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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]
Amérique du Nord, 29 mai 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 : Vrai/Faux \hfill 20 points}}

\medskip

\begin{enumerate}
\item Prix ordonnés~: 7~; 10~; 12~; 13~; 15.

\textbf{Affirmation A}~: $\dfrac{7+10+12+13+15}{5} = \dfrac{57}{5} = 11{,}4$~\euro. \textbf{Vrai}.

\textbf{Affirmation B}~: La médiane est le 3\up{e} terme = 12~\euro{} (pas 10). \textbf{Faux}.

\item 20~m en 6~s → $\dfrac{20}{6} \approx 3{,}33$~m/s = $3{,}33 \times 3{,}6 = 12$~km/h (pas 14). \textbf{Faux}.

\item Nombres premiers de 1 à 15~: 2, 3, 5, 7, 11, 13 = 6 nombres. Probabilité $= \dfrac{6}{15} = \dfrac{2}{5}$ (pas $\dfrac{7}{15}$). \textbf{Faux}.

\item L'aire est multipliée par $3^2 = 9$ (pas 3). \textbf{Faux}.
\end{enumerate}

\bigskip

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

\medskip

\begin{center}
\begin{tikzpicture}[>=stealth,node distance=1.8cm]
    \node[draw,align=center] (start) at (0,7) {Nombre choisi au départ};
    \node[draw,align=center] (addL)  at (-3,5) {Ajouter 2};
    \node[draw,align=center] (mulR)  at (3,5)  {Multiplier par 5};
    \node[draw,align=center] (mul4L) at (-3,3) {Multiplier par 4};
    \node[draw,align=center] (sub3R) at (3,3)  {Soustraire 3};
    \node[draw,align=center] (mult)  at (0,1)  {Multiplier les deux nombres};
    \node[draw,align=center] (res)   at (0,-0.5){Résultat obtenu à l'arrivée};
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    \draw[->] (mul4L) -- (mult);
    \draw[->] (sub3R) -- (mult);
    \draw[->] (mult)  -- (res);
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\end{center}

\begin{enumerate}
\item Gauche~: $2 \to 4 \to 16$~; droite~: $2 \to 10 \to 7$~; résultat~: $16 \times 7 = 112$.
\item Gauche~: $-3 \to -1 \to -4$~; droite~: $-3 \to -15 \to -18$~; résultat~: $(-4)\times(-18) = 72$.
\item $x \to (x+2) \to 4(x+2)$ et $x \to 5x \to 5x-3$. Résultat~: $4(x+2)(5x-3) = (4x+8)(5x-3)$ : expressions C et D.
\item $(4x+8)(5x-3) = 0 \iff 4x+8 = 0$ ou $5x-3 = 0 \iff x = -2$ ou $x = \dfrac{3}{5} = 0{,}6$.
\item $B = (4x+2)(5x-3) = 20x^2 - 12x + 10x - 6 = 20x^2 - 2x - 6$.
\end{enumerate}

\bigskip

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

\medskip

\begin{enumerate}
\item $3 \times 11 = 33$~\euro.
\item $50 + 8 \times 5 = 90$~\euro.
\item $f \leftrightarrow$ Essentiel~; $g \leftrightarrow$ Liberté~; $h \leftrightarrow$ Classique.

\begin{center}
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    \draw[line width=1.25pt,->] (0,-10) -- (0,295) node[above,font=\scriptsize] {\small Prix (\euro)};
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    \draw[cyan,->,>=stealth,dashed] (0,200) -- (30,200) -- (30,0);
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\end{center}

$(d_1)~\leftrightarrow$ Classique~; $(d_2)~\leftrightarrow$ Essentiel~; $(d_3)~\leftrightarrow$ Liberté.

\item Le tarif Classique offre un prix proportionnel (fonction linéaire $h$).

\item
    \begin{enumerate}
        \item $(d_2)$ à $y=150$~: $x=20$ entrées au maximum.
        \item $(d_2)$ à $y=240$~: $x=38$~; à partir de 39 entrées, Liberté est le plus intéressant.
        \item Pour $y=200$, c'est $(d_2)$ qui donne le plus d'entrées (30) : tarif Essentiel.
    \end{enumerate}
\end{enumerate}

\bigskip

{\large \textbf{Exercice 4 \hfill 21 points}}

\medskip

\begin{center}
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    \node[above left] at (1,4) {H};
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\begin{center}
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\begin{enumerate}
\item EFGH rectangle, donc EF = 6. FJ = EJ $-$ EF $= 10 - 6 = 4$~m.
\item Triangle GFJ rectangle en F~: $\text{GJ}^2 = 3^2 + 4^2 = 25$, d'où GJ = 5~m.

