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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]
Polynésie française, 9 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 21 points}}

\medskip

\begin{enumerate}
\item Divisibles par $21 = 3 \times 7$~:
\begin{itemize}
\item Nombre 4~: $2 \times 3 \times 5 \times 7 = 2 \times 5 \times 21$, divisible par 21.
\item Nombre 5~: $2^3 \times 3^2 \times 7 = 2^3 \times 3 \times 21$, divisible par 21.
\end{itemize}

\item $\np{0,00000276} = 2{,}76 \times 10^{-6}$.

\item $\np{2640}~\text{km/min} = \dfrac{\np{2640}}{60} \times \np{1000}~\text{m/s} = \np{44000}~\text{m/s}$.

\item $(2x-7)(3x+1) = 0 \iff x = \dfrac{7}{2}$ ou $x = -\dfrac{1}{3}$.

\item $f(-3) = 5 \times (-3)^2 + 2 = 45 + 2 = 47$.

\item
\end{enumerate}

\begin{minipage}{0.52\linewidth}
Les droites (AB) et (ED) sont parallèles, avec A, E, C alignés et B, D, C alignés~: Thalès~:

$\dfrac{\text{CD}}{\text{CB}} = \dfrac{\text{DE}}{\text{BA}}$, soit $\dfrac{3}{4} = \dfrac{\text{DE}}{5}$, d'où $\text{DE} = \dfrac{15}{4} = 3{,}75$~cm.
\end{minipage}\hfill
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\end{minipage}

\bigskip

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

\medskip

\begin{enumerate}
\item Formule en B4~: \fbox{\texttt{= SOMME(B2:J2)/9}}

\item $\dfrac{29 + 23{,}1 + 22{,}6 + 17{,}4 + 23{,}4 + 25{,}7 + 25{,}2 + 26 + 24}{9} = \dfrac{216{,}4}{9} \approx 24$~\degres C.

\item La médiane est 24~\degres C : la moitié des températures sont inférieures ou égales à 24, et la moitié sont supérieures ou égales à 24.

\item Température la plus basse : 17,4~\degres C. Étendue 18,5~: température la plus élevée = $17{,}4 + 18{,}5 = 35{,}9$~\degres C le 25 juin 2019.

\item
    \begin{enumerate}
        \item Probabilité d'obtenir 26~\degres C~: $\dfrac{1}{9}$.
        \item Probabilité $\leqslant 24$~\degres C~: 5 valeurs sur 9, soit $\dfrac{5}{9}$.
        \item Probabilité $> 25$~\degres C~: 4 valeurs ($25{,}2$~; $25{,}7$~; $26$~; $29$) sur 9, soit $\dfrac{4}{9} \approx 44{,}4~\%> 40~\%$~: affirmation vraie.
    \end{enumerate}
\end{enumerate}

\bigskip

{\large \textbf{Exercice 3 \hfill 17 points}}

\medskip

\begin{minipage}{0.55\linewidth}
\begin{enumerate}
\item Dans le triangle OMN rectangle en N~:

$\text{MN} = \text{ON} \times \tan\left(\widehat{\text{MON}}\right) = 6 \times \tan(32\degres) \approx 3{,}7$~cm.

\item Pythagore dans OPQ rectangle en P~:

$\text{OP}^2 = \text{OQ}^2 - \text{PQ}^2 = 6{,}5^2 - 2{,}5^2 = 36$, donc $\text{OP} = 6$~cm.

\item $\text{ON} = \text{OP} = 6$ mais $\text{MN} \approx 3{,}7 \neq \text{PQ} = 2{,}5$ : les triangles ne sont pas égaux.

\item $\text{OQ} = 2 \times \text{OS}$ : OPQ est un agrandissement de ORS de facteur 2. L'aire est multipliée par 4.

$\mathcal{A}(\text{OPQ}) = \dfrac{6 \times 2{,}5}{2} = 7{,}5~\text{cm}^2$, donc $\mathcal{A}(\text{ORS}) = \dfrac{7{,}5}{4} = \np{1,875}~\text{cm}^2$.
\end{enumerate}
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\bigskip

{\large \textbf{Exercice 4 \hfill 19 points}}

\medskip

\begin{enumerate}
\item
\begin{multicols}{2}
$\widehat{\text{DEF}} + \widehat{\text{FE}y} = 180\degres$~; or $\widehat{\text{DEF}} = 108\degres$, donc $\widehat{\text{FE}y} = 72\degres$.

