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\begin{center}\section*{\textcolor{red}{\decofourleft~Corrigé du baccalauréat S Centres étrangers 10 juin 2016~\decofourright \\}}\end{center}

\medskip

\subsection{\hfill \textcolor{blue}{(4 points)}}

\begin{enumerate}
\item \textbf{Affirmation 1} Dans une boulangerie industrielle, on prélève au hasard une baguette de pain  dans la production.

Soit M la variable aléatoire exprimant sa masse, en gramme. M suit la loi normale d'espérance $200$ et d'écart-type $10$.

On a $p(M\geqslant 187)=p(187\leqslant X\leqslant 200)+p(M>200)=p(187\leqslant X\leqslant 200)+0,5$.

À la calculatrice, on a $p(187\leqslant X\leqslant 200)>0,4$ et par conséquent 
$p(M\geqslant 187)>0,9$.

La probabilité que la masse de la baguette soit supérieure à $187$~g est supérieure à $0,9$.

\textbf{Affirmation 1: VRAIE}



\smallskip

\item \textbf{Affirmation 2}



Considérons la fonction $f$ définie sur $\left[0~;~\dfrac{\pi}{2}\right]$ par $f(x)=x-\cos(x)$.

$f$ est dérivable sur $\left[0;\dfrac{\pi}{2}\right]$ et $f'(x)=1+\sin(x)$.

Pour tout $x\in \left[0~;~\dfrac{\pi}{2}\right]$, $0\leqslant \sin(x)\leqslant 1$ et $1+\sin(x)>0$.

On en déduit que $f$ est strictement croissante sur $\left[0~;~\dfrac{\pi}{2}\right]$.

Comme $f$ est dérivable sur $\left[0;\dfrac{\pi}{2}\right]$, $f$ est continue sur $\left[0~;~\dfrac{\pi}{2}\right]$.

On a de plus $f(0)=-1$ et $f\left(\dfrac{\pi}{2}\right) =\dfrac{\pi}{2}$.

On a $0\in \left[-1~;~\dfrac{\pi}{2}\right]$.

On applique à $f$ le théorème des valeurs intermédiaires pour les fonctions continues et strictement monotones sur  un intervalle et on peut affirmer que l'équation $x-\cos x = 0$ admet une unique solution dans l'intervalle $\left[0~;~\dfrac{\pi}{2}\right]$.

\textbf{Affirmation 2: VRAIE}

\emph{Dans les questions $3$. et $4$., l'espace est rapporté à un repère orthonormal et l'on considère les droites $\mathscr{D}_1$ et $\mathscr{D}_2$ qui admettent pour représentations paramétriques respectives :}

\[\begin{cases}x=1+2t\\y=2-3t\\z=\phantom{1 +}4t\end{cases},\: t\in\mathbb{R}\hspace{1cm}\text{et}\hspace{1cm}\begin{cases}x=-5t'+3\\y=\phantom{-}2t'\\z=\phantom{-5}t'+4\end{cases},\: t' \in \mathbb{R}\]

\item\textbf{Affirmation 3}

Déterminons un vecteur directeur de chacune des ces deux droites.

Soit $\vect{u}_1$ un vecteur directeur de $\mathcal{D}_1$ et Soit $\vect{u}_2$ un vecteur directeur de $\mathcal{D}_2$

On a $\vect{u}_1
 \left(
\begin{array}{c}
-5 \\ 
2 \\ 
1
\end{array} \right)$ et $\vect{u}_2
 \left(
\begin{array}{c}
2 \\ 
-3 \\ 
4
\end{array} \right)$.

$\vect{u}_1$ et  $\vect{u}_2$ ne sont pas colinéaires. Les droites ne sont ni parallèles ni confondues.

On a:  $
 \left\{
\begin{array}{c}
1+2t=-5t'+3 \\ 
2-3t=2t' \\ 
4t=t'+4
\end{array}\right.\iff \left\{
\begin{array}{c}
1+2t=-20t+20+3 \\ 
2-3t=8t-8\\ 
t'=4t-4
\end{array}\right.\left\{
\begin{array}{c}
t=1 \\ 
t=10/11 \\ 
4t=t'+4
\end{array}\right.$

Le  système n'a pas de solution. les droites ne sont pas sécantes.

\textbf{Affirmation 3: FAUSSE}

\item \textbf{Affirmation 4}

Le plan $P$, d'équation $x+2y+z-3=0$ admet le vecteur $\vect{n}
 \left(
\begin{array}{c}
1 \\ 
2 \\ 
1
\end{array} \right)$ comme vecteur normal.

On a d'une part $\vect{n} \cdot \vect{u}_1=0$ .

D'après la représentation paramétrique de $\mathcal{D}_1$,  le point A de coordonnées $\text{A}(1~;~2~;~4)$ est  un point de $\mathcal{D}_1$.

Ses coordonnées ne  vérifient pas l'équation de $P$.

On en déduit que $\mathcal{D}_1 $ est parallèle à $ P$



\textbf{Affirmation 4: VRAIE}
\end{enumerate}

\subsection{\hfill \textcolor{blue}{(6 points)}}

\parbox{0.48\linewidth}{Soit $f$ une fonction définie sur l'intervalle $[0~;~1]$, continue et positive sur cet intervalle, et $a$ une réel tel que $0<a<1$.

On note  :

\begin{itemize}
\item $\mathscr{C}$ la courbe représentative de la fonction $f$ dans un repère orthogonal :

\item $\mathscr{A}_1$ l'aire du domaine plan limité par l'axe des abscisses et la courbe $\mathscr{C}$ d'une part, les droites d'équations $x=0$ et $x=a$ d'autre part.

