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\begin{document}


\emph{Une histoire d'encerclement : Quelques éléments de correction.}

\begin{description}
\item[Partie A.] 
\end{description}

\begin{enumerate}
\item La droite $(AB)$ coupe les cercles $C_{1}$ et $C_{2}$ en deux points,
les plus éloignés du milieu de $[AB]$ étant : $A_{1}$ et $B_{1}$. Le cercle $%
C_{0}$ de diamètre $A_{1}B_{1}$ et de centre $O_{1}$, contient $C_{1}$ et $%
C_{2}$. Son rayon est $r_{0}=\dfrac{A_{1}B_{1}}{2}.$ Soit $C$ un cercle de
centre $O$ et de rayon minimal $r$ contenant $C_{1}$ et $C_{2}$. On a donc $%
r\leqslant r_{0}.$ $C$ contient donc $A_{1}$ et $B_{1}$. Alors $%
OA_{1}+OB_{1}\geqslant A_{1}B_{1}\Longrightarrow $ $2r\geqslant
2r_{0}\Longrightarrow $ $r\geqslant r_{0}\Longrightarrow r=r_{0}$. De plus $%
OA_{1}=OB_{1}=r_{0}$ donc $O$ est le milieu de $\left[ A_{1}B_{1}\right] $
et aussi de $[AB]$. L'unique cercle répondant à la question est donc $C_{0}$%
. (Figure 1)

\item \textbf{Cas où }$n=3$\textbf{\ figure 2} : $(EF)$ coupe $C_{1}$ et $%
C_{2}$ en $E_{1}$ et $F_{1}$.

Le cercle de diamètre $\left[ E_{1}F_{1}\right] $ contient $C_{3}$ en plus
de $C_{1}$et $C_{2}$ . Son rayon est minimal d'après \textbf{1. ( }Figure 2).
\end{enumerate}

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$ \\ 
Figure 1 & Figure 2 & Figure 3%
\end{tabular}%
\bigskip

\qquad \qquad \textbf{Cas où }$n=3$\textbf{\ figure 2 : }Soit $C$ un cercle
de centre $O$ et de rayon minimal $r_{0}$ contenant les trois cercles $C_{1}$%
,$C_{2}$ et $C_{3}.$ ( Figure 3 )

\begin{itemize}
\item Les médiatrices de $\left[ IK\right] $ et $\left[ JK\right] $ se
coupent en $O_{1}$.

\item Les droites $\left( O_{1}I_{1}\right) ,$ $\left( O_{1}J_{1}\right) $
et $\left( O_{1}K_{1}\right) $ coupent $C_{1}$,$C_{2}$ et $C_{3}$
respectivement en $I_{1}$, $J_{1}$ et $K_{1}$ (points les plus éloignés de $%
O_{1}$).

\item $C$ contient $I_{1}$, $J_{1}$ et $K_{1}$. Son centre $O$ est tel que $%
OI_{1}\leqslant r_{0}$ , $OJ_{1}\leqslant r_{0}$ et $OK_{1}\leqslant r_{0}.$
Notons $r_{1}\ $le rayon du cercle circonscrit à $I_{1}J_{1}K_{1}.$ Comme $%
r_{0}$ est minimal on a $r_{1}=r_{0}.$ Or dans le cas de la figure les trois
disques de centres respectifs $I_{1}$, $J_{1}$ et $K_{1}$ et de rayon $r_{1}$
n'ont qu'un point commun $O_{1}$ donc $O=O_{1}$ . Il y a donc un unique
cercle de rayon minimal contenant les trois cercles $C_{1}$,$C_{2}$ et $%
C_{3}.$

\item Remarquons que la différence entre les deux configurations (Figures 2
et 3) vient de la présence ou non d'un angle obtus dans le triangle formé
par les centres des cercles. Cela vient du fait que la médiane issue du
sommet de l'angle obtus (respectivement droit, aigu) est inférieure
(respectivement égale,supérieure) à la moitié du coté opposé. Preuve dans
cette figure :%
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\end{equation*}
\end{itemize}

\begin{description}
\item[Partie B.] 
\end{description}

Conjectures de configurations :

\begin{tabular}{ccc}
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\\ 
$n=2$ & $n=3$ & $n=4$%
\end{tabular}

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\\ 
$n=5$ & $n=6$ & $n=7$%
\end{tabular}

\bigskip

\begin{description}
\item[Partie C.] 
\end{description}

\begin{enumerate}
\item Le triangle $A_{1}OA_{2}$ a pour cotés $3,3,2$. L'angle $\alpha =$ $%
\widehat{A_{1}OA_{2}}$ est donc tel que $\sin \dfrac{\alpha }{2}=\dfrac{1}{2}
$ d'où $\alpha \approx \allowbreak 38°,\allowbreak 942$ puis $\allowbreak 
\dfrac{360}{\alpha }\approx \allowbreak \allowbreak 9,244\,5.$On peut donc
loger au plus $9$ cercles à l'intérieur de la couronne.

\item Pour pouvoir loger exactement $5$ cercles on doit avoir : $\alpha
\leqslant \dfrac{2\pi }{5}$. Or $\sin \dfrac{\alpha }{2}=\dfrac{1}{1+r}$
donc $\dfrac{1}{1+r}\leqslant \sin \dfrac{\pi }{5}$, On en déduit $%
r\geqslant \dfrac{1}{\sin \dfrac{\pi }{5}}-1\approx \allowbreak 0.701\,3.$

\item Notons $R_{i}$ le rayon du grand cercle correspondant au rayon minimal
d'un cercle contenant les $i$ pièces de la configuration conjecturée. On a
alors $:$

\begin{itemize}
\item $R_{2}=2$

\item $R_{3}=\dfrac{1}{\sin \dfrac{\pi }{3}}+1=\allowbreak \dfrac{2}{3}\sqrt{%
3}+1\approx \allowbreak 2.\,\allowbreak 154\,7$

\item $R_{4}=\dfrac{1}{\sin \dfrac{\pi }{4}}+1=\allowbreak \sqrt{2}+1\approx
\allowbreak 2.\,\allowbreak 414\,2$

\item $R_{5}=\dfrac{1}{\sin \dfrac{\pi }{5}}+1\approx \allowbreak
2.\,\allowbreak 701\,3$

\item $R_{6}=\dfrac{1}{\sin \dfrac{\pi }{6}}+1=\allowbreak 3$

\item Et pour aller plus loin on peut montrer que $R_{7}=\dfrac{1}{\sin 
\dfrac{\pi }{6}}+1=\allowbreak 3,$ car à partir de $n=7$ l'espace intérieur
peut contenir une autre pièce.
\end{itemize}
\end{enumerate}

\bigskip

\bigskip

\label{fin}

\end{document}