\columnbreak

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\end{multicols}

\item
    \begin{enumerate}
        \item Le bloc \og pentagone\fg{} avec les cases complétées ({\color{red}en rouge})~:

\hfill
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{
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\blockmove{tourner \turnleft{} de \ovalnum{\textbf{72}} degrés}
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\blockpen{relever le stylo}
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\hfill~

        \item Associations programme–copie–transformation~:

\begin{center}
{\renewcommand{\arraystretch}{1.5}
\begin{tabular}{|c|c|c|}
\hline
Élève & Copie d'écran & Transformation \\ \hline
Camille & {\color{blue}3} & {\color{blue}translation} \\ \hline
Lou & {\color{blue}1} & {\color{blue}rotation} \\ \hline
Zoé & {\color{blue}2} & {\color{blue}symétrie centrale} \\ \hline
\end{tabular}}
\end{center}

\begin{center}
\begin{tabularx}{\linewidth}{*{3}{>{\centering\arraybackslash}X}}
Copie 1 (Lou) & Copie 2 (Zoé) & Copie 3 (Camille) \\[4pt]
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&
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\\
\end{tabularx}
\end{center}

        \item Ordre des instructions du programme de Sofia~:

\begin{center}
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\hline
\hfill Instruction \hfill~ & Ordre \\ \hline
\begin{scratch}[scale=0.9]\blockpen{effacer tout}\end{scratch} & 2\up{e} \\ \hline
\begin{scratch}[scale=0.9]\blockmove{s'orienter à \ovalnum{90}}\end{scratch} & 4\up{e} \\ \hline
\begin{scratch}[scale=0.9]\blockmoreblocks{pentagone}\end{scratch} & 6\up{e} \\ \hline
\begin{scratch}[scale=0.9]\blockinit{Quand \greenflag est cliqué}\end{scratch} & 1\up{re} \\ \hline
\begin{scratch}[scale=0.9]\blockvariable{mettre \selectmenu{longueur} à \ovalnum{60}}\end{scratch} & 5\up{e} \\ \hline
\begin{scratch}[scale=0.9]\blockmove{aller à x : \ovalnum{0} y : \ovalnum{0}}\end{scratch} & 3\up{e} \\ \hline
\begin{scratch}[scale=0.9]\blockmoreblocks{pentagone}\end{scratch} & 8\up{e} \\ \hline
\begin{scratch}[scale=0.9]\blockvariable{mettre \selectmenu{longueur} à \ovaloperator{\ovalvariable{longueur} * \ovalnum{1.5}}}\end{scratch} & 7\up{e} \\ \hline
\end{tabular}
\end{center}

    \end{enumerate}
\end{enumerate}

\newpage

{\large \textbf{Exercice 5 \hfill 23 points}}

\medskip

\begin{enumerate}
\item $V = \pi \times 3{,}6^2 \times 1{,}5 \approx 61{,}1$~m$^3$.

\item En 2 heures, la pompe vide $2 \times 14{,}1 = 28{,}2$~m$^3$. Volume restant~: $61{,}1 - 28{,}2 = 32{,}9$~m$^3$.

\item
    \begin{enumerate}
        \item La pompe vide $\dfrac{14{,}1}{60} = 0{,}235$~m$^3$/min. En $t$ minutes, volume restant : $61{,}1 - 0{,}235t = V(t)$.
        \item $V(t) = 30 \iff 61{,}1 - 0{,}235t = 30 \iff t = \dfrac{31{,}1}{0{,}235} \approx 132$~min.
    \end{enumerate}

\item

\begin{center}
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\end{center}

    \begin{enumerate}
        \item D'après le graphique, l'antécédent de 40 par $V$ est environ 90 : au bout de 90~minutes, il reste 40~m$^3$ à vider.
        \item La pompe vide complètement le bassin au bout d'environ 260~minutes.
    \end{enumerate}
\end{enumerate}

\end{document}
