Japan Mathematical Olympiad

222 Let $XY$ represent the length of segment $XY$. In the figure, let $A$, $B$, $C$ and $D$ be the vertices of the rectangle, and $P$, $Q$, $R$, $S$, $T$, $U$ and $O$ the vertices and the center of the regular hexagon. Since $\angle QAP=\angle PQS=90^{\circ }$, we can conclude that $\angle APQ=90^{\circ }-\angle AQP=\angle BQS$. Moreover, $\angle PAQ=\angle QBS=90^{\circ }$. Hence, triangles $PAQ$ and $QBS$ are similar. From $\angle PQS=90^{\circ }$ and $\angle QPS=60^{\circ }$, the ratio of similarity between triangles $PAQ$ and $QBS$ is $PQ:QS=1:\sqrt{3}$. Therefore, the area of triangle $QBS$ is $20\cdot (\sqrt{3}{)}^2=60$, and the area of triangle $QRS$ is $60-23=37$. Since $OQ\parallel SR$, the areas of triangles $QRS$ and $ORS$ are equal, and this is $1/6$ of the area of hexagon $PQRSTU$. Hence, the answer is $37\cdot 6=222$.