\item $\mathscr{A}_2$ l'aire du domaine plan limité par l'axe des abscisses et la courbe $\mathscr{C}$ d'une part, les droites d'équations $x=a$ et $x=1$ d'autre part.

\end{itemize}}\hfill
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\bigskip
%Le but de cet exercice est de déterminer, pour différentes fonctions $f$, une valeur du réel $a$ vérifiant la condition (E) : \og{}les aires $\mathscr{A}_1$ et $\mathscr{A}_2$ sont égales\fg{}.

%On admet l'existence d'un tel réel $a$ pour chacune des fonctions considérées.

\bigskip

\textbf{Partie A : Étude de quelques exemples}

\medskip

\begin{enumerate}
\item %Vérifier que dans les cas suivants, la condition (E) est remplie pour un unique réel $a$ et déterminer sa valeur.
\begin{enumerate}
\item Soit $f$  une fonction constante strictement positive sur [0;1].

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\begin{minipage}{0.8\textwidth}
Alors il existe un réel $k>0$ tel que pour tout $x\in [0;1]$, $f(x)=k$.

Pour tout réel $a\in]0~;~1[$, on a : $\mathscr{A}_1 = ka$ et $\mathscr{A}_2 = (1-a)k$.

On a: $\mathscr{A}_1=\mathscr{A}_2\iff ka=k(1-a)$. Et comme $k>0$,  $\mathscr{A}_1=\mathscr{A}_2\iff a=\dfrac{1}{2}$.
\end{minipage}

Ce qui prouve que si la fonction $f$ est continue , constante et strictement positive sur $[0~;~1]$, les aires $\mathscr{A}_1$ et $\mathscr{A}_2$ sont égales pour une seule valeur de $a$ et $a=\dfrac{1}{2}$.

\item Soit $f$ la fonction définie sur $[0~;~1]$ par $f(x)=x$.

$f$ est continue sur $[0~;~1]$ et la fonction $F$ définie sur $[0~;~1]$ par $F(x)=\dfrac{1}{2}x^2$ est une primitive de $f$.

On a: $\mathscr{A}_1=\displaystyle\int_0^a{x\text{d}x}= \left[\dfrac{1}{2}x^2\right]_0^a=\dfrac{1}{2}a^2$


et: $\mathscr{A}_2=\displaystyle\int_a^1{x\text{d}x}= \left[\dfrac{1}{2}x^2\right]_a^1=1-\dfrac{1}{2}a^2$.

On en déduit pour $a\in]0~;~1[$: $\mathscr{A}_1=\mathscr{A}_2\iff \dfrac{1}{2}a^2=1-\dfrac{1}{2}a^2 \iff a=\dfrac{\sqrt{2}}{2}$.


\begin{minipage}{0.5\textwidth}
Remarque: on peut aussi tout simplement calculer les aires du triangle et du trapèze sur le graphique ci-contre et égaliser

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\end{enumerate}

\item 
\begin{enumerate}
\item À l'aide d'intégrales, exprimons, en unités d'aires, les aires $\mathscr{A}_1$ et $\mathscr{A}_2$.

Comme $f\geqslant 0$ sur $[0;1]$, on a: 
$\mathscr{A}_1=\displaystyle\int_0^a{f(x)\text{d}x}$ et $\mathscr{A}_2=\displaystyle\int_a^1{f(x)\text{d}x}$.


\item On note $F$ une primitive de la fonction $f$ sur l'intervalle $[0~;~1]$.

Soit  $a$ un réel de $]0;1[$ qui satisfait la condition (E), c'est à dire tel que $\mathscr{A}_1=\mathscr{A}_2$.

Si F est une primitive de $f$ sur $[0;1]$, alors 

$(\mathscr{A}_1=\mathscr{A}_2 \Rightarrow F(a)-F(0)= F(1)-F(a))$ et finalement:

Si $a$ un réel de $]0;1[$ qui satisfait la condition (E)
 alors $F(a)=\dfrac{F(0)+F(1)}{2}$.

Supposons maintenant que F est une primitive de $f$ telle que $F(a)=\dfrac{F(0)+F(1)}{2}$.

On a: $\mathscr{A}_1=F(a)-F(0)=\dfrac{F(0)+F(1)}{2}-F(0)= \dfrac{F(1)-F(0)}{2}$ 

et:
 $\mathscr{A}_2=F(1)-F(a)=F(1)+\dfrac{F(0)+F(1)}{2}= \dfrac{F(1)-F(0)}{2}$ 

Si F est une primitive telle que $F(a)=\dfrac{F(0)+F(1)}{2}$ alors $\mathscr{A}_1=\mathscr{A}_2$ et la condition (E) est vérifiée: la réciproque est vraie.



\end{enumerate}

\item Dans cette question, on  envisage deux autres fonctions particulières.
\begin{enumerate}
\item Considérons la fonction $f$  définie pour tout réel $x$ de $[0~;~1]$ par $f(x)=\mathrm{e}^{x}$.

Montrons que la condition (E) est vérifiée pour un unique réel $a$.

$f$ est continue sur $[0~;~1]$ et admet pour primitive $F(x)=\mathrm{e}^x$.

On a $\mathscr{A}_1=\displaystyle\int_0^a{\mathrm{e}^{x}\text{d}x}=\left[\mathrm{e}^{x}\right]_0^a=\mathrm{e}^{a}-1$ et $\mathscr{A}_2=\displaystyle\int_a^1{\mathrm{e}^{x}\text{d}x}=\left[\mathrm{e}^{x}\right]_a^1=\mathrm{e}-\mathrm{e}^{a}$.

