### 10.395 The centers of four circles are located at the vertices of a square

 with side $a$. The radii of these circles are $a$. Determine the area of their common part.
## Solution.

$C$

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-240.jpg?height=274&width=354&top_left_y=1294&top_left_x=128)
B
A

$$
S_{A M N}=\frac{a^{2}}{2} \sin 30^{\circ}=\frac{a^{2}}{4} \Rightarrow S_{\dot{M} N N}=S_{A M P N}-S_{A M N}=\frac{a^{2}}{4}\left(\frac{\pi}{3}-1\right)
$$

The common part MPNQKTLRM of four circles of radius $a$ with centers at the vertices of the given square $A B C D$ consists of four identical segments $M P N, N Q K, K T L, L R M$ and the square $M N K L$. Since $M, N, K, L$ are the points of intersection of the given circles, $\triangle A B N$ and $\triangle A M D$ are equilateral triangles (their sides are equal to $a$) $\Rightarrow$

$$
\begin{aligned}
& \Rightarrow \angle B A N=\angle D A M=60^{\circ} \Rightarrow \\
& \Rightarrow \angle M A N=30^{\circ} \Rightarrow S_{A M P N}=\frac{\pi a^{2}}{12}, \Rightarrow
\end{aligned}
$$

$$
M N^{2}=2 a^{2}-2 a^{2} \cos 30^{\circ}=a^{2}(2-\sqrt{3})=S_{M N K L}
$$

In the end, the area of the desired common part is

$$
4 \cdot S_{M P N}+S_{M N K L}=a^{2} \cdot \frac{\pi+3-3 \sqrt{3}}{3}
$$

Answer: $a^{2} \cdot \frac{\pi+3-3 \sqrt{3}}{3}$.