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\begin{center}
    \small Les 5 exercices ci-dessous balaient les transformations du cycle 4 : symétrie axiale, symétrie centrale, translation, rotation d'un quart de tour et homothétie. Tout se passe dans un repère : chaque transformation se lit aussi comme une opération sur les coordonnées, et se vérifie donc au chiffre près.
\end{center}
\vspace{0.5em}
\sloppy
\needspace{5\baselineskip}
\subsection*{\textcolor{red}{Exercice 1 — Symétrie axiale}}
\noindent Dans le repère ci-dessous, le triangle ABC a pour sommets \textbf{A(1 ; 4), B(1 ; 2), C(3 ; 1)}.\par\addvspace{0.35em}
\needspace{3\baselineskip}
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\end{center}
\begin{enumerate}
  \item Construire le triangle A'B'C', symétrique de ABC par rapport à \textbf{l'axe des ordonnées}.
  \item Donner les \textbf{coordonnées} de A', B' et C'.
  \item Que peut-on dire des \textbf{longueurs} AB et A'B' ? Justifier.
\end{enumerate}
\needspace{5\baselineskip}
\subsection*{\textcolor{red}{Exercice 2 — Symétrie centrale}}
\noindent Le triangle ABC a pour sommets \textbf{A(4 ; 4), B(1 ; 3), C(4 ; 1)}.\par\addvspace{0.35em}
\needspace{3\baselineskip}
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\begin{enumerate}
  \item Construire le symétrique A'B'C' de ABC par rapport à \textbf{le point O(0 ; 0)}.
  \item Donner les coordonnées de A', B' et C'.
  \item Une symétrie centrale est aussi une \textbf{rotation}. Laquelle ? Justifier.
\end{enumerate}
\needspace{5\baselineskip}
\subsection*{\textcolor{red}{Exercice 3 — Translation}}
\noindent Le triangle ABC a pour sommets \textbf{A(\(-\)1 ; 2), B(\(-\)3 ; 1), C(\(-\)1 ; 0)}.\par\addvspace{0.35em}
\needspace{3\baselineskip}
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\noindent On applique la translation qui déplace tout point de \textbf{3 carreaux vers la gauche} et \textbf{3 carreaux vers le haut}.\par\addvspace{0.35em}
\begin{enumerate}
  \item Construire l'image A'B'C' du triangle ABC par cette translation.
  \item Donner les coordonnées de A', B' et C'.
  \item Que peut-on dire des droites (AB) et (A'B') ? Justifier.
\end{enumerate}
\needspace{5\baselineskip}
\subsection*{\textcolor{red}{Exercice 4 — Rotation d'un quart de tour}}
\noindent Le triangle ABC a pour sommets \textbf{A(4 ; 4), B(2 ; 4), C(4 ; 1)}.\par\addvspace{0.35em}
\needspace{3\baselineskip}
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\begin{enumerate}
  \item Construire l'image A'B'C' de ABC par la \textbf{rotation de centre O et d'angle 90\textdegree{}}, dans le \textbf{sens des aiguilles d'une montre}.
  \item Donner les coordonnées de A', B' et C'.
  \item Quelle serait l'image de A par la \textbf{même rotation appliquée quatre fois} ? Justifier sans calcul.
\end{enumerate}
\needspace{5\baselineskip}
\subsection*{\textcolor{red}{Exercice 5 — Homothétie}}
\noindent Le triangle ABC a pour sommets \textbf{A(2 ; 1), B(2 ; 4), C(4 ; 1)}.\par\addvspace{0.35em}
\needspace{3\baselineskip}
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\begin{enumerate}
  \item Construire l'image A'B'C' de ABC par l'\textbf{homothétie de centre O et de rapport 2}.
  \item Donner les coordonnées de A', B' et C'.
  \item L'aire du triangle ABC vaut \textbf{3 carreaux}. En déduire l'aire de A'B'C', en carreaux, \textbf{sans la recalculer}.
