In a square grid, whose squares have a side length of $a$, two circles are drawn (see the image). Both have their center at point $S$ and each passes through four grid points. The gray-shaded figure is bounded by parts of these circles and one grid line. Express the area of the gray-shaded figure in terms of the length $a$.

![](https://cdn.mathpix.com/cropped/2024_04_17_7a9cd767764e1257df19g-2.jpg?height=585&width=498&top_left_y=995&top_left_x=781)

(L. Šiminek)
The grid points through which the smaller circle passes are denoted as $A, B, C$ and $D$. On the larger circle, we mark points $K, L, M, N, O$ and $P$ as shown in the figure. The hexagon $K L M N O P$ is regular, which follows from the fact that all its vertices lie on one circle, the sides $K L$ and $N O$ have a length evidently equal to the radius of this circle, and the other four sides have the same length.

![](https://cdn.mathpix.com/cropped/2024_04_17_7a9cd767764e1257df19g-2.jpg?height=583&width=588&top_left_y=2023&top_left_x=736)

To calculate the area of the gray figure, we will need the area $S_{1}$ of the larger circle and the area $S_{2}$ of its segment defined by the chord $L N$, as well as the area $S_{3}$ of the smaller circle and the area $S_{4}$ of its segment defined by the chord $B C$.

The larger circle has a radius of $2a$, and its area is

$$
S_{1}=\pi(2a)^{2}=4\pi a^{2}
$$

The area $S_{2}$ of the circular segment is equal to the difference between the area of the circular sector $L S N$ and the triangle $L S N$. The angle $L S N$ clearly defines one-third of the hexagon $K L M N O P$, so the area of the circular sector $L S N$ is one-third of the area $S_{1}$ of the larger circle, i.e., $\frac{4}{3} \pi a^{2}$. Now we express the area of the triangle $L S N$. The midpoint of the segment $L N$ is denoted as $Z$. Using the Pythagorean theorem, we determine the length of the side $Z N$ of the triangle $S Z N$:

$$
|Z N|=\sqrt{(2a)^{2}-a^{2}}=a \sqrt{3}.
$$

Thus, $|L N|=2a \sqrt{3}$ and the area of the triangle $L S N$ is $\frac{1}{2} \cdot 2a \sqrt{3} \cdot a = a^{2} \sqrt{3}$. Finally, we can express the area $S_{2}$ of the circular segment:

$$
S_{2}=\frac{4}{3} \pi a^{2} - a^{2} \sqrt{3} = \left(\frac{4}{3} \pi - \sqrt{3}\right) a^{2}.
$$

The radius of the smaller circle corresponds to the diagonal of a square with side $a$, i.e., $a \sqrt{2}$. The area of this circle is

$$
S_{3}=\pi(a \sqrt{2})^{2}=2\pi a^{2}.
$$

If we subtract the area of the square $A B C D$ from $S_{3}$ and divide the difference by four, we get the area $S_{4}$ of the circular segment:

$$
S_{4}=\frac{1}{4}\left(2\pi a^{2} - 4a^{2}\right) = \left(\frac{\pi}{2} - 1\right) a^{2}.
$$

The desired area is equal to

$$
\left(S_{1} - S_{2}\right) - \left(S_{3} - S_{4}\right) = 4\pi a^{2} - \left(\frac{4}{3} \pi - \sqrt{3}\right) a^{2} - 2\pi a^{2} + \left(\frac{\pi}{2} - 1\right) a^{2},
$$

which simplifies to the expression

$$
\left(\frac{7}{6} \pi + \sqrt{3} - 1\right) a^{2}
$$

Evaluation. 1 point for the areas $S_{1}, S_{3}, S_{4}; 2$ points for the area $S_{2}; 1$ point for the correct conclusion; the final simplification is not mandatory.