}%
+ }%
+}%
+
+%%%
% Labyrinthe Nombre
%%%
@@ -3042,6 +3360,76 @@
}%
%%%
+% Rapido
+%%%
+%% D'après https://www.facebook.com/groups/994675223903586/user/100017057226847
+%% et une programmation de Laurent Lassale-Carrere
+\newcounter{nexo}
+\newtcolorbox[use counter=nexo,number format=\arabic]{RapidoBox}{%
+ % Titre
+ colbacktitle=white,
+ fonttitle=\color{black}\Large\bfseries,
+ toptitle=1mm,
+ bottomtitle=1mm,
+ bottom=1mm,
+ title={Rapido n°\thetcbcounter\hfill Date :\hspace*{2.5cm}},
+ %% Cadre principal
+ enhanced,
+ %nobeforeafter,
+ width=\WidthRapido,
+ colback=white,
+ valign=top,
+ drop lifted shadow%,
+ %grow to left by=5mm
+}
+\newtcolorbox{QuestionBox}{enhanced,nobeforeafter,size=small,sidebyside adapt=left}
+\newtcolorbox{QuestionReponse}{enhanced,nobeforeafter,upperbox=invisible,colback=white,width=1.5cm,grow to left by=3mm,grow to right by=3mm,height=10mm}
+
+\setKVdefault[ClesRapido]{Debut=false,Largeur=0.9\linewidth}%
+\defKV[ClesRapido]{Numero=\setKV[ClesRapido]{Debut=true}}
+
+\newlength{\WidthRapido}
+
+\newcommand\Rapido[2][]{% numéro
+\useKVdefault[ClesRapido]%
+\setKV[ClesRapido]{#1}%
+%
+\ifboolKV[ClesRapido]{Debut}{%
+ \setcounter{nexo}{\fpeval{\useKV[ClesRapido]{Numero}-1}}
+}{}%
+\setlength{\WidthRapido}{\useKV[ClesRapido]{Largeur}}%
+%
+\setsepchar[*]{§*/}%
+\readlist*\ListeRapido{#2}%
+\begin{RapidoBox}
+ \xintFor* ##1 in {\xintSeq {1}{\ListeRapidolen}}\do{%
+ \tcbsidebyside[
+ sidebyside adapt=right,
+ bicolor,
+ colback=white,colbacklower=yellow!10!white,
+ nobeforeafter,
+ top=0mm,left=1mm,
+ grow to left by=3mm,
+ grow to right by=3mm,
+ bottom=0mm,
+ ]{%
+ \ListeRapido[##1,1]
+ }{%
+ \ListeRapido[##1,2]
+ }
+}
+\end{RapidoBox}
+}
+
+\newcommand\BoiteRapido[1]{%
+ \ifx\bla#1\bla%
+ \tcbox[BoiteExpression]{\phantom{100000}}%
+ \else
+ \tcbox[BoiteExpression]{#1}%
+ \fi
+}
+
+%%%
% Fractions
%%%
\setKVdefault[ClesFraction]{Rayon=2cm,Disque,Regulier=false,Segment=false,Rectangle=false,Longueur=5cm,Largeur=2cm,Cotes=5,Triangle=false,Parts=3,Couleur=green,Reponse=false,Multiple=1,Hachures=false,Epaisseur=1}
@@ -4454,7 +4842,7 @@
%%%
% Le th\'eor\`eme de Pythagore
%%%
-\setKVdefault[ClesPythagore]{Exact=false,AvantRacine=false,Racine=false,Entier=false,Egalite=false,Precision=2,Soustraction=false,Figure=false,FigureSeule=false,Angle=0,Echelle=1cm,Reciproque=false,ReciColonnes=false,Faible=false,Unite=cm,EnchaineA=false,EnchaineB=false,EnchaineC=false,ValeurA=0,ValeurB=0,ValeurC=0,Perso=false}
+\setKVdefault[ClesPythagore]{Exact=false,AvantRacine=false,Racine=false,Entier=false,Egalite=false,Precision=2,Soustraction=false,Figure=false,FigureSeule=false,Angle=0,Echelle=1cm,Reciproque=false,ReciColonnes=false,Faible=false,Unite=cm,EnchaineA=false,EnchaineB=false,EnchaineC=false,ValeurA=0,ValeurB=0,ValeurC=0,Perso=false,AllPerso=false}
% On d\'efinit les figures \`a utiliser
\def\MPFigurePytha#1#2#3#4#5#6{%
@@ -4635,9 +5023,13 @@
\fi
}
-\newcommand\RedactionPythagore{}
+\newcommand\RedactionPythagore{}%
+\newcommand\RedactionReciPythagore{}%
+\newcommand\RedactionCalculsPythagore{}%
+\newcommand\RedactionCalculsReciPythagore{}%
+\newcommand\RedactionConclusionReciPythagore{}%
-\newcommand{\Pythagore}[5][]{%
+\newcommand\Pythagore[5][]{%
% #1 Param\`etres sous forme de cl\'es
% #2 Nom "complet" du triangle : ABC par exemple
% #3 Premi\`ere longueur
@@ -4650,10 +5042,14 @@
\StrMid{#2}{1}{1}[\NomA]%
\StrMid{#2}{2}{2}[\NomB]%
\StrMid{#2}{3}{3}[\NomC]%
+ \xdef\NomTriangle{\NomA\NomB\NomC}%
% on stocke les valeurs donn\'ees
\opcopy{#3}{A1}%
\opcopy{#4}{A2}%
\opcopy{#5}{A3}%
+ \xdef\GrandCote{#3}%
+ \xdef\PetitCote{#4}%
+ \xdef\MoyenCote{#5}%
% On trace une figure ou pas ?
\ifboolKV[ClesPythagore]{FigureSeule}{%
\MPFigureReciPytha{\NomA}{\NomB}{\NomC}{#3}{#4}{#5}{\useKV[ClesPythagore]{Angle}}%
@@ -4663,8 +5059,56 @@
{\em La figure est donn\'ee \`a titre indicatif.}%
\[\MPFigureReciPytha{\NomA}{\NomB}{\NomC}{#3}{#4}{#5}{\useKV[ClesPythagore]{Angle}}\]%
\par\columnbreak\par%
- % on r\'edige
- Dans le triangle $#2$, $[\NomA\NomC]$ est le plus grand c\^ot\'e.%
+ \ifboolKV[ClesPythagore]{AllPerso}{%
+ \RedactionReciPythagore%
+ \RedactionCalculsReciPythagore%
+ \RedactionConclusionReciPythagore%
+ }{%
+ % on r\'edige
+ \ifboolKV[ClesPythagore]{Perso}{%
+ \RedactionReciPythagore%
+ }{%
+ Dans le triangle $#2$, $[\NomA\NomC]$ est le plus grand c\^ot\'e.%
+ }
+ \ifboolKV[ClesPythagore]{ReciColonnes}{%
+ \[
+ \begin{array}{cccc|cccc}
+ &&\NomA\NomC^2&&&\NomA\NomB^2&+&\NomB\NomC^2\\
+ &&\opexport{A1}{\Aun}\num{\Aun}^2&&&\opexport{A2}{\Adeux}\num{\Adeux}^2&+&\opexport{A3}{\Atrois}\num{\Atrois}^2\\
+ &&\opmul*{A1}{A1}{a1}&&&\opmul*{A2}{A2}{a2}\opexport{a2}{\Adeux}\num{\Adeux}&+&\opmul*{A3}{A3}{a3}\opexport{a3}{\Atrois}\num{\Atrois}\\
+ &&\opexport{a1}{\Aun}\num{\Aun}&&&\multicolumn{3}{c}{\opadd*{a2}{a3}{a4}\opexport{a4}{\Aquatre}\num{\Aquatre}}\\
+ \end{array}
+ \]
+ }{%
+ \[\left.