\end{enumerate}
\newpage
\section*{\textcolor{red}{Correction}}
\needspace{5\baselineskip}
\subsection*{\textcolor{red}{Exercice 1 — Correction}}
\noindent{\color{green!45!black}Par symétrie d'axe \textbf{l'axe des ordonnées}, un point de coordonnées (x ; y) a pour image (\textbf{\(-\)x} ; y) : l'ordonnée ne change pas, l'abscisse change de signe.}\par\addvspace{0.4em}
\noindent \textbf{1. et 2.} On applique la règle à chaque sommet :\par\addvspace{0.35em}
\begin{center}
\renewcommand{\arraystretch}{1.35}
\begin{tabular}{|c|c|c|c|}
\hline
\cellcolor{blue!7}\bfseries Point & \cellcolor{blue!7}\bfseries Coordonnées & \cellcolor{blue!7}\bfseries Image & \cellcolor{blue!7}\bfseries Coordonnées \\ \hline
A & (1 ; 4) & A' & \colorbox{yellow!35}{(\(-\)1 ; 4)} \\ \hline
B & (1 ; 2) & B' & \colorbox{yellow!35}{(\(-\)1 ; 2)} \\ \hline
C & (3 ; 1) & C' & \colorbox{yellow!35}{(\(-\)3 ; 1)} \\ \hline
\end{tabular}
\end{center}
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\draw[line width=0.5pt] (3.6,5.6) -- (3.6,3.6) -- (1.6,2.6) -- cycle;
\fill[red!75!black] (3.6,5.6) circle (0.11);
\node at (3.804,6.111) {\scalebox{1.533}{\small\bfseries A'}};
\fill[red!75!black] (3.6,3.6) circle (0.11);
\node at (4.092,3.354) {\scalebox{1.533}{\small\bfseries B'}};
\fill[red!75!black] (1.6,2.6) circle (0.11);
\node at (1.211,2.211) {\scalebox{1.533}{\small\bfseries C'}};
\end{tikzpicture}
\end{center}
\noindent{\small\itshape Le triangle ABC et son symétrique A'B'C'.}\par\addvspace{0.4em}
\noindent{\color{red!80!black}\bfseries Les images sont A'(\(-\)1 ; 4), B'(\(-\)1 ; 2), C'(\(-\)3 ; 1).}\par\addvspace{0.4em}
\noindent \textbf{3.}\par\addvspace{0.35em}
\noindent{\color{red!80!black}\bfseries AB = A'B'.}\par\addvspace{0.4em}
\noindent{\color{green!45!black}Une symétrie axiale \textbf{conserve les longueurs} : c'est une isométrie. Elle conserve aussi les mesures d'angles, les aires et l'alignement. En revanche elle \textbf{change l'orientation} de la figure — c'est ce que fait un miroir.}\par\addvspace{0.4em}
\needspace{5\baselineskip}
\subsection*{\textcolor{red}{Exercice 2 — Correction}}
\noindent{\color{green!45!black}Par symétrie de centre \textbf{O(0 ; 0)}, O est le \textbf{milieu} de [MM']. On en déduit x' = 2 \(\times\) 0 \(-\) x et y' = 2 \(\times\) 0 \(-\) y.}\par\addvspace{0.4em}
\noindent \textbf{1. et 2.}\par\addvspace{0.35em}
\begin{center}
\renewcommand{\arraystretch}{1.35}
\begin{tabular}{|c|c|c|c|}
\hline
\cellcolor{blue!7}\bfseries Point & \cellcolor{blue!7}\bfseries Coordonnées & \cellcolor{blue!7}\bfseries Image & \cellcolor{blue!7}\bfseries Coordonnées \\ \hline
A & (4 ; 4) & A' & \colorbox{yellow!35}{(\(-\)4 ; \(-\)4)} \\ \hline
B & (1 ; 3) & B' & \colorbox{yellow!35}{(\(-\)1 ; \(-\)3)} \\ \hline
C & (4 ; 1) & C' & \colorbox{yellow!35}{(\(-\)4 ; \(-\)1)} \\ \hline
\end{tabular}
\end{center}
\needspace{3\baselineskip}
\begin{center}
\begin{tikzpicture}[scale=0.5357, line join=round, line cap=round]
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\fill[red!75!black] (9.6,6.6) circle (0.11);
\node at (9.883,6.128) {\scalebox{1.867}{\small\bfseries C}};
\draw[line width=0.5pt] (1.6,1.6) -- (4.6,2.6) -- (1.6,4.6) -- cycle;
\fill[red!75!black] (1.6,1.6) circle (0.11);
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\fill[red!75!black] (5.6,5.6) circle (0.11);
\node at (5.15,6.05) {\scalebox{1.867}{\small\bfseries O}};
\end{tikzpicture}
\end{center}