Périmètre~: EF + FJ + JG + GH + HE $= 6 + 4 + 5 + 6 + 3 = 24$~m.

\item
    \begin{enumerate}
        \item $\mathcal{A}(\text{EFGJH}) = 3 \times 6 + \dfrac{3 \times 4}{2} = 18 + 6 = 24$~m$^2$.

Volume~: $\mathcal{V} = 24 \times 0{,}15 = 3{,}6$~m$^3 < 4$~m$^3$. ✓

        \item $4 \times 250 = \np{1000}$~kg de ciment.
        \item Ratio 2:7:5 → multiplié par 500~: gravier~= $\np{3500}$~kg~; sable~= $\np{2500}$~kg.
    \end{enumerate}
\item 2 couches sur 24~m$^2$ = 48~m$^2$. Peinture~: $\dfrac{48}{5} = 9{,}6$~L.

Deux pots A~: $79{,}90 + 39{,}95 = 119{,}85$~\euro{} (2\up{e} à 50\,\%). Un pot B~: 129,90~\euro. Les deux pots A coûtent moins cher.
\end{enumerate}

\bigskip

{\large \textbf{Exercice 5 \hfill 19 points}}

\medskip

\textbf{Partie A}

\medskip

\begin{minipage}{0.52\linewidth}
\begin{enumerate}
\item Triangle ABC~: $\widehat{A} = \widehat{B} = 60\degres$ (figure), donc $\widehat{C} = 180 - 60 - 60 = 60\degres$. Les trois angles sont égaux~: triangle ABC équilatéral.
\item CDE équilatéral (par symétrie/codage)~: $\widehat{\text{CED}} = 60\degres = \widehat{\text{CAB}}$. Ce sont des angles alternes-internes~: (AB) $\parallel$ (DE).
\end{enumerate}
\end{minipage}\hfill
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    \begin{scope}[shift={(1.9,0)}]
        \draw (120:0.5) arc (120:180:0.5);
    \end{scope}
    \node[below left] at (-1.9,0) {A};
    \node[below right] at (1.9,0) {B};
    \node[left] at (-0.6,4.2) {D};
    \node[right] at (0.6,4.2) {E};
    \node[right] at (0.1,3.15) {C};
    \fill (0,4.2) circle (2pt);
    \fill (-0.3,3.7) circle (2pt);
    \fill (0.3,3.7) circle (2pt);
    \node[font=\small] at (-1.1,0.35) {$60\degres$};
\end{tikzpicture}
\end{minipage}

\medskip

\textbf{Partie B}

\medskip

\begin{center}
\begin{tabularx}{\linewidth}{|X|X|}\hline
Programme & Bloc \textbf{triangle} \\[2pt]
\begin{scratch}[num blocks, scale=0.9]
\blockinit{Quand \greenflag est cliqué}
\blockmove{aller à x: \ovalnum{-180} y: \ovalnum{-150}}
\blockmove{s'orienter à \ovalnum{90} degrés}
\blockvariable{mettre \selectmenu{côté} à \ovalnum{240}}
\blockmoreblocks{triangle}
\blockmove{tourner \turnleft{} de \ovalnum{60} degrés}
\blockmove{avancer de \ovalnum{240} pas}
\blockvariable{mettre \selectmenu{côté} à \ovaloperator{\ovalvariable{côté} / \ovalnum{3}}}
\end{scratch}
&
\begin{scratch}[scale=0.9]
\initmoreblocks{définir \namemoreblocks{triangle}}
\blockpen{stylo en position d'écriture}
\blockrepeat{répéter \ovalnum{3} fois}
{
\blockmove{avancer de \ovalvariable{côté} pas}
\blockmove{tourner \turnleft{} de \ovalnum{120} degrés}
}
\blockpen{relever le stylo}
\end{scratch}
\\ \hline
\end{tabularx}
\end{center}

\begin{center}
\begin{tikzpicture}[x=1cm,y=1cm]
    \draw[orange,line width=1.5pt] (0,0) grid (11,9);
    \foreach [count=\i] \l in {A,B,C,D,E,F,G,H,I,J} {
        \node at (\i-0.5,8.5) {\small \l};
    }
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        \node at (0.5,9-\r-0.5) {\small \r};
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    \draw[->,line width=1pt] (3.5,0.5) -- (4.4,0.5);
\end{tikzpicture}
\end{center}

\begin{enumerate}
\item Point de départ~: coordonnées $(-180~;~-150)$ = case D8.
\item À la ligne 4, on écrit \textbf{240}.
\item Après la ligne 7, le lutin est en case \textbf{G3}.
\item `côté / 3`~: les côtés du petit triangle sont $\frac{1}{3}$ de ceux du grand ($240/3 = 80$~pas).
\end{enumerate}

\end{document}