+ \begin{array}{l}
+ \NomA\NomC^2=\opexport{A1}{\Aun}\num{\Aun}^2=\opmul*{A1}{A1}{a1}\opexport{a1}{\Aun}\num{\Aun}\\
+ \\
+ \NomA\NomB^2+\NomB\NomC^2=\opexport{A2}{\Adeux}\num{\Adeux}^2+\opexport{A3}{\Atrois}\num{\Atrois}^2=\opmul*{A2}{A2}{a2}\opexport{a2}{\Adeux}\num{\Adeux}+\opmul*{A3}{A3}{a3}\opexport{a3}{\Atrois}\num{\Atrois}=\opadd*{a2}{a3}{a4}\opexport{a4}{\Aquatre}\num{\Aquatre}\\
+ \end{array}
+ \right\}\opcmp{a1}{a4}\ifopeq\NomA\NomC^2=\NomA\NomB^2+\NomB\NomC^2\fi\opcmp{a1}{a4}\ifopneq\NomA\NomC^2\not=\NomA\NomB^2+\NomB\NomC^2\fi
+ \]
+ }
+ \ifboolKV[ClesPythagore]{Egalite}{%
+ \opcmp{a1}{a4}\ifopeq Comme $\NomA\NomC^2=\NomA\NomB^2+\NomB\NomC^2$, alors l'\'egalit\'e de Pythagore est v\'erifi\'ee. Donc le triangle $#2$ est rectangle en $\NomB$.\fi%
+ \opcmp{a1}{a4}\ifopneq Comme $\NomA\NomC^2\not=\NomA\NomB^2+\NomB\NomC^2$, alors l'\'egalit\'e de Pythagore n'est pas v\'erifi\'ee. Donc le triangle $#2$ n'est pas rectangle.\fi%
+ }{%
+ \opcmp{a1}{a4}\ifopeq Comme $\NomA\NomC^2=\NomA\NomB^2+\NomB\NomC^2$, alors le triangle $#2$ est rectangle
+ en $\NomB$ d'apr\`es la r\'eciproque du th\'eor\`eme de Pythagore.\fi%
+ \opcmp{a1}{a4}\ifopneq Comme $\NomA\NomC^2\not=\NomA\NomB^2+\NomB\NomC^2$, alors le
+ triangle $#2$ n'est pas rectangle\ifboolKV[ClesPythagore]{Faible}{.}{ d'apr\`es la contrapos\'ee du th\'eor\`eme de Pythagore.}\fi%
+ }
+ }
+ \end{multicols}
+ }{%
+ \ifboolKV[ClesPythagore]{AllPerso}{%
+ \RedactionReciPythagore%
+ \RedactionCalculsReciPythagore%
+ \RedactionConclusionReciPythagore%
+ }{%
+ \ifboolKV[ClesPythagore]{Perso}{\RedactionReciPythagore}{%
+ Dans le triangle $#2$, $[\NomA\NomC]$ est le plus grand c\^ot\'e.%
+ }
\ifboolKV[ClesPythagore]{ReciColonnes}{%
\[
\begin{array}{cccc|cccc}
@@ -4683,7 +5127,7 @@
\end{array}
\right\}\opcmp{a1}{a4}\ifopeq\NomA\NomC^2=\NomA\NomB^2+\NomB\NomC^2\fi\opcmp{a1}{a4}\ifopneq\NomA\NomC^2\not=\NomA\NomB^2+\NomB\NomC^2\fi
\]
- }
+ }%
\ifboolKV[ClesPythagore]{Egalite}{%
\opcmp{a1}{a4}\ifopeq Comme $\NomA\NomC^2=\NomA\NomB^2+\NomB\NomC^2$, alors l'\'egalit\'e de Pythagore est v\'erifi\'ee. Donc le triangle $#2$ est rectangle en $\NomB$.\fi%
\opcmp{a1}{a4}\ifopneq Comme $\NomA\NomC^2\not=\NomA\NomB^2+\NomB\NomC^2$, alors l'\'egalit\'e de Pythagore n'est pas v\'erifi\'ee. Donc le triangle $#2$ n'est pas rectangle.\fi%
@@ -4692,38 +5136,8 @@
en $\NomB$ d'apr\`es la r\'eciproque du th\'eor\`eme de Pythagore.\fi%
\opcmp{a1}{a4}\ifopneq Comme $\NomA\NomC^2\not=\NomA\NomB^2+\NomB\NomC^2$, alors le
triangle $#2$ n'est pas rectangle\ifboolKV[ClesPythagore]{Faible}{.}{ d'apr\`es la contrapos\'ee du th\'eor\`eme de Pythagore.}\fi%
- }
- \end{multicols}
- }{%
- Dans le triangle $#2$, $[\NomA\NomC]$ est le plus grand c\^ot\'e.%
- \ifboolKV[ClesPythagore]{ReciColonnes}{%
- \[
- \begin{array}{cccc|cccc}
- &&\NomA\NomC^2&&&\NomA\NomB^2&+&\NomB\NomC^2\\
- &&\opexport{A1}{\Aun}\num{\Aun}^2&&&\opexport{A2}{\Adeux}\num{\Adeux}^2&+&\opexport{A3}{\Atrois}\num{\Atrois}^2\\
- &&\opmul*{A1}{A1}{a1}&&&\opmul*{A2}{A2}{a2}\opexport{a2}{\Adeux}\num{\Adeux}&+&\opmul*{A3}{A3}{a3}\opexport{a3}{\Atrois}\num{\Atrois}\\
- &&\opexport{a1}{\Aun}\num{\Aun}&&&\multicolumn{3}{c}{\opadd*{a2}{a3}{a4}\opexport{a4}{\Aquatre}\num{\Aquatre}}\\
- \end{array}
- \]
- }{%
- \[\left.
