
\subsection{Analyse a priori} 
Le problème est posé dans un cadre purement géométrique. C’est à l’élève de choisir saméthode, de plus la longueur du côté du carré n’est pas indiquée, certes en mesurant, ilpourrait s’engager dans des essais numériques, longs et fastidieux. \\
En posant $a$ la longueur du côté du carré et $x$ la distance d’un sommet de la croix à un sommet du carré (situés sur le même côté), on arrive à une équation du second degré, que l’élève ne peut résoudre.  \\
Si l’on choisit $x$ la distance du milieu du côté au sommet de la croix,
on obtient une équation produit, mais ce choix n’est pas naturel. \\
Il peut également s’engager dans une méthode géométrique. Après avoir tracé une croix dans le carré, on attend à ce que l’élève s’approprie la configuration, la complète. L’élève devrait tracer tous les axes de
symétrie et repérer des figures extraites de même aire. Deux figures symétriques par rapport à une droite ayant la même aire, le problème se ramène à construire un point $M$ situé sur le côté $BC$ d’un triangle de sommet $A$ tel que l’aire du triangle $ABM$ soit le double de l’aire du triangle $ACM$. 

\subsection{Éléments de solution.}
\begin{enumerate}
  \item Figure 1 : Tracé d'une croix dans un carré, pour étude de cette configuration.
  \item Figure 2 : A partir de la figure 1, tracé des quatre axes de symétrie et du carré central intersection des deux rectangles de la croix. Cette figure est composée de rectangles et de triangles rectangles isocèles (des petits et des grands). \\
  A l’intérieur de chacune de ces figures, on inscrit une lettre, tel que deux figures de même aire contiennent la même lettre.
  La traduction de la condition demandée est : $8\alpha + 8\beta = 4(2\alpha + \beta + \gamma ) $ \\
  Après simplification on obtient : $\beta = \gamma $.
  \item Figure 3 : Figure extraite de la figure 2. Comment construire un rectangle inscrit dans un triangle rectangle isocèle tel que $\beta = \gamma $ ?  
  \item  Figure 4 : On note $ABCD$ le rectangle d’aire \small$\beta$ , $BCD$ est le triangle rectangle isocèle de hauteur
  $[CH]$ et d’aire \small$\gamma$.
  Comment placer B pour que l’on ait $\beta = \gamma $ ?\\
  
  L'aire de $ABCD$ est $BD \times AB$, l'aire de $BCD$ est $\dfrac{BD\times CH}{2}$,\\ $\beta = \gamma $   se traduit par $CH=2\times AB$.\\
  Si $K$ est le projeté de $H$ sur$(AE)$  , on a $CH=2\times HK$\\
  Ce qui permet d'en déduire que $B$ est au tiers sur le côté du petit carré ou au sixième sur le côté du carré de départ.\\
  Cela permet alors de réaliser les tracés sur le carré donné de la feuille de réponse.
\end{enumerate}
\medskip
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