Task B-2.4. A square and an equilateral triangle are inscribed in a circle of radius 1 so that they share one vertex. Determine the area of the common part of the square and the triangle.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_c2de659189058fc448a1g-09.jpg?height=830&width=870&top_left_y=539&top_left_x=237)

The desired area is

$$
P_{A B C D E}=2\left(P_{A H E}-P_{G H D}\right)
$$

The side of the inscribed equilateral triangle is

$$
a=\frac{3 r}{\sqrt{3}}=\sqrt{3}
$$

The base of the triangle $\triangle A H E$ is

$$
|A H|=2 r=2
$$

Determine the height $|E F|$ of this triangle. From the isosceles right triangle $\triangle F H E$ we have

$$
|E F|=|F H|=2-|A F|
$$

and from the triangle $\triangle A F E$ we have

$$
\operatorname{tg} 60^{\circ}=\frac{|A F|}{|E F|} \quad, \text { i.e. } \quad|A F|=\sqrt{3}|E F|
$$

From these two equations, we get $|E F|=2-\sqrt{3}|E F|$, so the height is

$$
|E F|=\sqrt{3}-1
$$

It follows that

$$
P_{\triangle A H E}=\frac{|A H| \cdot|E F|}{2}=\frac{2(\sqrt{3}-1)}{2}=\sqrt{3}-1
$$

The triangle $\triangle G H D$ is an isosceles right triangle, i.e.

$$
|D G|=|G H|=2 r-|A G|=2 r-\frac{a \sqrt{3}}{2}=2-\frac{3}{2}=\frac{1}{2}
$$

It follows that

$$
P_{\triangle G H D}=\frac{|G H| \cdot|G D|}{2}=\frac{\frac{1}{2} \cdot \frac{1}{2}}{2}=\frac{1}{8}
$$

Then the desired area is

$$
P_{A B C D E}=2\left(P_{A H E}-P_{G H D}\right)=2\left((\sqrt{3}-1)-\frac{1}{8}\right)=2 \sqrt{3}-\frac{9}{4}
$$