- \begin{array}{l}
- \NomA\NomC^2=\opexport{A1}{\Aun}\num{\Aun}^2=\opmul*{A1}{A1}{a1}\opexport{a1}{\Aun}\num{\Aun}\\
- \\
- \NomA\NomB^2+\NomB\NomC^2=\opexport{A2}{\Adeux}\num{\Adeux}^2+\opexport{A3}{\Atrois}\num{\Atrois}^2=\opmul*{A2}{A2}{a2}\opexport{a2}{\Adeux}\num{\Adeux}+\opmul*{A3}{A3}{a3}\opexport{a3}{\Atrois}\num{\Atrois}=\opadd*{a2}{a3}{a4}\opexport{a4}{\Aquatre}\num{\Aquatre}\\
- \end{array}
- \right\}\opcmp{a1}{a4}\ifopeq\NomA\NomC^2=\NomA\NomB^2+\NomB\NomC^2\fi\opcmp{a1}{a4}\ifopneq\NomA\NomC^2\not=\NomA\NomB^2+\NomB\NomC^2\fi
- \]
+ }%
}%
- \ifboolKV[ClesPythagore]{Egalite}{%
- \opcmp{a1}{a4}\ifopeq Comme $\NomA\NomC^2=\NomA\NomB^2+\NomB\NomC^2$, alors l'\'egalit\'e de Pythagore est v\'erifi\'ee. Donc le triangle $#2$ est rectangle en $\NomB$.\fi%
- \opcmp{a1}{a4}\ifopneq Comme $\NomA\NomC^2\not=\NomA\NomB^2+\NomB\NomC^2$, alors l'\'egalit\'e de Pythagore n'est pas v\'erifi\'ee. Donc le triangle $#2$ n'est pas rectangle.\fi%
- }{%
- \opcmp{a1}{a4}\ifopeq Comme $\NomA\NomC^2=\NomA\NomB^2+\NomB\NomC^2$, alors le triangle $#2$ est rectangle
- en $\NomB$ d'apr\`es la r\'eciproque du th\'eor\`eme de Pythagore.\fi%
- \opcmp{a1}{a4}\ifopneq Comme $\NomA\NomC^2\not=\NomA\NomB^2+\NomB\NomC^2$, alors le
- triangle $#2$ n'est pas rectangle\ifboolKV[ClesPythagore]{Faible}{.}{ d'apr\`es la contrapos\'ee du th\'eor\`eme de Pythagore.}\fi%
- }%
}%
}%
}{%
@@ -4731,6 +5145,13 @@
\opcopy{#3}{A1}%
\opcopy{#4}{A2}%
\opcopy{\useKV[ClesPythagore]{Precision}}{pres}%
+ \xintifboolexpr{#3<#4 || #3==#4}{
+ \xdef\PetitCote{#3}%
+ \xdef\MoyenCote{#4}%
+ }{%
+ \xdef\GrandCote{#3}%
+ \xdef\MoyenCote{#4}%
+ }
% On retient les noms des sommets
\StrMid{#2}{1}{1}[\NomA]%
\StrMid{#2}{2}{2}[\NomB]%
@@ -4754,12 +5175,55 @@
\[\MPFigurePytha{\NomA}{\NomB}{\NomC}{#3}{#4}{\useKV[ClesPythagore]{Angle}}\]
\par\columnbreak\par%
% On d\'emarre la r\'esolution
+ \ifboolKV[ClesPythagore]{AllPerso}{%
+ \RedactionPythagore%
+ \RedactionCalculsPythagore%
+ }{%
+ \ifboolKV[ClesPythagore]{Perso}{%
+ \RedactionCalculsPythagore%
+ }{%
+ \ifboolKV[ClesPythagore]{Egalite}{Comme le triangle $#2$ est rectangle en $\NomB$, alors l'\'egalit\'e de Pythagore est v\'erifi\'ee :}{Dans le triangle $#2$ rectangle en $\NomB$, le th\'eor\`eme de Pythagore permet d'\'ecrire :%
+ }%
+ }
+ \xintifboolexpr{#3<#4 || #3==#4}{%\ifnum#3<#4%
+ \xdef\ResultatPytha{\fpeval{round(sqrt(#3^2+#4^2),\useKV[ClesPythagore]{Precision})}}%
+ % \xdef\ResultatPytha{\fpeval{round(sqrt(#3^2+#4^2),\useKV[ClesPythagore]{Precision})}}%
+ \begin{align*}
+ \NomA\NomC^2&=\NomA\NomB^2+\NomB\NomC^2\\
+ \NomA\NomC^2&=\ifboolKV[ClesPythagore]{EnchaineA}{\opcopy{\useKV[ClesPythagore]{ValeurA}}{a1}\opexport{a1}{\Aun}\num{\Aun}}{\opexport{A1}{\Aun}\num{\Aun}^2}+\ifboolKV[ClesPythagore]{EnchaineB}{\opcopy{\useKV[ClesPythagore]{ValeurB}}{a2}\opexport{a2}{\Adeux}\num{\Adeux}}{\opexport{A2}{\Adeux}\num{\Adeux}^2}\\
+ \NomA\NomC^2&=\ifboolKV[ClesPythagore]{EnchaineA}{\opexport{a1}{\Aun}\num{\Aun}}{\opmul*{A1}{A1}{a1}\opexport{a1}{\Aun}\num{\Aun}}+\ifboolKV[ClesPythagore]{EnchaineB}{\opexport{a2}{\Adeux}\num{\Adeux}}{\opmul*{A2}{A2}{a2}\opexport{a2}{\Adeux}\num{\Adeux}}\\
+ \NomA\NomC^2&=\opadd*{a1}{a2}{a3}\opexport{a3}{\Atrois}\num{\Atrois}%\\
+ \ifboolKV[ClesPythagore]{AvantRacine}{}{%
+ \ifboolKV[ClesPythagore]{Entier}{}{\\\NomA\NomC&=\sqrt{\opexport{a3}{\Atrois}\num{\Atrois}}}
+ \ifboolKV[ClesPythagore]{Racine}{}{\\\ifboolKV[ClesPythagore]{Exact}{\NomA\NomC&=\opsqrt[maxdivstep=3]{a3}{a4}\opunzero{a4}\opexport{a4}{\Aquatre}\num{\Aquatre}~\text{\useKV[ClesPythagore]{Unite}}}{\NomA\NomC&\approx\opsqrt[maxdivstep=5]{a3}{a4}\opround{a4}{pres}{a4}\opunzero{a4}\opexport{a4}{\Aquatre}\num{\Aquatre}~\text{\useKV[ClesPythagore]{Unite}}}}%\\
+ }%
+ \end{align*}
+ }{%\else%
+ \xdef\ResultatPytha{\fpeval{round(sqrt(#3^2-#4^2),\useKV[ClesPythagore]{Precision})}}%
+ \begin{align*}
+ \NomA\NomC^2&=\NomA\NomB^2+\NomB\NomC^2\\
+ \ifboolKV[ClesPythagore]{EnchaineC}{\opcopy{\useKV[ClesPythagore]{ValeurC}}{a1}\opexport{a1}{\Aun}\num{\Aun}}{\opexport{A1}{\Aun}\num{\Aun}^2}&=\NomA\NomB^2+\ifboolKV[ClesPythagore]{EnchaineB}{\opcopy{\useKV[ClesPythagore]{ValeurB}}{a2}\opexport{a2}{\Adeux}\num{\Adeux}}{\opexport{A2}{\Adeux}\num{\Adeux}^2}\\