On en déduit que:

$\mathscr{A}_1=\mathscr{A}_2 \Rightarrow \mathrm{e}^{a}-1=\mathrm{e}-\mathrm{e}^{a}\iff  \mathrm{e}^{a} = \dfrac{1+\mathrm{e}}{2}\iff a= \ln\left(\dfrac{1+\mathrm{e}}{2}\right)$ .





\item Soit $f$ la fonction définie pour tout réel $x$ de $\left[0~;~1\right]$ par $f(x)=\dfrac{1}{(x+2)^2}$.



Vérifions que la valeur $a=\dfrac{2}{5}$ convient.


On a d'une part:  $\displaystyle\int_0^{\frac{2}{5}}{f(x)\text{d}x}= \left[\dfrac{-1}{x+2}\right]_0^{\frac{2}{5}}=\dfrac{-1}{\frac{2}{5}+2}+\dfrac{1}{2}= \dfrac{1}{12}$.


D'autre part: $\displaystyle\int_{\frac{2}{5}}^1{f(x)\text{d}x}= \left[\dfrac{-1}{x+2}\right]_{\frac{2}{5}}^1=\dfrac{-1}{3}+\dfrac{1}{\frac{2}{5}+2}= \dfrac{1}{12}$

On a pour $a=\dfrac{2}{5}$, $\mathscr{A}_1=\mathscr{A}_2$. La condition (E) est vérifiée.
\end{enumerate}

Remarque:  pour ces deux questions on peut utiliser A. 2. b., c'est à dire
$a$ est solution de (E) si et seulement si $F(a) = \dfrac{F(0)+F(1)}{2}$
\end{enumerate}

\bigskip

\textbf{Partie B : Utilisation d'une suite pour déterminer une valeur approchée de $a$}

\bigskip
Dans cette partie, on considère la fonction $f$ définie pour tout réel $x$ de $[0~;~1]$ par $f(x)= 4 - 3x^2$.

\medskip

\begin{enumerate}
\item Démontrons que si $a$ est un réel satisfaisant la condition (E), alors $a$ est solution de l'équation : \[x=\dfrac{x^3}{4}+\dfrac{3}{8}.\]

Si $a$ est un réel satisfaisant la condition (E),on a vu au A. 2. b que si F est une primitive de $f$ alors: \[F(a) = \dfrac{F(0)+F(1)}{2}.\]

Pour tout réel $x$ de $[0~;~1]$ , la fonction $f$ définie par  $f(x)=4-3x^2$ admet comme primitive sur $[0~;~1]$, la fonction $F$ définie par $F(x)= 4x - x^3.$

On a alors: $F(0)=0$ et $F(1)=3$ et:

$F(a)=\dfrac{F(0)+F(1)}{2} \iff 4a-3a^3=3 \iff a=\dfrac{a^3}{4}+\dfrac{3}{8}$.

Par conséquent, si $a$ est un réel satisfaisant la condition (E), $a$ est solution de l'équation : \[x=\dfrac{x^3}{4}+\dfrac{3}{8}.\]








Dans la suite de l'exercice, on admettra que cette équation a une unique solution dans l'intervalle $[0~;~1]$. On note $a$ cette solution.

\item On considère la fonction $g$ définie pour tout réel $x$ de $[0~;~1]$ par $g(x)=\dfrac{x^3}{4}+\dfrac{3}{8}$ et la suite $\left(u_n\right)$ définie par : $u_0=0$ et, pour tout entier naturel $n$, $u_{n+1}=g\left(u_n\right)$.
\begin{enumerate}
\item Calculons $u_1$.

On a $u_1=g(u_0)=g(0)=\dfrac{3}{8}$

\item Démontrons que la fonction $g$ est croissante sur l'intervalle $[0~;~1]$.

$g$ est dérivable sur $[0;1]$ et pour tout $x\in [0;1]$, $g'(x)=\dfrac{3}{4}x^2\geqslant 0$. On en déduit que $g$ est croissante sur $[0;1]$.

\item Démontrons par récurrence que, pour tout entier naturel $n$, on a $0\leqslant u_n\leqslant u_{n+1}\leqslant 1$.

Considérons la proposition $(P_n)$ définie sur $\mathbb{N}$ par :
\[0\leqslant u_n\leqslant u_{n+1}\leqslant 1.\]

\textbf{Initialisation}
$(0\leqslant u_0\leqslant u_{1}\leqslant 1)$ est vraie.



\textbf{Hérédité}
Comme $g$ est croissante sur $[0;1]$, 

\[(0\leqslant u_n \leqslant u_{n+1}\leqslant 1)\Rightarrow (g(0)\leqslant g(u_n)\leqslant g(u_{n+1})\leqslant g( 1)).\]



Mais $g(0)=\dfrac{3}{8}>0$ et $g(1)=\dfrac{5}{8}<1$.

On en déduit:

\[(0\leqslant u_n\leqslant u_{n+1}\leqslant 1)\Rightarrow (0\leqslant u_{n+1}\leqslant u_{n+1}\leqslant g( 1)).\]



\textbf{Conclusion}

$P_0$ est vraie.

Pour tout entier naturel $n$, $(P_n \Rightarrow P_{n+1})$ est vraie.

D'après l'axiome de récurrence, $P_n$ est vraie quelque soit $n\in \mathbb{N}$.


\item Prouvons que la suite $\left(u_n\right)$ est convergente.

On vient de démontrer que pour tout entier naturel $n$, on a: $0\leqslant u_n\leqslant u_{n+1}\leqslant 1$, autrement dit que la suite $(u_n)$ est croissante et majorée par 1. On peut en déduire qu'elle converge. Soit L sa limite.

