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    \small Un facteur veut parcourir toutes les rues de son quartier en n'empruntant chacune qu'une seule fois. Selon le plan, c'est tantôt facile, tantôt impossible — et l'on peut le dire d'avance, sans essayer, en regardant simplement combien de rues aboutissent à chaque carrefour.
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\subsection*{\textcolor{red}{La tournée du facteur}}
\noindent Voici le plan d'un quartier.\par\addvspace{0.35em}
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\noindent \textbf{Le facteur part de la poste et doit emprunter chaque rue exactement une fois. Il n'est pas obligé d'y revenir : il termine sa tournée où il veut.}\par\addvspace{0.35em}
\begin{center}
\renewcommand{\arraystretch}{1.35}
\begin{tabular}{|c|c|c|c|}
\hline
\cellcolor{blue!7}\bfseries Les rues & \cellcolor{blue!7}\bfseries  & \cellcolor{blue!7}\bfseries Les carrefours & \cellcolor{blue!7}\bfseries Les bâtiments \\ \hline
① rue des Lilas & ⑦ rue du Pont & Ⓐ Rond-point du Stade & \textbf{a} la poste \\ \hline
② venelle du Chat & ⑧ rue Basse & Ⓑ Place des Tanneurs & \textbf{b} la gendarmerie \\ \hline
③ passage du Marché & ⑨ rue du Four & Ⓒ Carrefour du Lavoir & \textbf{c} le gymnase \\ \hline
④ rue Verte & ⑩ rue du Moulin & Ⓓ Place aux Herbes &  \\ \hline
⑤ allée des Peupliers & ⑪ rue des Écoles & Ⓔ Rond-point de la Poste &  \\ \hline
⑥ avenue de la Gare & ⑫ chemin des Vignes & Ⓕ Carrefour du Moulin &  \\ \hline
 &  & Ⓖ Carrefour Saint-Roch &  \\ \hline
 &  & Ⓗ Carrefour des Tilleuls &  \\ \hline
\end{tabular}
\end{center}
\noindent{\small\itshape Légende du plan}\par\addvspace{0.4em}
\noindent Les questions se suivent : la première n'est qu'un comptage, la dernière demande une preuve.\par\addvspace{0.45em}
\begin{enumerate}
  \item Pour \textbf{chaque carrefour}, compter le nombre de rues qui y aboutissent. Présenter les résultats dans un tableau.
  \item Le facteur imagine sa tournée. À chaque fois qu'il \textbf{traverse} un carrefour, combien de rues utilise-t-il ? Qu'en déduire pour les carrefours où aboutit un nombre \textbf{impair} de rues ?
  \item \textbf{Le facteur peut-il faire sa tournée en partant de la poste ?} Si oui, décrire un trajet possible ; sinon, démontrer que c'est impossible.
  \item Si ce n'est pas possible : une tournée existe-t-elle \textbf{malgré tout} dans ce quartier, en partant d'ailleurs ? Si oui, où faudrait-il installer la poste — et le facteur y reviendrait-il ?
\end{enumerate}
\newpage
\section*{\textcolor{red}{Correction}}
\needspace{5\baselineskip}
\subsection*{\textcolor{red}{La tournée du facteur — Correction}}
\needspace{3\baselineskip}
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\end{center}
\noindent \textbf{1.} On compte les rues aboutissant à chaque carrefour :\par\addvspace{0.35em}
\begin{center}
\renewcommand{\arraystretch}{1.35}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline
\cellcolor{blue!7}\bfseries Carrefour & \cellcolor{blue!7}\bfseries A & \cellcolor{blue!7}\bfseries B & \cellcolor{blue!7}\bfseries C & \cellcolor{blue!7}\bfseries D & \cellcolor{blue!7}\bfseries E & \cellcolor{blue!7}\bfseries F & \cellcolor{blue!7}\bfseries G & \cellcolor{blue!7}\bfseries H \\ \hline
Nombre de rues & 3 & 2 & 3 & 2 & 3 & 4 & 4 & 3 \\ \hline
Parité & \textbf{impair} & pair & \textbf{impair} & pair & \textbf{impair} & pair & pair & \textbf{impair} \\ \hline
\end{tabular}
\end{center}
\noindent\hspace{1.2em}Vérification : la somme de ces nombres vaut 24 = 2 \(\times\) 12, car chaque rue est comptée à ses \textbf{deux} extrémités.\par\addvspace{0.3em}
\noindent{\color{red!80!black}\bfseries Les carrefours à nombre \textbf{impair} de rues sont : \textbf{A}, \textbf{C}, \textbf{E}, \textbf{H} — il y en a 4.}\par\addvspace{0.4em}
\noindent \textbf{2.}\par\addvspace{0.35em}
\noindent{\color{green!45!black}\textbf{Le raisonnement clé.} Quand le facteur \textbf{traverse} un carrefour, il y arrive par une rue et en repart par une autre : il consomme les rues \textbf{deux par deux}. Un carrefour traversé 3 fois voit donc 6 de ses rues utilisées — toujours un nombre \textbf{pair}.}\par\addvspace{0.4em}
\noindent{\color{red!80!black}\bfseries Un carrefour où aboutit un nombre \textbf{impair} de rues ne peut donc pas être seulement traversé : il doit être le \textbf{départ} ou l'\textbf{arrivée} de la tournée. Là, une rue reste « non appariée ».}\par\addvspace{0.4em}
\noindent{\color{green!45!black}Ici la poste est \textbf{sur un carrefour}, en \textbf{A}, et le facteur n'a pas à y revenir. Ce carrefour est donc le \textbf{départ} — il a droit d'être impair. Mais il n'y a qu'\textbf{une seule arrivée} : tous les autres carrefours doivent être pairs, sauf un au plus.}\par\addvspace{0.4em}
\noindent \textbf{3.}\par\addvspace{0.35em}
\noindent{\color{red!80!black}\bfseries \textbf{Non.} Le quartier compte \textbf{4 carrefours impairs} (A, C, E, H).}\par\addvspace{0.4em}
\noindent{\color{green!45!black}Chacun d'eux exigerait d'être un départ ou une arrivée. Une tournée n'a qu'un départ et qu'une arrivée : au moins 2 d'entre eux resteraient sans solution.}\par\addvspace{0.4em}
\noindent \textbf{4.}\par\addvspace{0.35em}
\noindent{\color{red!80!black}\bfseries \textbf{Non plus.} Où que l'on installe la poste, il reste 4 carrefours impairs, et une tournée n'en tolère que deux : un pour partir, un pour arriver.}\par\addvspace{0.4em}
\noindent{\color{green!45!black}C'est la différence avec un quartier à deux carrefours impairs, où déplacer la poste suffisait. Ici, le problème ne vient pas du point de départ mais du plan lui-même.}\par\addvspace{0.4em}
\noindent{\color{green!45!black}\textbf{Combien de rues en double, au minimum ?} Il faut apparier les carrefours impairs deux à deux, ce qui demande 1 passage(s) supplémentaire(s) au mieux pour une tournée ouverte, un de plus pour revenir à la poste. C'est le « problème du postier chinois », étudié bien après Euler.}\par\addvspace{0.4em}
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