+ \ifboolKV[ClesPythagore]{EnchaineC}{\opcopy{\useKV[ClesPythagore]{ValeurC}}{a1}\opexport{a1}{\Aun}\num{\Aun}}{\opmul*{A1}{A1}{a1}\opexport{a1}{\Aun}\num{\Aun}}&=\NomA\NomB^2+\ifboolKV[ClesPythagore]{EnchaineB}{\opexport{a2}{\Adeux}\num{\Adeux}}{\opmul*{A2}{A2}{a2}\opexport{a2}{\Adeux}\num{\Adeux}}\\
+ \NomA\NomB^2&=\ifboolKV[ClesPythagore]{EnchaineC}{\opcopy{\useKV[ClesPythagore]{ValeurC}}{a1}\opexport{a1}{\Aun}\num{\Aun}}{\opmul*{A1}{A1}{a1}\opexport{a1}{\Aun}\num{\Aun}}-\ifboolKV[ClesPythagore]{EnchaineB}{\opexport{a2}{\Adeux}\num{\Adeux}}{\opmul*{A2}{A2}{a2}\opexport{a2}{\Adeux}\num{\Adeux}}\\
+ \NomA\NomB^2&=\opsub*{a1}{a2}{a3}\opexport{a3}{\Atrois}\num{\Atrois}%\\
+ \ifboolKV[ClesPythagore]{AvantRacine}{}{%
+ \ifboolKV[ClesPythagore]{Entier}{}{\\\NomA\NomB&=\sqrt{\opexport{a3}{\Atrois}\num{\Atrois}}}
+ \ifboolKV[ClesPythagore]{Racine}{}{\\\ifboolKV[ClesPythagore]{Exact}{\NomA\NomB&=\opsqrt[maxdivstep=3]{a3}{a4}\opunzero{a4}\opexport{a4}{\Aquatre}\num{\Aquatre}~\text{\useKV[ClesPythagore]{Unite}}}{\NomA\NomB&\approx\opsqrt[maxdivstep=5]{a3}{a4}\opround{a4}{pres}{a4}\opunzero{a4}\opexport{a4}{\Aquatre}\num{\Aquatre}~\text{\useKV[ClesPythagore]{Unite}}}}%\\
+ }%
+ \end{align*}
+ }%\fi%
+ }
+ \end{multicols}
+ }{%
+ % On d\'emarre la r\'esolution
+ \ifboolKV[ClesPythagore]{AllPerso}{%
+ \RedactionPythagore%
+ \RedactionCalculsPythagore%
+ }{%
\ifboolKV[ClesPythagore]{Perso}{\RedactionPythagore}{\ifboolKV[ClesPythagore]{Egalite}{Comme le triangle $#2$ est rectangle en $\NomB$, alors l'\'egalit\'e de Pythagore est v\'erifi\'ee :}{Dans le triangle $#2$ rectangle en $\NomB$, le th\'eor\`eme de Pythagore permet d'\'ecrire :%
- }%
- }%
+ }}%
\xintifboolexpr{#3<#4 || #3==#4}{%\ifnum#3<#4%
\xdef\ResultatPytha{\fpeval{round(sqrt(#3^2+#4^2),\useKV[ClesPythagore]{Precision})}}%
- %\xdef\ResultatPytha{\fpeval{round(sqrt(#3^2+#4^2),\useKV[ClesPythagore]{Precision})}}%
\begin{align*}
\NomA\NomC^2&=\NomA\NomB^2+\NomB\NomC^2\\
\NomA\NomC^2&=\ifboolKV[ClesPythagore]{EnchaineA}{\opcopy{\useKV[ClesPythagore]{ValeurA}}{a1}\opexport{a1}{\Aun}\num{\Aun}}{\opexport{A1}{\Aun}\num{\Aun}^2}+\ifboolKV[ClesPythagore]{EnchaineB}{\opcopy{\useKV[ClesPythagore]{ValeurB}}{a2}\opexport{a2}{\Adeux}\num{\Adeux}}{\opexport{A2}{\Adeux}\num{\Adeux}^2}\\
@@ -4767,67 +5231,37 @@
\NomA\NomC^2&=\opadd*{a1}{a2}{a3}\opexport{a3}{\Atrois}\num{\Atrois}%\\
\ifboolKV[ClesPythagore]{AvantRacine}{}{%
\ifboolKV[ClesPythagore]{Entier}{}{\\\NomA\NomC&=\sqrt{\opexport{a3}{\Atrois}\num{\Atrois}}}
- \ifboolKV[ClesPythagore]{Racine}{}{\\\ifboolKV[ClesPythagore]{Exact}{\NomA\NomC&=\opsqrt[maxdivstep=3]{a3}{a4}\opunzero{a4}\opexport{a4}{\Aquatre}\num{\Aquatre}~\text{\useKV[ClesPythagore]{Unite}}}{\NomA\NomC&\approx\opsqrt[maxdivstep=5]{a3}{a4}\opround{a4}{pres}{a4}\opunzero{a4}\opexport{a4}{\Aquatre}\num{\Aquatre}~\text{\useKV[ClesPythagore]{Unite}}}}%\\
- }%
+ \ifboolKV[ClesPythagore]{Racine}{}{\\\ifboolKV[ClesPythagore]{Exact}{\NomA\NomC&=\opsqrt[maxdivstep=3]{a3}{a4}\opunzero{a4}\opexport{a4}{\Aquatre}\num{\Aquatre}~\text{\useKV[ClesPythagore]{Unite}}}{\NomA\NomC&\approx\opsqrt[maxdivstep=5]{a3}{a4}\opround{a4}{pres}{a4}\opunzero{a4}\opexport{a4}{\Aquatre}\num{\Aquatre}~\text{\useKV[ClesPythagore]{Unite}}}}%\\
+ }
\end{align*}
- }{%\else%
+ }{%\else
\xdef\ResultatPytha{\fpeval{round(sqrt(#3^2-#4^2),\useKV[ClesPythagore]{Precision})}}%
- \begin{align*}
- \NomA\NomC^2&=\NomA\NomB^2+\NomB\NomC^2\\
- \ifboolKV[ClesPythagore]{EnchaineC}{\opcopy{\useKV[ClesPythagore]{ValeurC}}{a1}\opexport{a1}{\Aun}\num{\Aun}}{\opexport{A1}{\Aun}\num{\Aun}^2}&=\NomA\NomB^2+\ifboolKV[ClesPythagore]{EnchaineB}{\opcopy{\useKV[ClesPythagore]{ValeurB}}{a2}\opexport{a2}{\Adeux}\num{\Adeux}}{\opexport{A2}{\Adeux}\num{\Adeux}^2}\\
- \ifboolKV[ClesPythagore]{EnchaineC}{\opcopy{\useKV[ClesPythagore]{ValeurC}}{a1}\opexport{a1}{\Aun}\num{\Aun}}{\opmul*{A1}{A1}{a1}\opexport{a1}{\Aun}\num{\Aun}}&=\NomA\NomB^2+\ifboolKV[ClesPythagore]{EnchaineB}{\opexport{a2}{\Adeux}\num{\Adeux}}{\opmul*{A2}{A2}{a2}\opexport{a2}{\Adeux}\num{\Adeux}}\\
- \NomA\NomB^2&=\ifboolKV[ClesPythagore]{EnchaineC}{\opcopy{\useKV[ClesPythagore]{ValeurC}}{a1}\opexport{a1}{\Aun}\num{\Aun}}{\opmul*{A1}{A1}{a1}\opexport{a1}{\Aun}\num{\Aun}}-\ifboolKV[ClesPythagore]{EnchaineB}{\opexport{a2}{\Adeux}\num{\Adeux}}{\opmul*{A2}{A2}{a2}\opexport{a2}{\Adeux}\num{\Adeux}}\\
