## Task A-1.2.

Let $S$ be the center of the circle $k$ with radius of length 1. Vertices $A$ and $B$ of the square $ABCD$ lie on the circle $k$, and the side $\overline{CD}$ passes through the point $S$. Determine the length of the side of the square $ABCD$.
## Solution.

Since points $A$ and $B$ are on the circle $k$, its center $S$ lies on the perpendicular bisector of the segment $\overline{A B}$. This means that point $S$ is the midpoint of side $\overline{C D}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_f5c8a7635f49ecc51cacg-02.jpg?height=816&width=831&top_left_y=660&top_left_x=544)

By the Pythagorean theorem, we have $|A S|^{2}=|A D|^{2}+|D S|^{2}$.

Let $a$ be the length of the side of the square $A B C D$. Then $|S D|=\frac{1}{2} a$ and $|A D|=a$. Since $|A S|=1$, we have

$$
1=1^{2}=a^{2}+\left(\frac{a}{2}\right)^{2}=\frac{5}{4} a^{2}
$$

Thus, the length of the side of the square $A B C D$ is $a=\frac{2}{\sqrt{5}}$.