Par définition de $u_n$, et par unicité de la limite, on en déduit que $g(L)=L$ et L est solution de l'équation $ x=\dfrac{x^3}{4}+\dfrac{3}{8}$ d'où $L=a$.




\item On admet que le réel $a$ vérifie l'inégalité $0< a-u_{10}<10^{-9}$. 


A la calculatrice, on obtient; $u_{10}=0,38980784$ à $10^{-8}$ près.

\end{enumerate}
\end{enumerate}

\subsection{\hfill \textcolor{blue}{(5 points)}}
Un institut effectue un sondage pour connaître, dans une population donnée, la proportion de personnes qui sont favorables à un projet d'aménagement du territoire. Pour cela, on interroge un échantillon aléatoire de personnes de cette population, et l'on pose une question à chaque personne.

\bigskip

%\emph{Les trois parties sont relatives à cette même situation, mais peuvent être traitées de manière indépendante.}

\bigskip

\textbf{Partie A : Nombre de personnes qui acceptent de répondre au sondage}


\bigskip

On admet dans cette partie que la probabilité qu'une personne interrogée accepte de répondre à la question est égale à 0,6.
\begin{enumerate}
\item L'institut de sondage interroge 700 personnes. On note $X$ la variable aléatoire correspondant  au nombre de personnes interrogées qui acceptent de répondre à la question posée.





\begin{enumerate}
\item Pour chaque personne il n'y a que deux issues possibles: répondre/ne pas répondre.

Si de plus on fait l'hypothèse que les choix des personnes sont indépendants, on peut affirmer que $X$ suit une loi binomiale de paramètres $n=700$ et $p=0,6$.

\item 

On a :  $P(X\geqslant 400)=1-P(X< 400)=1- P(X\leqslant 399)$.


  La calculatrice donne $P(X\leqslant 399)\simeq 0,0573$ et $P(X\geqslant 400)\simeq 0,9427$.
  
  La meilleure valeur approchée parmi les solutions proposées est 0,94. 
\end{enumerate}

\item Déterminons combien de personnes l'institut doit interroger au minimum pour garantir, avec une probabilité supérieure à 0,9, que le nombre de personnes répondant au sondage soit supérieur ou égal à 400.

On peut déjà affirmer que $n\leqslant 700$.

Une exploration à la calculatrice donne $n=694$ car si Y désigne la variable aléatoire de loi binomiale de paramètres $n$ et $p=0,6$ on a pour $n=694$: $p(Y\geqslant 400) \geqslant 0,9$ et 
pour $n=693$: $p(Y\geqslant 400) < 0,9$.

Il suffit donc d'interroger 694 personnes pour garantir, avec une probabilité supérieure à 0,9, que le nombre de personnes répondant au sondage soit supérieur ou égal à 400.

\end{enumerate}


\bigskip

\textbf{Partie B : Proportion de personnes favorables au projet dans la population}

\bigskip

Dans cette partie, on suppose que $n$ personnes ont répondu à la question ,et on admet que ces personnes constituent un échantillon aléatoire de taille $n$ (où $n$ est un entier naturel supérieur à 50).

Parmi ces personnes, 29\:\% sont favorables au projet d'aménagement.

\medskip

\begin{enumerate}
\item Un intervalle de confiance, au niveau de confiance de 95\:\%, de la proportion de personnes qui sont favorables au projet dans la population totale est de la forme:

$\left[f_{obs}-\dfrac{1}{\sqrt{n}}~;f_{obs}+\dfrac{1}{\sqrt{n}}\right]$.

Et ici avec $f_{obs}=0,29$, on a  $\left[0,29-\dfrac{1}{\sqrt{n}};0,29+\dfrac{1}{\sqrt{n}}\right]$.

\item L'intervalle de confiance, au niveau de confiance de 95\:\% a pour amplitude $\dfrac{2}{\sqrt{n}}$.


$\dfrac{2}{\sqrt{n}}\leqslant 0,04 \iff \sqrt{n}\geqslant \dfrac{2}{0,04}\iff n \geqslant \np{2500}$.

Il suffit d'interroger plus de \np{2500} personnes pour que l'amplitude de l'intervalle de confiance au niveau de 95\,\% soit inférieure à $0,04$.
\end{enumerate}
\bigskip

\textbf{Partie C : Correction due à l'insincérité de certaines réponses}

\medskip

Dans cette partie, on suppose que, parmi les personnes sondées qui ont accepté de répondre à la question posée, 29\:\% affirment qu'elles sont favorables au projet.

L'institut de sondage sait par ailleurs que la question posée pouvant être gênante pour les personnes interrogées, certaines d'entre elles ne sont pas sincères et répondent le contraire de leur opinion véritable. Ainsi, une personne qui se dit favorable peut :

\begin{itemize}
\item soit être en réalité favorable au projet si elle est sincère.
\item soit être en réalité défavorable au projet si elle n'est pas sincère.
\end{itemize}

Par expérience, l'institut estime à 15\:\% le taux de réponses non sincères parmi les personnes ayant répondu, et admet que ce taux est le même quelle que soit l'opinion de la personne interrogée.