- \NomA\NomB^2&=\opsub*{a1}{a2}{a3}\opexport{a3}{\Atrois}\num{\Atrois}%\\
- \ifboolKV[ClesPythagore]{AvantRacine}{}{%
- \ifboolKV[ClesPythagore]{Entier}{}{\\\NomA\NomB&=\sqrt{\opexport{a3}{\Atrois}\num{\Atrois}}}
- \ifboolKV[ClesPythagore]{Racine}{}{\\\ifboolKV[ClesPythagore]{Exact}{\NomA\NomB&=\opsqrt[maxdivstep=3]{a3}{a4}\opunzero{a4}\opexport{a4}{\Aquatre}\num{\Aquatre}~\text{\useKV[ClesPythagore]{Unite}}}{\NomA\NomB&\approx\opsqrt[maxdivstep=5]{a3}{a4}\opround{a4}{pres}{a4}\opunzero{a4}\opexport{a4}{\Aquatre}\num{\Aquatre}~\text{\useKV[ClesPythagore]{Unite}}}}%\\
- }%
- \end{align*}
+ \ifboolKV[ClesPythagore]{Soustraction}{%
+ \begin{align*}
+ \NomA\NomB^2&=\NomA\NomC^2-\NomB\NomC^2\\
+ \NomA\NomB^2&=\ifboolKV[ClesPythagore]{EnchaineC}{\opcopy{\useKV[ClesPythagore]{ValeurC}}{a1}\opexport{a1}{\Aun}\num{\Aun}}{\opexport{A1}{\Aun}\num{\Aun}^2}-\ifboolKV[ClesPythagore]{EnchaineB}{\opcopy{\useKV[ClesPythagore]{ValeurB}}{a2}\opexport{a2}{\Adeux}\num{\Adeux}}{\opexport{A2}{\Adeux}\num{\Adeux}^2}\\
+ \NomA\NomB^2&=\ifboolKV[ClesPythagore]{EnchaineC}{\opcopy{\useKV[ClesPythagore]{ValeurC}}{a1}\opexport{a1}{\Aun}\num{\Aun}}{\opmul*{A1}{A1}{a1}\opexport{a1}{\Aun}\num{\Aun}}-\ifboolKV[ClesPythagore]{EnchaineB}{\opexport{a2}{\Adeux}\num{\Adeux}}{\opmul*{A2}{A2}{a2}\opexport{a2}{\Adeux}\num{\Adeux}}\\
+ \NomA\NomB^2&=\opsub*{a1}{a2}{a3}\opexport{a3}{\Atrois}\num{\Atrois}%\\
+ \ifboolKV[ClesPythagore]{AvantRacine}{}{%
+ \ifboolKV[ClesPythagore]{Entier}{}{\\\NomA\NomB&=\sqrt{\opexport{a3}{\Atrois}\num{\Atrois}}}
+ \ifboolKV[ClesPythagore]{Racine}{}{\\\ifboolKV[ClesPythagore]{Exact}{\NomA\NomB&=\opsqrt[maxdivstep=3]{a3}{a4}\opunzero{a4}\opexport{a4}{\Aquatre}\num{\Aquatre}~\text{\useKV[ClesPythagore]{Unite}}}{\NomA\NomB&\approx\opsqrt[maxdivstep=5]{a3}{a4}\opround{a4}{pres}{a4}\opunzero{a4}\opexport{a4}{\Aquatre}\num{\Aquatre}~\text{\useKV[ClesPythagore]{Unite}}}}%\\
+ }
+ \end{align*}
+ }{%
+ \begin{align*}
+ \NomA\NomC^2&=\NomA\NomB^2+\NomB\NomC^2\\
+ \ifboolKV[ClesPythagore]{EnchaineC}{\opcopy{\useKV[ClesPythagore]{ValeurC}}{a1}\opexport{a1}{\Aun}\num{\Aun}}{\opexport{A1}{\Aun}\num{\Aun}^2}&=\NomA\NomB^2+\ifboolKV[ClesPythagore]{EnchaineB}{\opcopy{\useKV[ClesPythagore]{ValeurB}}{a2}\opexport{a2}{\Adeux}\num{\Adeux}}{\opexport{A2}{\Adeux}\num{\Adeux}^2}\\
+ \ifboolKV[ClesPythagore]{EnchaineC}{\opcopy{\useKV[ClesPythagore]{ValeurC}}{a1}\opexport{a1}{\Aun}\num{\Aun}}{\opmul*{A1}{A1}{a1}\opexport{a1}{\Aun}\num{\Aun}}&=\NomA\NomB^2+\ifboolKV[ClesPythagore]{EnchaineB}{\opexport{a2}{\Adeux}\num{\Adeux}}{\opmul*{A2}{A2}{a2}\opexport{a2}{\Adeux}\num{\Adeux}}\\
+ \NomA\NomB^2&=\ifboolKV[ClesPythagore]{EnchaineC}{\opcopy{\useKV[ClesPythagore]{ValeurC}}{a1}\opexport{a1}{\Aun}\num{\Aun}}{\opmul*{A1}{A1}{a1}\opexport{a1}{\Aun}\num{\Aun}}-\ifboolKV[ClesPythagore]{EnchaineB}{\opexport{a2}{\Adeux}\num{\Adeux}}{\opmul*{A2}{A2}{a2}\opexport{a2}{\Adeux}\num{\Adeux}}\\
+ \NomA\NomB^2&=\opsub*{a1}{a2}{a3}\opexport{a3}{\Atrois}\num{\Atrois}%\\
+ \ifboolKV[ClesPythagore]{AvantRacine}{}{%
+ \ifboolKV[ClesPythagore]{Entier}{}{\\\NomA\NomB&=\sqrt{\opexport{a3}{\Atrois}\num{\Atrois}}}%
+ \ifboolKV[ClesPythagore]{Racine}{}{\\\ifboolKV[ClesPythagore]{Exact}{\NomA\NomB&=\opsqrt[maxdivstep=3]{a3}{a4}\opunzero{a4}\opexport{a4}{\Aquatre}\num{\Aquatre}~\text{\useKV[ClesPythagore]{Unite}}}{\NomA\NomB&\approx\opsqrt[maxdivstep=5]{a3}{a4}\opround{a4}{pres}{a4}\opunzero{a4}\opexport{a4}{\Aquatre}\num{\Aquatre}~\text{\useKV[ClesPythagore]{Unite}}}}%\\
+ }
+ \end{align*}
+ }%
}%\fi%
- \end{multicols}
- }{%
- % On d\'emarre la r\'esolution
- \ifboolKV[ClesPythagore]{Perso}{\RedactionPythagore}{\ifboolKV[ClesPythagore]{Egalite}{Comme le triangle $#2$ est rectangle en $\NomB$, alors l'\'egalit\'e de Pythagore est v\'erifi\'ee :}{Dans le triangle $#2$ rectangle en $\NomB$, le th\'eor\`eme de Pythagore permet d'\'ecrire :%
- }}%
- \xintifboolexpr{#3<#4 || #3==#4}{%\ifnum#3<#4%
- \xdef\ResultatPytha{\fpeval{round(sqrt(#3^2+#4^2),\useKV[ClesPythagore]{Precision})}}%
- \begin{align*}
- \NomA\NomC^2&=\NomA\NomB^2+\NomB\NomC^2\\
- \NomA\NomC^2&=\ifboolKV[ClesPythagore]{EnchaineA}{\opcopy{\useKV[ClesPythagore]{ValeurA}}{a1}\opexport{a1}{\Aun}\num{\Aun}}{\opexport{A1}{\Aun}\num{\Aun}^2}+\ifboolKV[ClesPythagore]{EnchaineB}{\opcopy{\useKV[ClesPythagore]{ValeurB}}{a2}\opexport{a2}{\Adeux}\num{\Adeux}}{\opexport{A2}{\Adeux}\num{\Adeux}^2}\\