\noindent{\small\itshape ABC et son symétrique A'B'C' par rapport à O.}\par\addvspace{0.4em}
\noindent{\color{red!80!black}\bfseries Les images sont A'(\(-\)4 ; \(-\)4), B'(\(-\)1 ; \(-\)3), C'(\(-\)4 ; \(-\)1).}\par\addvspace{0.4em}
\noindent \textbf{3.}\par\addvspace{0.35em}
\noindent{\color{red!80!black}\bfseries Une symétrie centrale de centre O est exactement la \textbf{rotation de centre O et d'angle 180\textdegree{}} (un demi-tour).}\par\addvspace{0.4em}
\noindent{\color{green!45!black}La symétrie centrale conserve les longueurs, les angles et les aires, et — contrairement à la symétrie axiale — elle \textbf{conserve l'orientation}. Elle transforme aussi toute droite en une droite qui lui est \textbf{parallèle}.}\par\addvspace{0.4em}
\needspace{5\baselineskip}
\subsection*{\textcolor{red}{Exercice 3 — Correction}}
\noindent{\color{green!45!black}Une \textbf{translation} fait glisser la figure sans la tourner ni la déformer. Ici, chaque point voit son abscisse augmenter de \textbf{\(-\)3} et son ordonnée de \textbf{3} : l'image de (x ; y) est (x + \(-\)3 ; y + 3).}\par\addvspace{0.4em}
\noindent \textbf{1. et 2.}\par\addvspace{0.35em}
\begin{center}
\renewcommand{\arraystretch}{1.35}
\begin{tabular}{|c|c|c|c|}
\hline
\cellcolor{blue!7}\bfseries Point & \cellcolor{blue!7}\bfseries Coordonnées & \cellcolor{blue!7}\bfseries Image & \cellcolor{blue!7}\bfseries Coordonnées \\ \hline
A & (\(-\)1 ; 2) & A' & \colorbox{yellow!35}{(\(-\)4 ; 5)} \\ \hline
B & (\(-\)3 ; 1) & B' & \colorbox{yellow!35}{(\(-\)6 ; 4)} \\ \hline
C & (\(-\)1 ; 0) & C' & \colorbox{yellow!35}{(\(-\)4 ; 3)} \\ \hline
\end{tabular}
\end{center}
\needspace{3\baselineskip}
\begin{center}
\begin{tikzpicture}[scale=0.6522, line join=round, line cap=round]
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\fill[red!75!black] (6.6,1.6) circle (0.11);
\node at (6.905,1.142) {\scalebox{1.533}{\small\bfseries C}};
\draw[line width=0.5pt] (3.6,6.6) -- (1.6,5.6) -- (3.6,4.6) -- cycle;
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\node at (3.905,7.058) {\scalebox{1.533}{\small\bfseries A'}};
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\end{tikzpicture}
\end{center}
\noindent{\small\itshape ABC et son image A'B'C' par la translation.}\par\addvspace{0.4em}
\noindent{\color{red!80!black}\bfseries Les images sont A'(\(-\)4 ; 5), B'(\(-\)6 ; 4), C'(\(-\)4 ; 3).}\par\addvspace{0.4em}
\noindent \textbf{3.}\par\addvspace{0.35em}
\noindent{\color{red!80!black}\bfseries Les droites (AB) et (A'B') sont \textbf{parallèles}, et AB = A'B'.}\par\addvspace{0.4em}
\noindent{\color{green!45!black}Une translation conserve les longueurs, les angles, les aires \textbf{et l'orientation}. Elle transforme toute droite en une droite \textbf{parallèle} : la figure est simplement déplacée, jamais retournée ni tournée. Le quadrilatère ABA'B' est d'ailleurs un \textbf{parallélogramme}.}\par\addvspace{0.4em}
\needspace{5\baselineskip}
\subsection*{\textcolor{red}{Exercice 4 — Correction}}
\noindent{\color{green!45!black}Un quart de tour de centre O dans le sens des aiguilles d'une montre transforme (x ; y) en (\textbf{y} ; \textbf{\(-\)x}). On peut le retrouver en comptant les carreaux : le point tourne d'un quart de tour autour de O, la distance à O étant conservée.}\par\addvspace{0.4em}
\noindent \textbf{1. et 2.}\par\addvspace{0.35em}
\begin{center}
\renewcommand{\arraystretch}{1.35}
\begin{tabular}{|c|c|c|c|}
\hline