Le but de cette partie est, à partir de ces données, de déterminer le taux réel de personnes favorables au projet, à l'aide d'un modèle probabiliste. On prélève au hasard la fiche d'une personne ayant répondu, et on définit :
 
\begin{itemize}
\item[$\bullet$] $F$  l'évènement \og{}la personne est en réalité favorable au projet\fg{} ;
\item[$\bullet$] $\overline{F}$  l'évènement \og{}la personne est en réalité défavorable au projet\fg{} ;
\item[$\bullet$] $A$ l'évènement \og{}la personne affirme qu'elle est favorable au projet\fg{};
\item[$\bullet$] $\overline{A}$ l'évènement \og{}la personne affirme qu'elle est défavorable au projet\fg{}.
\end{itemize}
Ainsi, d'après les données, on a $p(A) = 0,29$.

\begin{enumerate}
\item  L'institut estime à 15\:\% le taux de réponses non sincères parmi les personnes ayant répondu, et admet que ce taux est le même quelle que soit l'opinion de la personne interrogée.

On en déduit que $P_{\overline{F}}(A)=P_F\left(\overline{A}\right)=0,15$.

Et comme $P_F(A)+P_F\left(\overline{A}\right)=1$, on a: $p_F(A) = 0,85$.
\item On pose $x = P(F)$.
\begin{enumerate}
\item On a l'arbre de probabilité ci-dessous.

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\item D'après la formule des probabilités totales, 

$P(A) = P(A\cap F) + P\left(A\cap \overline{F}\right)$.

Mais $P(A) = 0,29$, $P(A\cap F) = 0,85x$ et $P\left(A\cap \overline{F}\right) =  0,15(1 - x)$.

On en déduit que $x$ est solution de $0,29 = 0,85x + 0,15(1 - x)$.
\end{enumerate}

\medskip

\item Déterminons, parmi les personnes ayant répondu au sondage, la proportion de celles qui sont réellement favorables au projet.

Il suffit de résoudre $0,29 = 0,85x + 0,15(1 - x)$. On trouve $x = \dfrac{14}{70} = 0,2$.

20\,\% des personnes ayant répondu au sondage sont favorables au projet.

\end{enumerate}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{\hfill \textcolor{blue}{(5 points)}}

\textbf{Candidat/e/s n'ayant pas choisi la spécialité mathématique}

\medskip

On veut modéliser dans le plan la coquille d'un nautile à l'aide d'une ligne brisée en forme de spirale. On s'intéresse à l'aire délimitée par cette ligne.

\bigskip

On munit le plan d'un repère orthonormal direct $\left(O~;~\overrightarrow{u}~;~\overrightarrow{v}\right)$.

Soit $n$ un entier supérieur ou égal à 2. Pour tout entier $k$ allant de 0 à $n$, on définit les nombres complexes $z_k=\left(1+\dfrac{k}{n}\right)\mathrm{e}^{\mathrm{i}\frac{2k\pi}{n}}$ et on note $M_k$ le point d'affixe $z_k$.

Dans ce modèle, le pourtour du nautile est la ligne brisée reliant tous les points $M_k$ avec $0\leqslant k\leqslant n$.

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%\bigskip
\textbf{Partie A : Ligne brisée formée à partir de sept points}

\bigskip
Dans cette partie, on suppose que $n=6$. %Ainsi, pour $0\leqslant k\leqslant 6$, on a $z_k=\left(1+\dfrac{k}{6}\right)\mathrm{e}^{\mathrm{i}\frac{2k\pi}{6}}$.
\begin{enumerate}
\item Déterminons la forme algébrique de $z_1$.

On a : $z_1=\left(1+\dfrac{1}{6}\right)\mathrm{e}^{\mathrm{i}\frac{2\pi}{6}}=
\left(\dfrac{7}{6}\right)\left(\cos\left(\frac{\pi}{3}\right)+\text{i}\sin \left(\frac{\pi}{3}\right)\right)=
\dfrac{7}{6}\left(\dfrac{1}{2}+\text{i}\dfrac{\sqrt{3}}{2}\right)= \dfrac{7}{12}+\text{i}\dfrac{7\sqrt{3}}{12}
$.

\item On a $z_0=\mathrm{e}^{0}=1$ et $z_6=\left(1+\dfrac{6}{6}\right)\mathrm{e}^{\mathrm{i}{2\pi}}=2$.

\item Calculons la longueur de la hauteur issue de $M_1$ dans le triangle $OM_0M_1$ .

Soit H le pied de la hauteur issue de $M_1$ dans le triangle $OM_0M_1$. Comme O et $M_0$ sont pour abscisse 0, on a: $M_1H=\dfrac{7\sqrt{3}}{12}$.


L'aire du triangle $OM_0M_1$ est égale à:  $\dfrac{1}{2}OM_0\times M_1H= \dfrac{7\sqrt{3}}{24}$.
\end{enumerate}

\bigskip
\textbf{Partie B : Ligne brisée formée à partir de \boldmath$n+1$\unboldmath\: points}

Dans cette partie, $n$ est un entier supérieur ou égal à 2.

\begin{enumerate}
\item Pour tout entier $k$ tel que $0\leqslant k\leqslant n$, déterminons la longueur $OM_k$.

On a $OM_k=|z_k|=1+\dfrac{k}{n}$, car pour tout entier naturel $n$ est un entier supérieur ou égal à 2,  tout entier $k$ tel que $0\leqslant k\leqslant n$,  le module de $\mathrm{e}^{\mathrm{i}\frac{2k\pi}{n}}$ vaut 1.

\item Soit $k$ entier tel que $0\leqslant k\leqslant n-1$.

  $\left(\vect{u}~;~\vect{OM_k}\right)= \arg(z_k)=\dfrac{2k\pi}{n}$ et $\left(\vect{u}~;~\vect{OM_{k+1}}\right)=\arg(z_k)=\dfrac{2(k+1)\pi}{n}$ .

Or $\left(\vect{OM_k}~;~\vect{OM_{k+1}}\right)= \left(\vect{u}~;~\vect{OM_{k+1}}\right)-\left(\vect{u}~;~\vect{OM_k}\right)$ à $2\pi$ près.