- \NomA\NomC^2&=\ifboolKV[ClesPythagore]{EnchaineA}{\opexport{a1}{\Aun}\num{\Aun}}{\opmul*{A1}{A1}{a1}\opexport{a1}{\Aun}\num{\Aun}}+\ifboolKV[ClesPythagore]{EnchaineB}{\opexport{a2}{\Adeux}\num{\Adeux}}{\opmul*{A2}{A2}{a2}\opexport{a2}{\Adeux}\num{\Adeux}}\\
- \NomA\NomC^2&=\opadd*{a1}{a2}{a3}\opexport{a3}{\Atrois}\num{\Atrois}%\\
- \ifboolKV[ClesPythagore]{AvantRacine}{}{%
- \ifboolKV[ClesPythagore]{Entier}{}{\\\NomA\NomC&=\sqrt{\opexport{a3}{\Atrois}\num{\Atrois}}}
- \ifboolKV[ClesPythagore]{Racine}{}{\\\ifboolKV[ClesPythagore]{Exact}{\NomA\NomC&=\opsqrt[maxdivstep=3]{a3}{a4}\opunzero{a4}\opexport{a4}{\Aquatre}\num{\Aquatre}~\text{\useKV[ClesPythagore]{Unite}}}{\NomA\NomC&\approx\opsqrt[maxdivstep=5]{a3}{a4}\opround{a4}{pres}{a4}\opunzero{a4}\opexport{a4}{\Aquatre}\num{\Aquatre}~\text{\useKV[ClesPythagore]{Unite}}}}%\\
- }
- \end{align*}
- }{%\else
- \xdef\ResultatPytha{\fpeval{round(sqrt(#3^2-#4^2),\useKV[ClesPythagore]{Precision})}}%
- \ifboolKV[ClesPythagore]{Soustraction}{%
- \begin{align*}
- \NomA\NomB^2&=\NomA\NomC^2-\NomB\NomC^2\\
- \NomA\NomB^2&=\ifboolKV[ClesPythagore]{EnchaineC}{\opcopy{\useKV[ClesPythagore]{ValeurC}}{a1}\opexport{a1}{\Aun}\num{\Aun}}{\opexport{A1}{\Aun}\num{\Aun}^2}-\ifboolKV[ClesPythagore]{EnchaineB}{\opcopy{\useKV[ClesPythagore]{ValeurB}}{a2}\opexport{a2}{\Adeux}\num{\Adeux}}{\opexport{A2}{\Adeux}\num{\Adeux}^2}\\
- \NomA\NomB^2&=\ifboolKV[ClesPythagore]{EnchaineC}{\opcopy{\useKV[ClesPythagore]{ValeurC}}{a1}\opexport{a1}{\Aun}\num{\Aun}}{\opmul*{A1}{A1}{a1}\opexport{a1}{\Aun}\num{\Aun}}-\ifboolKV[ClesPythagore]{EnchaineB}{\opexport{a2}{\Adeux}\num{\Adeux}}{\opmul*{A2}{A2}{a2}\opexport{a2}{\Adeux}\num{\Adeux}}\\
- \NomA\NomB^2&=\opsub*{a1}{a2}{a3}\opexport{a3}{\Atrois}\num{\Atrois}%\\
- \ifboolKV[ClesPythagore]{AvantRacine}{}{%
- \ifboolKV[ClesPythagore]{Entier}{}{\\\NomA\NomB&=\sqrt{\opexport{a3}{\Atrois}\num{\Atrois}}}
- \ifboolKV[ClesPythagore]{Racine}{}{\\\ifboolKV[ClesPythagore]{Exact}{\NomA\NomB&=\opsqrt[maxdivstep=3]{a3}{a4}\opunzero{a4}\opexport{a4}{\Aquatre}\num{\Aquatre}~\text{\useKV[ClesPythagore]{Unite}}}{\NomA\NomB&\approx\opsqrt[maxdivstep=5]{a3}{a4}\opround{a4}{pres}{a4}\opunzero{a4}\opexport{a4}{\Aquatre}\num{\Aquatre}~\text{\useKV[ClesPythagore]{Unite}}}}%\\
- }
- \end{align*}
- }{%
- \begin{align*}
- \NomA\NomC^2&=\NomA\NomB^2+\NomB\NomC^2\\
- \ifboolKV[ClesPythagore]{EnchaineC}{\opcopy{\useKV[ClesPythagore]{ValeurC}}{a1}\opexport{a1}{\Aun}\num{\Aun}}{\opexport{A1}{\Aun}\num{\Aun}^2}&=\NomA\NomB^2+\ifboolKV[ClesPythagore]{EnchaineB}{\opcopy{\useKV[ClesPythagore]{ValeurB}}{a2}\opexport{a2}{\Adeux}\num{\Adeux}}{\opexport{A2}{\Adeux}\num{\Adeux}^2}\\
- \ifboolKV[ClesPythagore]{EnchaineC}{\opcopy{\useKV[ClesPythagore]{ValeurC}}{a1}\opexport{a1}{\Aun}\num{\Aun}}{\opmul*{A1}{A1}{a1}\opexport{a1}{\Aun}\num{\Aun}}&=\NomA\NomB^2+\ifboolKV[ClesPythagore]{EnchaineB}{\opexport{a2}{\Adeux}\num{\Adeux}}{\opmul*{A2}{A2}{a2}\opexport{a2}{\Adeux}\num{\Adeux}}\\
- \NomA\NomB^2&=\ifboolKV[ClesPythagore]{EnchaineC}{\opcopy{\useKV[ClesPythagore]{ValeurC}}{a1}\opexport{a1}{\Aun}\num{\Aun}}{\opmul*{A1}{A1}{a1}\opexport{a1}{\Aun}\num{\Aun}}-\ifboolKV[ClesPythagore]{EnchaineB}{\opexport{a2}{\Adeux}\num{\Adeux}}{\opmul*{A2}{A2}{a2}\opexport{a2}{\Adeux}\num{\Adeux}}\\
- \NomA\NomB^2&=\opsub*{a1}{a2}{a3}\opexport{a3}{\Atrois}\num{\Atrois}%\\
- \ifboolKV[ClesPythagore]{AvantRacine}{}{%
- \ifboolKV[ClesPythagore]{Entier}{}{\\\NomA\NomB&=\sqrt{\opexport{a3}{\Atrois}\num{\Atrois}}}%
- \ifboolKV[ClesPythagore]{Racine}{}{\\\ifboolKV[ClesPythagore]{Exact}{\NomA\NomB&=\opsqrt[maxdivstep=3]{a3}{a4}\opunzero{a4}\opexport{a4}{\Aquatre}\num{\Aquatre}~\text{\useKV[ClesPythagore]{Unite}}}{\NomA\NomB&\approx\opsqrt[maxdivstep=5]{a3}{a4}\opround{a4}{pres}{a4}\opunzero{a4}\opexport{a4}{\Aquatre}\num{\Aquatre}~\text{\useKV[ClesPythagore]{Unite}}}}%\\
- }
- \end{align*}
- }%
- }%\fi%
+ }%
}%
}%
}%
@@ -6853,7 +7287,7 @@
\ppcm=\numexpr#1*#2/\pgcd\relax
}
-\setKVdefault[ClesThales]{Calcul=true,Droites=false,Propor=false,Segment=false,Figure=false,FigureSeule=false,Figurecroisee=false,FigurecroiseeSeule=false,Angle=0,Precision=2,Entier=false,Unite=cm,Reciproque=false,Produit=false,ChoixCalcul=0,Simplification,Redaction=false,Remediation=false,Echelle=1cm}