\cellcolor{blue!7}\bfseries Point & \cellcolor{blue!7}\bfseries Coordonnées & \cellcolor{blue!7}\bfseries Image & \cellcolor{blue!7}\bfseries Coordonnées \\ \hline
A & (4 ; 4) & A' & \colorbox{yellow!35}{(4 ; \(-\)4)} \\ \hline
B & (2 ; 4) & B' & \colorbox{yellow!35}{(4 ; \(-\)2)} \\ \hline
C & (4 ; 1) & C' & \colorbox{yellow!35}{(1 ; \(-\)4)} \\ \hline
\end{tabular}
\end{center}
\needspace{3\baselineskip}
\begin{center}
\begin{tikzpicture}[scale=0.8333, line join=round, line cap=round]
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\end{tikzpicture}
\end{center}
\noindent{\small\itshape ABC et son image A'B'C' par le quart de tour.}\par\addvspace{0.4em}
\noindent{\color{red!80!black}\bfseries Les images sont A'(4 ; \(-\)4), B'(4 ; \(-\)2), C'(1 ; \(-\)4).}\par\addvspace{0.4em}
\noindent \textbf{3.}\par\addvspace{0.35em}
\noindent{\color{red!80!black}\bfseries Quatre quarts de tour font un \textbf{tour complet} (4 \(\times\) 90\textdegree{} = 360\textdegree{}) : l'image de A est \textbf{A lui-même}, soit (4 ; 4).}\par\addvspace{0.4em}
\noindent{\color{green!45!black}Une rotation conserve les longueurs, les angles et les aires, et conserve l'orientation. Attention : contrairement à la translation, elle ne conserve \textbf{pas} le parallélisme des droites avec leurs images — une droite et son image font entre elles un angle égal à l'angle de la rotation.}\par\addvspace{0.4em}
\needspace{5\baselineskip}
\subsection*{\textcolor{red}{Exercice 5 — Correction}}
\noindent{\color{green!45!black}L'\textbf{homothétie de centre O et de rapport k} transforme le point (x ; y) en (k \(\times\) x ; k \(\times\) y). Ici k = 2, donc l'image de (x ; y) est (2x ; 2y). Le rapport est \textbf{positif} : l'image est du \textbf{même côté} de O que la figure de départ, et la figure est \textbf{agrandie} 2 fois.}\par\addvspace{0.4em}
\noindent \textbf{1. et 2.}\par\addvspace{0.35em}
\begin{center}
\renewcommand{\arraystretch}{1.35}
\begin{tabular}{|c|c|c|c|}
\hline
\cellcolor{blue!7}\bfseries Point & \cellcolor{blue!7}\bfseries Coordonnées & \cellcolor{blue!7}\bfseries Image & \cellcolor{blue!7}\bfseries Coordonnées \\ \hline
A & (2 ; 1) & A' & \colorbox{yellow!35}{(4 ; 2)} \\ \hline
B & (2 ; 4) & B' & \colorbox{yellow!35}{(4 ; 8)} \\ \hline
C & (4 ; 1) & C' & \colorbox{yellow!35}{(8 ; 2)} \\ \hline
\end{tabular}
\end{center}
\needspace{3\baselineskip}
\begin{center}
\begin{tikzpicture}[scale=0.5357, line join=round, line cap=round]
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\end{center}
\noindent{\small\itshape ABC et son image A'B'C' par l'homothétie de rapport 2.}\par\addvspace{0.4em}
\noindent{\color{red!80!black}\bfseries Les images sont A'(4 ; 2), B'(4 ; 8), C'(8 ; 2).}\par\addvspace{0.4em}
\noindent \textbf{3.} Les longueurs sont multipliées par |k| = 2, donc les aires par k\textsuperscript{2} :\par\addvspace{0.35em}
\noindent\hspace{1.2em}k\textsuperscript{2} = 2\textsuperscript{2} = 4\par\addvspace{0.3em}
\noindent\hspace{1.2em}Aire de A'B'C' = 3 \(\times\) 4 = 12\par\addvspace{0.3em}
\noindent{\color{red!80!black}\bfseries L'aire de A'B'C' est de 12 carreaux.}\par\addvspace{0.4em}
\noindent{\color{green!45!black}\textbf{Le point à retenir :} une homothétie \textbf{ne conserve pas les longueurs} — c'est la seule des cinq transformations dans ce cas. Les longueurs sont multipliées par |k|, les \textbf{aires par k\textsuperscript{2}}. Les angles, eux, sont conservés : la figure garde sa forme, elle change seulement de taille.}\par\addvspace{0.4em}
\end{document}