On en déduit que $\left(\vect{OM_k}~;~\vect{OM_{k+1}}\right)=\dfrac{2(k+1)\pi}{n}-\dfrac{2k\pi}{n}=\dfrac{2\pi}{n}$ à $2\pi$ près.

\item Pour $k$ entier tel que $0\leqslant k\leqslant n-1$, calculons la longueur de la hauteur issue de $M_{k+1}$ dans le triangle $OM_kM_{k+1}$.

\begin{minipage}{0.6\textwidth}
Soit $H_{k+1}$ le pied de la hauteur issue de $M_{k+1}$ dans le triangle $OM_kM_{k+1}$. 
\medskip

On a:
$\dfrac{M_{k+1}H_{k+1}}{OM_{k+1}}=\sin\left(\vect{OM_k}~;~\vect{OM_{k+1}}\right)$.
\medskip

Mais $OM_{k+1}=1+\dfrac{k+1}{n}$ et $\sin\left(\vect{OM_k}~;~\vect{OM_{k+1}}\right)= \sin\left(\dfrac{2\pi}{n}\right).$

On en déduit que pour $k$ entier tel que $0\leqslant k\leqslant n-1$,

$M_{k+1}H_{k+1}=\left(1+\dfrac{k+1}{n}\right)\sin\left(\dfrac{2\pi}{n}\right)$
\end{minipage}
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\item On admet que l'aire du triangle $OM_kM_{k+1}$ est égale à $a_k=\dfrac{1}{2}\sin\left(\dfrac{2\pi}{n}\right)\times \left(1+\dfrac{k}{n}\right)\left(1+\dfrac{k+1}{n}\right)$ et que l'aire totale délimitée par la ligne brisée est  égale à $A_n = a_0 + a_1 + \cdots + a_n$.

L'algorithme suivant permet de calculer l'aire $A_n$ lorsqu'on entre l'entier $n$ :

\begin{center}
\fbox{\begin{tabular}{ll}
VARIABLES&$A$ est un nombre réel\\
&$k$ est un entier\\
&$n$ est un entier\\
TRAITEMENT&Lire la valeur de $n$\\
&$A$ prend la valeur 0\\
&Pour $k$ allant de $0$ à $n - 1$\\
&\hspace{0,5cm} $A$ prend la valeur $A+\dfrac{1}{2}\sin\left(\dfrac{2\pi}{n}\right)\times \left(1+\dfrac{k}{n}\right)\left(1+\dfrac{k+1}{n}\right)$\\
&Fin Pour\\
SORTIE&Afficher $n$
\end{tabular}}

\end{center}
On entre dans l'algorithme $n = 10$

On obtient le tableau ci-dessous qui illustre le fonctionnement de l'algorithme.

\begin{tabularx}{\linewidth}{|l|*{11}{>{\centering \arraybackslash}X|}} \hline
$k$&0&1&2&3&4&5&6&7&8&9\\
\hline
$A$&0,323&0,711&1,170&10705&2,322&3,027&3,826&4,726&5,731&6,848\\
\hline
\end{tabularx}

\item On admet que $A_2=0$ et que la suite $\left(A_n\right)$ converge et que $\lim_{n\rightarrow +\infty}A_n=\dfrac{7\pi}{3}\approx 7,3$.

En L6 tant que $A < 7,2$
 
En L13 Afficher $n$.

%\medskip
%\begin{center}
%\fbox{\begin{tabular}{lll}
%$L1$&VARIABLES :&$A$ est un nombre réel\\
%$L2$&&$k$ est un entier\\
%$L3$&&$n$ est un entier\\
%$L4$&TRAITEMENT :&$n$ prend la valeur 2\\
%$L5$&&$A$ prend la valeur 0\\
%$\textbf{L6}$&&\textbf{Tant que}\dots\dots\dots\dots\\
%$L7$&&\hspace{0,8cm}$n$ prend la valeur $n+1$\\
%$L8$&&\hspace{0,8cm}$A$ prend la valeur 0\\
%$L9$&&\hspace{0,8cm}Pour $k$ allant de 0 à $n-1$\\
%$L10$&&\hspace{1.6cm}$A$ prend la valeur $A+\dfrac{1}{2}\sin\left(\dfrac{2\pi}{n}\right)\times \left(1+\dfrac{k}{n}\right)\left(1+\dfrac{k+1}{n}\right)$\\
%&&\hspace{0,8cm}Fin Pour\\
%$L12$&&Fin Tant que\\
%$\textbf{L13}$&\textbf{SORTIE :}&\textbf{Afficher} \dots\\
%\end{tabular}}
%\end{center}
\end{enumerate}


\label{fin}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{\hfill \textcolor{blue}{(5 points)}}

\textbf{Candidat/e/s ayant choisi la spécialité mathématique}

\medskip

Le but de cet exercice est d'étudier, sur un exemple, une méthode de chiffrement publiée en 1929
par le mathématicien et cryptologue Lester Hill. Ce chiffrement repose sur la donnée d'une matrice $A$, connue  uniquement de l'émetteur et du destinataire.


Dans tout l'exercice,  on note $A$ la matrice définie par : $A = \begin{pmatrix}5&2\\7&7\end{pmatrix}$.