+\setKVdefault[ClesThales]{Calcul=true,Droites=false,Propor=false,Segment=false,Figure=false,FigureSeule=false,Figurecroisee=false,FigurecroiseeSeule=false,Angle=0,Precision=2,Entier=false,Unite=cm,Reciproque=false,Produit=false,ChoixCalcul=0,Simplification,Redaction=false,Remediation=false,Echelle=1cm,Perso=false,CalculsPerso=false}
%On d\'efinit la figure \`a utiliser
\def\MPFigThales#1#2#3#4#5#6{
@@ -6862,6 +7296,7 @@
% #3 Troisi\`eme sommet
% #4 point sur le segment #1#2
% #5 point sur le segment #1#3
+ % #6 angle de rotation
\ifluatex
\mplibcodeinherit{enable}
\mplibforcehmode
@@ -7263,33 +7698,39 @@
\fi
}
+\newcommand\RedactionThales{}%
+\newcommand\EcritureCalculs{}%
+\newcommand\EcritureQuotients{}%
+
%%%
\newcommand{\TTThales}[6][]{%
\useKVdefault[ClesThales]%
\setKV[ClesThales]{#1}%
- \ifboolKV[ClesThales]{Droites}{%
- Les droites \ifboolKV[ClesThales]{Remediation}{\pointilles[2cm]}{$(#3#5)$} et \ifboolKV[ClesThales]{Remediation}{\pointilles[2cm]}{$(#4#6)$} sont s\'ecantes en \ifboolKV[ClesThales]{Remediation}{\pointilles[2cm]}{$#2$}.%
- }{%
- Dans le triangle \ifboolKV[ClesThales]{Remediation}{\pointilles[2cm]}{$#2#3#4$}, \ifboolKV[ClesThales]{Remediation}{\pointilles[1cm]}{$#5$} est un point \ifboolKV[ClesThales]{Segment}{du segment}{de la
- droite}
- \ifboolKV[ClesThales]{Remediation}{\pointilles[2cm]}{\ifboolKV[ClesThales]{Segment}{$[#2#3]$}{$(#2#3)$}},
- \ifboolKV[ClesThales]{Remediation}{\pointilles[1cm]}{$#6$} est un
- point \ifboolKV[ClesThales]{Segment}{du segment}{de la droite}
- \ifboolKV[ClesThales]{Remediation}{\pointilles[2cm]}{\ifboolKV[ClesThales]{Segment}{$[#2#4]$}{$(#2#4)$}}.%
- }
- \\Comme les droites \ifboolKV[ClesThales]{Remediation}{\pointilles[2cm]}{$(#5#6)$} et \ifboolKV[ClesThales]{Remediation}{\pointilles[2cm]}{$(#3#4)$} sont parall\`eles, alors \ifboolKV[ClesThales]{Propor}{le tableau%
- \[\begin{array}{c|c|c}
- \ifboolKV[ClesThales]{Remediation}{\pointilles[1cm]}{#2#5}&\ifboolKV[ClesThales]{Remediation}{\pointilles[1cm]}{#2#6}&\ifboolKV[ClesThales]{Remediation}{\pointilles[1cm]}{#5#6}\\
- \hline
- \ifboolKV[ClesThales]{Remediation}{\pointilles[1cm]}{#2#3}&\ifboolKV[ClesThales]{Remediation}{\pointilles[1cm]}{#2#4}&\ifboolKV[ClesThales]{Remediation}{\pointilles[1cm]}{#3#4}\\
- \end{array}
- \]
- est un tableau de proportionnalit\'e\ifboolKV[ClesThales]{Segment}{.}{ d'apr\`es le th\'eor\`eme de Thal\`es.}%
- }{%
- \ifboolKV[ClesThales]{Segment}{on a :}{le th\'eor\`eme de Thal\`es permet d'\'ecrire :}%
- \[\frac{\ifboolKV[ClesThales]{Remediation}{\pointilles[1cm]}{#2#5}}{\ifboolKV[ClesThales]{Remediation}{\pointilles[1cm]}{#2#3}}=\frac{\ifboolKV[ClesThales]{Remediation}{\pointilles[1cm]}{#2#6}}{\ifboolKV[ClesThales]{Remediation}{\pointilles[1cm]}{#2#4}}=\frac{\ifboolKV[ClesThales]{Remediation}{\pointilles[1cm]}{#5#6}}{\ifboolKV[ClesThales]{Remediation}{\pointilles[1cm]}{#3#4}}\]%
- }
-}
+ \ifboolKV[ClesThales]{Perso}{\RedactionThales}{%
+ \ifboolKV[ClesThales]{Droites}{%
+ Les droites \ifboolKV[ClesThales]{Remediation}{\pointilles[2cm]}{$(#3#5)$} et \ifboolKV[ClesThales]{Remediation}{\pointilles[2cm]}{$(#4#6)$} sont s\'ecantes en \ifboolKV[ClesThales]{Remediation}{\pointilles[2cm]}{$#2$}.%
+ }{%
+ Dans le triangle \ifboolKV[ClesThales]{Remediation}{\pointilles[2cm]}{$#2#3#4$}, \ifboolKV[ClesThales]{Remediation}{\pointilles[1cm]}{$#5$} est un point \ifboolKV[ClesThales]{Segment}{du segment}{de la
+ droite}
+ \ifboolKV[ClesThales]{Remediation}{\pointilles[2cm]}{\ifboolKV[ClesThales]{Segment}{$[#2#3]$}{$(#2#3)$}},
+ \ifboolKV[ClesThales]{Remediation}{\pointilles[1cm]}{$#6$} est un
+ point \ifboolKV[ClesThales]{Segment}{du segment}{de la droite}
+ \ifboolKV[ClesThales]{Remediation}{\pointilles[2cm]}{\ifboolKV[ClesThales]{Segment}{$[#2#4]$}{$(#2#4)$}}.%
+ }
+ \\Comme les droites \ifboolKV[ClesThales]{Remediation}{\pointilles[2cm]}{$(#5#6)$} et \ifboolKV[ClesThales]{Remediation}{\pointilles[2cm]}{$(#3#4)$} sont parall\`eles, alors \ifboolKV[ClesThales]{Propor}{le tableau%
+ \[\begin{array}{c|c|c}
+ \ifboolKV[ClesThales]{Remediation}{\pointilles[1cm]}{#2#5}&\ifboolKV[ClesThales]{Remediation}{\pointilles[1cm]}{#2#6}&\ifboolKV[ClesThales]{Remediation}{\pointilles[1cm]}{#5#6}\\
+ \hline
