8,9\end{aligned}

On the plane, there are two circles with radii 5 and 2, centered at points $S_{1}$ and $S_{2}$, touching a certain line at points $A_{1}$ and $A_{2}$ and lying on opposite sides of this line. The ratio of the segment $A_{1} A_{2}$ to the segment $S_{1} S_{2}$ is $\frac{\sqrt{2}}{2}$. Find $A_{1} A^{2}$.

Answer

## 7.

In an isosceles trapezoid $P Q R S$, the diagonals are perpendicular and are divided by their intersection point $O$ in the ratio 1 : $\sqrt{3}$. The larger base $P S$ of the trapezoid is equal to 1. Find the area of the common part of the circles circumscribed around triangles $P Q O$ and $P O S$.
The centers of the circles circumscribed around triangles $P Q O$ and $P O S$ are the midpoints of segments $P Q$ and $P S$.

## Solution

The center $A$ of the circle circumscribed around triangle $P O S$ is the midpoint of the base $P S$, and the radius of this circle is $\frac{1}{2} \cdot$ The area of sector $O A P$ is

$$
\frac{1}{4} \pi\left(\frac{1}{2}\right)^{2}=\frac{\pi}{16}
$$

Subtracting the area of triangle $O A P$ from it, we get

$$
\frac{\pi}{16}-\frac{1}{2}\left(\frac{1}{2}\right)^{2}=\frac{\pi}{16}-\frac{1}{8}
$$

Since

$$
\operatorname{ctg} \angle O P Q=\frac{P Q}{O Q}=\frac{P O}{O R}=\sqrt{3}
$$

then $\angle O P Q=30^{\circ}$. Therefore,

$$
P Q=\frac{P O}{\cos 30^{\circ}}=\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{3}}{2}}=\frac{2}{\sqrt{6}}
$$

The center $B$ of the circle circumscribed around the right triangle $O P Q$ is the midpoint of side $P Q$, its radius is $\frac{1}{2} P Q=\frac{1}{\sqrt{6}}, \angle O B P=120^{\circ}$. The area of sector $O B P$ is

$$
\frac{1}{3} \pi\left(\frac{1}{\sqrt{6}}\right)^{2}=\frac{\pi}{18}
$$

Subtracting the area of triangle $O B P$ from it, we get

$$
\frac{\pi}{18}-\frac{1}{2}\left(\frac{1}{\sqrt{6}}\right)^{2} \frac{\sqrt{3}}{2}=\frac{\pi}{18}-\frac{\sqrt{3}}{24}
$$

Therefore, the desired area is

$$
\left(\frac{\pi}{16}-\frac{\pi}{8}\right)+\left(\frac{\pi}{18}-\frac{\sqrt{3}}{24}\right)=\frac{17 \pi}{144}-\frac{3+\sqrt{3}}{24}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_b9b3326e3456e69a696eg-05.jpg?height=506&width=560&top_left_y=-1&top_left_x=752)

## Answer

$\frac{17 \pi}{144}-\frac{3+\sqrt{3}}{24}$.

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