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\begin{document}
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\begin{center}
{\Large\bfseries Diplôme national du brevet\\[4pt]
Métropole Antilles-Guyane, 18 septembre 2023}
\end{center}

\vspace{3mm}
\begin{center}
\colorbox{gray!15}{\parbox{0.85\textwidth}{\centering\Large\bfseries CORRIGÉ DE L'ÉPREUVE}}
\end{center}

\bigskip

{\large \textbf{Exercice 1 : QCM \hfill 20 points}}

\medskip

\begin{enumerate}
\item Diviseurs de 84~: 2, 3 et 4 sont diviseurs ($84 = 2^2 \times 3 \times 7$). Réponse \textbf{B}.
\item Agrandissement de coefficient 2~: le volume est multiplié par $2^3 = 8$. Réponse \textbf{B}.
\item $(-2)^2 + 3 \times (-2) - 5 = 4 - 6 - 5 = -7$. Réponse \textbf{C}.
\item Multiples de 2 dans \{1,...,8\}~: 2, 4, 6, 8 → 4 boules sur 8. $p = \dfrac{4}{8} = \dfrac{1}{2}$. Réponse \textbf{A}.
\item Rapport de l'homothétie de centre O~: les points de DEF sont à l'opposé (rapport négatif) et deux fois plus loin. Réponse \textbf{A} ($k = -2$).
\end{enumerate}

\bigskip

{\large \textbf{Exercice 2 : Losanges \hfill 14 points}}

\medskip

\begin{enumerate}
\item Le bloc losange tracé dans le script~: on répète 2 fois [avancer 50, tourner 60°, avancer 50, tourner 120°], ce qui donne bien un losange à angles 60° et 120°. Les lignes complétées en rouge sont \textbf{2} et \textbf{50}.

\item Association des scripts~: Script 1 $\leftrightarrow$ Figure C~; Script 2 $\leftrightarrow$ Figure A~; Script 3 $\leftrightarrow$ Figure B.

\item La transformation qui permet d'obtenir le losange ABCD à partir du losange EDCF est la \textbf{rotation de centre C et d'angle $60\degres$}.

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    \node[above left] at (120:1.4) {D};
    \node[above left] at (150:2.424) {E};
    \node[left] at (180:1.4) {F};
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\end{center}

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\blockmoreblocks{losange}
\blockmove{tourner \turnright{} de \ovalnum{60} degrés}
}
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\end{enumerate}

\bigskip

{\large \textbf{Exercice 3 : Tarifs piscine \hfill 22 points}}

\medskip

\begin{enumerate}
\item
    \begin{enumerate}
        \item Tarif A pour 10 entrées~: $5{,}90 \times 10 = 59$~\euro.
        \item Tarif B pour 10 entrées~: $4{,}40 \times 10 + 30 = 74$~\euro.
    \end{enumerate}
\item $f(x) = 5{,}9x$ ~;~ $g(x) = 4{,}4x + 30$.
\item
    \begin{enumerate}
        \item $5{,}9x = 4{,}4x + 30 \iff 1{,}5x = 30 \iff x = 20$.
        \item Les deux tarifs donnent le même prix pour \textbf{20 entrées}.
    \end{enumerate}
\item
    \begin{enumerate}
        \item $\overline{x} = \dfrac{\np{12500}+\np{13700}+\np{10400}+\np{13600}+\np{12300}+\np{11700}+\np{10400}+\np{11600}+\np{10200}+\np{13800}+\np{12600}+\np{11800}}{12} = \dfrac{\np{144600}}{12} = \np{12050}$.
        \item Étendue~: $\np{13800} - \np{10200} = \np{3600}$.
    \end{enumerate}
\item Volume~: $50 \times 25 \times 3 = \np{3750}$~m$^3$.
\end{enumerate}

\bigskip

{\large \textbf{Exercice 4 : Funiculaire \hfill 20 points}}

\medskip

\begin{center}
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\begin{enumerate}
\item
    \begin{enumerate}
        \item Tarif individuel~: $12 \times 10 + 8 \times 8 = 120 + 64 = 184$~\euro.
        \item Tarif groupe~: $12 \times 8{,}50 + 8 \times 7 = 102 + 56 = 158$~\euro.
        \item Réduction~: $\dfrac{184 - 158}{184} = \dfrac{26}{184} \approx 0{,}1413$, soit environ $14{,}13\,\%$.
    \end{enumerate}
\item Vitesse~: $v = \dfrac{448{,}5}{8 \times 60 + 45} = \dfrac{448{,}5}{525} \approx 0{,}854$~m/s, soit $\approx 0{,}85$~m/s au centième près.
\item Pente $=\dfrac{\text{Dénivelée}}{\text{Longueur horizontale}}$, donc $\dfrac{50}{\text{AC}} = \dfrac{25}{100}$, d'où AC $= \dfrac{50 \times 100}{25} = 200$~m.
\end{enumerate}

\bigskip

{\large \textbf{Exercice 5 : Triangle et Thalès \hfill 24 points}}

\medskip

\begin{minipage}{0.52\linewidth}
\begin{enumerate}
\item AB $=$ AM $+$ MB $= 2{,}7 + 2{,}5 = 5{,}2$~cm.
\item Pythagore dans ABH rectangle en H~:

$\text{AH}^2 = \text{AB}^2 - \text{BH}^2 = 5{,}2^2 - 2^2 = 27{,}04 - 4 = 23{,}04$.

AH $= \sqrt{23{,}04} = 4{,}8$~cm.

\item $\sin\widehat{\text{ACH}} = \dfrac{\text{AH}}{\text{AC}} = \dfrac{4{,}8}{8{,}5} \approx 0{,}5647$.

$\widehat{\text{ACH}} \approx 34\degres$.

\item $\cos 34\degres \approx 0{,}829$, d'où $\text{CH} = 8{,}5 \times \cos 34\degres \approx 7{,}04$~cm $\approx 7$~cm.

\item Thalès (MN $\parallel$ BC)~:

$\dfrac{\text{AN}}{\text{AC}} = \dfrac{\text{AM}}{\text{AB}} = \dfrac{2{,}7}{5{,}2}$.

$\text{AN} = \dfrac{2{,}7 \times 8{,}5}{5{,}2} \approx 4{,}4$~cm. L'élève a tort.

\item $\mathcal{A}_{\triangle AHC} = \dfrac{4{,}8 \times 7}{2} = 16{,}8$~cm$^2$.
\end{enumerate}
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