$\left.\begin{array}{c}\text { Circle, sector, segment, etc. } \\ {\left[\begin{array}{l}{[\text { Pythagorean Theorem (direct and inverse) }}\end{array}\right]}\end{array}\right]$

In a segment, the arc of which is $60^{\circ}$, a square is inscribed. Find the area of the square if the radius of the circle is $2 \sqrt{3}+$ $\sqrt{17}$.
Let $x$ be the side of the square and apply the Pythagorean theorem to the triangle with vertices: at the center of the circle, at the midpoint of the side of the square, and at the vertex of the square belonging to this side.

## Solution

Let $O$ be the center of the circle, $R$ its radius ($R=2 \sqrt{3}+\sqrt{17}$), and $x$ the side of the square $ABCD$, with points $C$ and $D$ lying on the chord $PQ$ of the given segment.

Draw a line through the center of the circle perpendicular to $PQ$. Let $K$ and $M$ be the points of intersection of this line with the chords $PQ$ and $AB$. Then

$$
O K=\frac{R \sqrt{3}}{2}, M B=\frac{x}{2}, K M=x
$$

Consider the right triangle $OMB$. By the Pythagorean theorem,

$$
(O K+K M)^{2}+M B^{2}=O B^{2}, \text { or }\left(\frac{R \sqrt{3}}{2}+x\right)^{2}+\frac{x^{2}}{4}=R^{2}
$$

From this, we find that

$$
x=\frac{R(\sqrt{17}-2 \sqrt{3})}{5}=\frac{(2 \sqrt{3}+\sqrt{17})(\sqrt{17}-2 \sqrt{3})}{5}=1
$$

Therefore, $S_{\mathrm{ABCD}}=1$.

![](https://cdn.mathpix.com/cropped/2024_05_06_16ab69fab1743849dba9g-33.jpg?height=469&width=755&top_left_y=2441&top_left_x=660)

## Problem