We draw circles over the legs of an isosceles right triangle, which intersect the hypotenuse at point $D$. What is the area of the common region determined by the circles, if each leg is $a$?
![](https://cdn.mathpix.com/cropped/2024_05_02_e33da7b728155819c44cg-1.jpg?height=274&width=314&top_left_y=126&top_left_x=900)

The sought area is obtained by subtracting the area of the right triangle \( A D B \) from the area of the semicircle \( A D B \). Since

$$
A O_{1}=\frac{a}{2}, \quad A D=B D=\frac{a}{\sqrt{2}}
$$

it follows that

$$
t=\frac{a^{2} \pi}{8}-\frac{a^{2}}{4}=\frac{a^{2}}{8}(\pi-2)
$$

(Jónás Wittál, Budapest.)

Number of solutions: 54.