\bigskip

\textbf{Partie A -- Chiffrement de Hill}

\medskip

Voici les différentes étapes de chiffrement pour un mot comportant un nombre pair de lettres :

\begin{center}
\begin{tabularx}{\linewidth}{|l|X|}\hline
Étape 1&On divise le mot en blocs de deux lettres consécutives puis, pour chaque bloc, on
effectue chacune des étapes suivantes.\\ \hline
Étape 2&On associe aux deux lettres du bloc les deux entiers $x_1$ et $x_2$ tous deux compris entre 0
et 25, qui correspondent aux deux lettres dans le même ordre, dans le tableau suivant :

\begin{tabular}{|*{13}{c|}}\hline
A	&B	&C 	&D 	&E 	&F 	&G 	&H 	&I 	&J 	&K 	&L 	&M\\ \hline
0	&1 	&2 	&3 	&4 	&5 	&6 	&7 	&8 	&9 	&10 &11 &12\\ \hline\hline
N	&O 	&P 	&Q	&R 	&S 	&T 	&U 	&V	&W 	&X	&Y 	&Z\\ \hline
13 	&14 &15 &16 &17 &18 &19 &20 &21	&22 &23 &24 &25\\ \hline
\end{tabular}
\\ \hline
Étape 3& On transforme la matrice $X = \begin{pmatrix}x_1\\x_2\end{pmatrix}$ en la matrice $Y = \begin{pmatrix}y_1\\y_2\end{pmatrix}$ vérifiant $Y = A X$.\\ \hline
Étape 4&On transforme la matrice $Y = \begin{pmatrix}y_1\\y_2\end{pmatrix}$ en la matrice $R = \begin{pmatrix}r_1\\r_2\end{pmatrix}$, où $r_1$ est le reste de la division euclidienne de $y_1$ par 26 et $r_2$ celui de la division euclidienne de $y_2$ par 26.\\ \hline
Étape 5&On associe aux entiers $r_1$ et $r_2$ les deux lettres correspondantes du tableau de l'étape 2.

Le bloc chiffré est le bloc obtenu en juxtaposant ces deux lettres.\\ \hline
\end{tabularx}
\end{center}

\textbf{Question :} utiliser la méthode de chiffrement exposée pour chiffrer le mot \og HILL \fg.



\begin{minipage}{\textwidth}
HI donne $X=\begin{pmatrix}7\\8\end{pmatrix}$

$AX=\begin{pmatrix}5&2\\7&7\end{pmatrix}\begin{pmatrix}7\\8\end{pmatrix}= \begin{pmatrix}51\\105\end{pmatrix}$ 

Mais $51\equiv 25 [26]$ et $105\equiv 1 [26]$.

On en déduit que HI est codé par $Y=\begin{pmatrix}25\\1\end{pmatrix}$, c'est à dire ZB.

\end{minipage}
\begin{minipage}{\textwidth}
LL donne $X=\begin{pmatrix}11\\11\end{pmatrix}$

$AX=\begin{pmatrix}5&2\\7&7\end{pmatrix}\begin{pmatrix}11\\11\end{pmatrix}= \begin{pmatrix}77\\154\end{pmatrix}$ 

Mais $77\equiv 25 [26]$ et $154\equiv 24 [26]$.

On en déduit que LL est codé par $Y=\begin{pmatrix}25\\24\end{pmatrix}$, c'est à dire ZY.

\end{minipage}

On en déduit que HILL est codé ZBZY.


\bigskip

\textbf{Partie B - Quelques outils mathématiques nécessaires au déchiffrement}

\begin{enumerate}
\item Soit $a$ un entier relatif premier avec 26.

Démontrons qu'il existe un entier relatif $u$ tel que $u \times a \equiv  1 \:\text{modulo}\: 26$.

Comme $a$ un entier relatif premier avec 26, d'après le théorème de Bézout, il existe deux entiers relatifs $u$ et $v$ tels que $au+26v=1$ et alors $au\equiv 1 [26]$


\item On considère l'algorithme suivant :

\begin{center}
\begin{tabularx}{\linewidth}{|l|X|}\hline
VARIABLES :	&$a, u$, et $r$ sont des nombres ($a$ est naturel et premier avec 26)\\
TRAITEMENT :&Lire $a$\\
			&$u$ prend la valeur 0, et $r$ prend la valeur 0\\
			&Tant que $r \ne  1$\\
			&\hspace{0,8cm}$u$ prend la valeur $u + 1$\\
			&$\hspace{0,8cm}r$ prend la valeur du reste de la division euclidienne de $u \times a$ par 26
\\
			&Fin du Tant que\\
SORTIE 		&Afficher $u$\\ \hline
\end{tabularx}
\end{center}

On entre la valeur $a = 21$ dans cet algorithme.
	\begin{enumerate}
		\item On a:
		étape 1  u=1; ua=21; r=21
		
		étape 2  u=2; ua=42; r=16
		\\
		étape 3  u=3; ua=63; r=11
		\\
		étape 4  u=4; ua=84; r=6
		\\
		étape 5  u=5; ua=105; r=1
				\\
		
		
		
		
	On a le tableau suivant, jusqu'à l'arrêt de l'algorithme.
		
\begin{center}
\begin{tabularx}{0.7\linewidth}{|*{7}{>{\centering \arraybackslash}X|}X}\hline
$u$&0&1&2&3&4&5\\ \hline
$r$&0&21&16&11&6&1\\ \hline
\end{tabularx}
\end{center}
		\item En déduire que $5 \times 21 \equiv 1 \:\:\text{modulo}\:\: 26$.
	\end{enumerate}
\item On a $A = \begin{pmatrix}5&2\\7&7\end{pmatrix}$ et on note $I$ la matrice : $I = \begin{pmatrix}1&0\\0&1\end{pmatrix}$.
	\begin{enumerate}
		\item  $12A - A^2=12\begin{pmatrix}5&2\\7&7\end{pmatrix}-\begin{pmatrix}5&2\\7&7\end{pmatrix}\begin{pmatrix}5&2\\7&7\end{pmatrix}=
		\begin{pmatrix}60&24\\84&84\end{pmatrix}-\begin{pmatrix}39&24\\84&63\end{pmatrix}=
		\begin{pmatrix}21&0\\0&21\end{pmatrix}=21I$.
		