+ \ifboolKV[ClesThales]{Remediation}{\pointilles[1cm]}{#2#3}&\ifboolKV[ClesThales]{Remediation}{\pointilles[1cm]}{#2#4}&\ifboolKV[ClesThales]{Remediation}{\pointilles[1cm]}{#3#4}\\
+ \end{array}
+ \]
+ est un tableau de proportionnalit\'e\ifboolKV[ClesThales]{Segment}{.}{ d'apr\`es le th\'eor\`eme de Thal\`es.}%
+ }{%
+ \ifboolKV[ClesThales]{Segment}{on a :}{le th\'eor\`eme de Thal\`es permet d'\'ecrire :}%
+ \[\frac{\ifboolKV[ClesThales]{Remediation}{\pointilles[1cm]}{#2#5}}{\ifboolKV[ClesThales]{Remediation}{\pointilles[1cm]}{#2#3}}=\frac{\ifboolKV[ClesThales]{Remediation}{\pointilles[1cm]}{#2#6}}{\ifboolKV[ClesThales]{Remediation}{\pointilles[1cm]}{#2#4}}=\frac{\ifboolKV[ClesThales]{Remediation}{\pointilles[1cm]}{#5#6}}{\ifboolKV[ClesThales]{Remediation}{\pointilles[1cm]}{#3#4}}\]%
+ }%
+ }%
+}%
\newcommand{\TThalesCalculsD}[8][]{%
\setKV[ClesThales]{#1}%
@@ -7422,23 +7863,30 @@
\ifboolKV[ClesThales]{Calcul}{%
%%%%%%%%%%%%%%%%%%%%%%%%%%%
On remplace par les longueurs connues :%
- \ifboolKV[ClesThales]{Propor}{%
- \[\begin{array}{c|c|c}
- \IfDecimal{#3}{\num{#3}}{#3}&\IfDecimal{#4}{\num{#4}}{#4}&\IfDecimal{#5}{\num{#5}}{#5}\\
- \hline
- \IfDecimal{#6}{\num{#6}}{#6}&\IfDecimal{#7}{\num{#7}}{#7}&\IfDecimal{#8}{\num{#8}}{#8}
- \end{array}
- \]
+ \ifboolKV[ClesThales]{CalculsPerso}{%
+ \EcritureQuotients%
}{%
- \[\frac{\IfDecimal{#3}{\num{#3}}{#3}}{\IfDecimal{#6}{\num{#6}}{#6}}=\frac{\IfDecimal{#4}{\num{#4}}{#4}}{\IfDecimal{#7}{\num{#7}}{#7}}=\frac{\IfDecimal{#5}{\num{#5}}{#5}}{\IfDecimal{#8}{\num{#8}}{#8}}\]
+ \ifboolKV[ClesThales]{Propor}{%
+ \[\begin{array}{c|c|c}
+ \IfDecimal{#3}{\num{#3}}{#3}&\IfDecimal{#4}{\num{#4}}{#4}&\IfDecimal{#5}{\num{#5}}{#5}\\
+ \hline
+ \IfDecimal{#6}{\num{#6}}{#6}&\IfDecimal{#7}{\num{#7}}{#7}&\IfDecimal{#8}{\num{#8}}{#8}
+ \end{array}
+ \]
+ }{%
+ \[\frac{\IfDecimal{#3}{\num{#3}}{#3}}{\IfDecimal{#6}{\num{#6}}{#6}}=\frac{\IfDecimal{#4}{\num{#4}}{#4}}{\IfDecimal{#7}{\num{#7}}{#7}}=\frac{\IfDecimal{#5}{\num{#5}}{#5}}{\IfDecimal{#8}{\num{#8}}{#8}}\]
+ }%
}%
% On choisit \'eventuellement le calcul \`a faire s'il y en a plusieurs.
\xdef\CompteurCalcul{\useKV[ClesThales]{ChoixCalcul}}%
\xintifboolexpr{\CompteurCalcul>0}{\xintifboolexpr{\CompteurCalcul==1}{\xdef\cmya{0}\xdef\cmza{0}}{\xintifboolexpr{\CompteurCalcul==2}{\xdef\cmxa{0}\xdef\cmza{0}}{\xdef\cmxa{0}\xdef\cmya{0}}}}{}%
- %%on fait les calculs
-\begin{align*}
- %Premier compteur \xxx
- \ifnum\cmxa>0
+ %% on fait les calculs
+ \ifboolKV[ClesThales]{CalculsPerso}{%
+ \EcritureCalculs%
+ }{%
+ \begin{align*}
+ % Premier compteur \xxx
+ \ifnum\cmxa>0
\Nomx\uppercase{&}=\frac{\opexport{valx}{\valx}\num{\valx}\times\opexport{Valx}{\Valx}\num{\Valx}}{\opexport{denox}{\denox}\num{\denox}}\relax%\global\numx=\numexpr\opprint{valx}*\opprint{Valx}\relax
\fi
% % Deuxi\`eme compteur \yyy
@@ -7517,7 +7965,8 @@
\uppercase{&}\Nomz\uppercase{&}\opdiv*{numz}{denoz}{resultatz}{restez}\opcmp{restez}{0}\ifopeq=\num{\ResultatThalesz}\else\approx\num{\fpeval{round(\ResultatThalesz,\useKV[ClesThales]{Precision})}}\fi~\text{\useKV[ClesThales]{Unite}}%
\fi
\fi
-\end{align*}
+ \end{align*}
+ }
}{}
}
@@ -7638,7 +8087,7 @@
\StrMid{\the\xxx}{1}{1}[\cmxa]%
\ifboolKV[ClesThales]{Calcul}{%
%%%%%%%%%%%%%%%%%%%%%%%%%%%
- On remplace par les longueurs connues :
+ On remplace par les longueurs connues :
\ifboolKV[ClesThales]{Propor}{%
\[\begin{array}{c|c|c}
\IfDecimal{#3}{\num{#3}}{#3}&\IfDecimal{#4}{\num{#4}}{#4}&\IfDecimal{#5}{\num{#5}}{#5}\\
@@ -7875,7 +8324,7 @@
}%
%%%%
-\newcommand{\ReciThales}[6][]{%
+\newcommand\ReciThales[6][]{%
\ifboolKV[ClesThales]{Droites}{%
Les droites $(#3#5)$ et $(#4#6)$ sont s\'ecantes en $#2$.
}{%
@@ -8004,9 +8453,18 @@
}%
}%
-\newcommand{\Thales}[8][]{%
+\newcommand\Thales[8][]{%
\useKVdefault[ClesThales]%
\setKV[ClesThales]{#1}%
+ %Définir les points pour une utilisation perso
+ \StrMid{#2}{1}{1}[\NomA]\StrMid{#2}{2}{2}[\NomB]\StrMid{#2}{3}{3}[\NomC]\StrMid{#2}{4}{4}[\NomM]\StrMid{#2}{5}{5}[\NomN]%
+ \xdef\NomPointA{\NomA}%
+ \xdef\NomPointB{\NomB}%
+ \xdef\NomPointC{\NomC}%
+ \xdef\NomTriangle{\NomA\NomB\NomC}%
+ \xdef\NomPointM{\NomM}%
+ \xdef\NomPointN{\NomN}%
+ %
\ifboolKV[ClesThales]{Reciproque}{%
\ReciproqueThales[#1]{#2}{#3}{#4}{#5}{#6}{#7}{#8}%
}{%
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