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\newcommand{\titre}{Brevet des collèges}
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\begin{document}
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\begin{center}
{\Large\bfseries Diplôme national du brevet\\[4pt]
Métropole -- Antilles-Guyane, 19 septembre 2024}
\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 \hfill 20 points}}

\medskip

\begin{enumerate}
\item Le nombre 4 n'est pas premier donc $4 \times 5 \times 13$ n'est pas une décomposition en produit de facteurs premiers.

\hfill \textbf{Affirmation 1 fausse}

\item Urne 1 : $3 + 4 + 8 = 15$ boules, dont 8 rouges → probabilité $\dfrac{8}{15}$.

Urne 2 : $1 + 1 + 3 = 5$ boules, dont 3 marquées C → probabilité $\dfrac{3}{5} = \dfrac{9}{15}$.

Comme $\dfrac{8}{15} < \dfrac{9}{15}$, la probabilité rouge est inférieure.

\hfill \textbf{Affirmation 2 fausse}

\item $7x + 5 = 2x - 2 \iff 5x = -7 \iff x = -\dfrac{7}{5} = -1{,}4$.

\hfill \textbf{Affirmation 3 vraie}

\item $R = 0{,}95$~cm, $h = 0{,}2$~cm. Volume d'une pièce : $\pi \times 0{,}95^2 \times 0{,}2 \approx 0{,}567$~cm$^3$.

Volume de 10 pièces : $0{,}567 \times 10 \approx 5{,}67 \approx 6$~cm$^3$.

\hfill \textbf{Affirmation 4 vraie}

\item $5~\text{m/s} = 5 \times \np{3600}~\text{m/h} = \np{18000}~\text{m/h} = 18~\text{km/h} > 17~\text{km/h}$.

\hfill \textbf{Affirmation 5 vraie}
\end{enumerate}

\bigskip

{\large \textbf{Exercice 2 \hfill 20 points}}

\medskip

\begin{minipage}{0.65\linewidth}
\begin{enumerate}
\item
    \begin{enumerate}
        \item $\text{BG} = 5 \times \text{BE} = 5 \times 12 = 60$~m.
        \item $\text{CG} = \text{BC} - \text{BG} = 150 - 60 = 90$~m.
    \end{enumerate}
\item (AB) $\parallel$ (FG), avec B, G, C alignés et A, F, C alignés. Thalès~:

$\dfrac{\text{GF}}{\text{BA}} = \dfrac{\text{CG}}{\text{CB}}$, soit $\dfrac{\text{GF}}{200} = \dfrac{90}{150}$, d'où $\text{GF} = \dfrac{90}{150} \times 200 = 120$~m.
\end{enumerate}
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\begin{enumerate}[start=3]
\item
    \begin{enumerate}
        \item $\mathcal{A}(\triangle\text{CGF}) = \dfrac{\text{GF} \times \text{CG}}{2} = \dfrac{120 \times 90}{2} = \np{5400}~\text{m}^2$.

        \item $\dfrac{\np{5400} \times 80}{\np{9600}} = 45$ : il faut 45 minutes pour moissonner CGF.
    \end{enumerate}
\item $\text{AC}^2 = \text{AB}^2 + \text{BC}^2 = 200^2 + 150^2 = \np{62500}$, d'où $\text{AC} = 250$~m.

Longueur totale : $200 + 150 + 250 = 600$~m.
\end{enumerate}

\bigskip

{\large \textbf{Exercice 3 \hfill 20 points}}

\medskip

\begin{enumerate}
\item
    \begin{enumerate}
        \item Maximum : samedi, 405~kWh.
        \item Étendue : $405 - 322 = 83$~kWh.
        \item Moyenne : $\dfrac{381 + 363 + 322 + 329 + 393 + 405 + 376}{7} = \dfrac{\np{2569}}{7} = 367$~kWh.
    \end{enumerate}
\item 15~\% de \np{2569} kWh = $\np{2569} \times \dfrac{15}{100} = 385{,}35$~kWh. Recette : $385{,}35 \times 0{,}08 = 30{,}83$~\euro.

\item
\begin{multicols}{2}
Dans le triangle OLV rectangle en V~:

$\sin\left(\widehat{\text{OLV}}\right) = \dfrac{\text{OV}}{\text{OL}} = \dfrac{7}{13{,}5}$

$\widehat{\text{OLV}} \approx 31{,}2\degres$, compris entre 30 et 35 : les panneaux ont une production maximale.

\columnbreak

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\bigskip

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

\medskip

\begin{enumerate}
\item $f(6) = 36 + 60 + 16 = 112$.

\item
    \begin{enumerate}
        \item La formule saisie en B2 est \fbox{\texttt{=B1*B1+10*B1+16}}.
        \item D'après la colonne D, $-2$ est un antécédent de 0 par $f$.
    \end{enumerate}
\item
    \begin{enumerate}
        \item $(x+2)(x+8) = x^2 + 10x + 16 = f(x)$.
        \item $f(x) = 0 \iff (x+2)(x+8) = 0 \iff x = -2$ ou $x = -8$. L'autre antécédent de 0 est $-8$.
    \end{enumerate}
\end{enumerate}

\bigskip

{\large \textbf{Exercice 5 \hfill 20 points}}

\medskip

\begin{multicols}{2}
\begin{enumerate}
\item
    \begin{enumerate}
        \item On reproduit ABCD en vraie grandeur.
        \item AB = BC = CD = DA (côtés des deux triangles équilatéraux) : ABCD est un losange.
        \item $\widehat{\text{BCA}} = 60\degres$ et $\widehat{\text{ACD}} = 60\degres$, donc $\widehat{\text{BCD}} = 120\degres$.
    \end{enumerate}
\end{enumerate}

\columnbreak

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\begin{enumerate}[start=2]
\item Lignes 5 et 6 complétées ({\color{red}en rouge}) :

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\item Le programme A permet de tracer la figure {\color{blue}2}.

Le programme B permet de tracer la figure {\color{blue}3}.

Le programme C permet de tracer la figure {\color{blue}1}.

\bigskip

\begin{center}
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\fbox{Programme A} & \fbox{Programme B} & \fbox{Programme C}\\[4pt]
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\\[4pt]
Figure 1 & Figure 2 & Figure 3 \\[4pt]
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\\
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\end{center}
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\end{document}
