10.392 In a circle of radius $R$, an equilateral triangle and a square are inscribed, sharing a common vertex. Calculate the area of the common part of the triangle and the square.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-239.jpg?height=371&width=393&top_left_y=262&top_left_x=143)

Let the square $ABCD$ and the equilateral $\triangle AEF$ be inscribed in a given circle of radius $R$. Let $M$, $N$, $K$, $L$ be the points of intersection of the sides of the square and the triangle. Clearly, $AC$ is the axis of symmetry $\Rightarrow$ $\Rightarrow MN \perp AC, AC \perp EF \Rightarrow$

$$
\Rightarrow AP = \frac{\sqrt{3}}{2} EF = \frac{3}{2} R, PC = 2R - AP = \frac{R}{2}
$$

Let $x = TP$, then $MT = TC = x + \frac{R}{2}, AT = MT \cdot \operatorname{ctg} 30^{\circ} = \left(x + \frac{R}{2}\right) \cdot \sqrt{3} \Rightarrow 2R = AT + TC = x(\sqrt{3} + 1) + (\sqrt{3} + 1) \frac{R}{2} \Rightarrow x = \frac{R \sqrt{3}}{4}(\sqrt{3} - 1)^{2}
$$

$S_{ANLKM} = S_{ANCM} - S_{CKL} = \frac{AC \cdot MN}{2} - PC^2 = \frac{2R(2x + R)}{2} - \frac{R^2}{4} = \frac{R^2(8\sqrt{3} - 9)}{4}$.

Answer: $\frac{R^2(8\sqrt{3} - 9)}{4}$.