## Task A-2.2.

Inside the circle $k$ with radius 20, there are a circle $k_{1}$ with radius 5 and a square $A B C D$. The circles $k$ and $k_{1}$ touch at point $P$, points $A$ and $B$ lie on the circle $k$, and the line $C D$ touches the circle $k_{1}$ at point $Q$ such that $\overline{P Q}$ is the diameter of that circle.

Determine the length of the side of the square $A B C D$.
## Solution.

Let $a$ be the length of the side of the square $ABCD$, $S$ the center of the circle, and $N$ the foot of the perpendicular from $S$ to $\overline{AB}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_a374c05c2f1967f14be6g-06.jpg?height=702&width=691&top_left_y=1885&top_left_x=614)

Since $S$ is the center of the circle with radius 20, we have $|SP| = |SB| = 20$.

Due to symmetry, $|BN| = \frac{a}{2}$.

Since $|PQ| = 10$, we have $|SQ| = |SP| - |PQ| = 10$ and $|SN| = |QN| - |SQ| = a - 10$.

Triangle $BSN$ is a right triangle with a right angle at vertex $N$, so by the Pythagorean theorem we have

$$
\begin{aligned}
(a-10)^{2}+\left(\frac{a}{2}\right)^{2} & =20^{2} \\
a^{2}-20a+100+\frac{a^{2}}{4} & =400 \\
a^{2}-16a-240 & =0
\end{aligned}
$$

The solutions to this quadratic equation are

$$
\frac{16 \pm \sqrt{16^{2}+4 \cdot 240}}{2}=\frac{16 \pm 8 \sqrt{19}}{2}=8 \pm 4 \sqrt{19}
$$

Since $4 \sqrt{19}$ is greater than 8, and $a$ is the length of a side, it must be a positive number, so the only solution to the problem is $a = 8 + 4 \sqrt{19}$.