		\item On vient de voir que $12A-A^2=21I$.
		
		Mais $12A-A^2=(12I-A)\times A$ 
		
		En posant $B=12I-A=12\begin{pmatrix}1&0\\0&1\end{pmatrix}-\begin{pmatrix}5&2\\7&7\end{pmatrix}=\begin{pmatrix}7&-2\\-7&5\end{pmatrix}$ on a $BA = 21I$.
		\item Démontrons que si $A X = Y$, alors $21 X = B Y$.
		
		$AX=Y \Rightarrow BAX=BY \Rightarrow 21IX=BY \Rightarrow 21X=BY$ 
	\end{enumerate}
\end{enumerate}
	
\bigskip
	
\textbf{Partie C - Déchiffrement}
	
\medskip

On veut déchiffrer le mot VLUP.

On note $X = \begin{pmatrix}x_1\\x_2\end{pmatrix}$ la matrice associée, selon le tableau de correspondance, à un bloc de deux lettres
avant chiffrement, et $Y = \begin{pmatrix}y_1\\y_2\end{pmatrix}$ la matrice définie par l'égalité : $Y =  A X = \begin{pmatrix}5&2\\7&7\end{pmatrix}X$.

Si $r_1$ et $r_2$ sont les restes respectifs de $y_1$ et $y_2$ dans la division euclidienne par 26, le bloc de deux
lettres après chiffrement est associé à la matrice $R = \begin{pmatrix}r_1\\r_2\end{pmatrix}$.

\medskip


\begin{enumerate}
\item Démontrons que : $\left\{\begin{array}{l c l}
21x_1 &=& \phantom{-}7y_1 - 2y_2\\
21x_2 &=&- 7y_1 + 5 y_2
\end{array}\right.$

On vient de voir que si  $A X = Y$, alors $21 X = B Y$.

Or $X = \begin{pmatrix}x_1\\x_2\end{pmatrix}$ , $B=\begin{pmatrix}7&-2\\-7&5\end{pmatrix}$ et $Y = \begin{pmatrix}y_1\\y_2\end{pmatrix}$ 

Si $21 X = B Y$ alors $21 \begin{pmatrix}x_1\\x_2\end{pmatrix} =\begin{pmatrix}7&-2\\-7&5\end{pmatrix} \begin{pmatrix}y_1\\y_2\end{pmatrix}$ ou encore

$\left\{\begin{array}{l c l}
21x_1 &=& \phantom{-}7y_1 - 2y_2\\
21x_2 &=&- 7y_1 + 5 y_2
\end{array}\right.$

\item 

$\left\{\begin{array}{l c l}
21x_1 &=& \phantom{-}7y_1 - 2y_2\\
21x_2 &=&- 7y_1 + 5 y_2
\end{array}\right.\Rightarrow \left\{\begin{array}{l c l}
5\times 21 x_1 &=& \phantom{-}35 y_1 - 10 y_2\\
5\times 21 x_2 &=&- 35 y_1 +25 y_2
\end{array}\right.$

Mais d'après 2: $21\times 5\equiv 1[26]$.

$\phantom{-}35\equiv 9[26]$\\
$-10\equiv 16[26]$\\
$-35\equiv 17[26]$\\
$\phantom{-}25\equiv 25[26]$\\
$\phantom{-}y_1\equiv r_1 \:[26]$\\
$\phantom{-}y_2\equiv r_2 \:[26]$\\

$\left\{\begin{array}{l c l}
21x_1 &=& \phantom{-}7 y_1 - 2 y_2\\
21x_2 &=&- 7 y_1 + 5 y_2
\end{array}\right.\Rightarrow \left\{\begin{array}{l c l}
 x_1 &\equiv& 9 r_1 + 16 r_2 \:[26]\\
 x_2 &\equiv& 17 r_1 +25 r_2 \:[26]
\end{array}\right.$







\item Déchiffrons le mot VLUP, associé aux matrices $\begin{pmatrix}21\\11\end{pmatrix}$ et $\begin{pmatrix}20\\15\end{pmatrix}$.

Pour VL associé à la matrice $\begin{pmatrix}21\\11\end{pmatrix}$.

On a: $y_1=21=r_1$ et $y_2=11=r_2$. On en déduit:

$\left\{\begin{array}{l c l}
 x_1 &\equiv& 9\times 21+16\times11\:[26]\\
 x_2 &\equiv& 17\times 21 +25 \times 11 \:[26]
\end{array}\right.\Rightarrow \left\{\begin{array}{l c l} x_1 &\equiv& 1 \:[26]\\
 x_2 &\equiv& 8 \:[26]
\end{array}\right.$

VL était le code BI


Pour UP associé à la matrice $\begin{pmatrix}20\\15\end{pmatrix}$.

On a: $y_1=20=r_1$ et $y_2=15=r_2$. On en déduit:

$\left\{\begin{array}{l c l}
 x_1 &\equiv& 9\times 20+16\times15 \:[26]\\
 x_2 &\equiv& 17\times 20 +25 \times 15 \:[26]
\end{array}\right.\Rightarrow \left\{\begin{array}{l c l} x_1 &\equiv& 4 \:[26]\\
 x_2 &\equiv& 13 \:[26]
\end{array}\right.$

UP était le code EN

VLUP est le code de BIEN.

\end{enumerate}
\end{document}
