Japan Junior Mathematical Olympiad

First, let us show that the areas of the triangles $AUP$, $BQR$ and $CST$ are all equal to the area of the triangle $ABC$. Since we have $\angle QBR+\angle ABC=180^{\circ }$, $QB=AB$ and $BR=BC$, if we rotate the triangle $BQR$ clockwise for $90^{\circ }$ around the axis through $B$, it lands onto a triangle having the length of the base and the height equal to those of the triangle $ABC$. Therefore, the triangle $BQR$ has the same area as that of the triangle $ABC$. The same argument shows that the triangles $AUP$ and $CST$ also have the same area as $ABC$.

Therefore, the area of the hexagon $PQRSTU$ equals the sum of the areas of the 3 squares $PQBA$, $RSCB$, $TUAC$ plus 4 times the area of the triangle $ABC$. The sum of the areas of the 3 squares is $3^2+3^2+4^2=34$.

It remains to compute the area of the triangle $ABC$. Since it is an isosceles triangle with $AB=AC$, the line segment $AM$ connecting the vertex $A$ to the midpoint $M$ of the base $BC$ satisfies $AM\perp BC$. By the Pythagorean theorem, we get $AM=\sqrt{3^2-2^2}=\sqrt{5}$. Therefore, the area of the triangle $ABC$ is $\frac{1}{2}BC\cdot AM=2\sqrt{5}$ and we conclude that the area of the hexagon $PQRSTU$ is $34+8\sqrt{5}$.