10.156. Two equilateral triangles are inscribed in a circle of radius $R$ such that their mutual intersection divides each side into three equal segments. Find the area of the intersection of these triangles.
## Solution.

The figure formed by the intersection of such triangles will be a regular hexagon. The side

of the inscribed equilateral triangle is $a_{3}=\frac{3 R}{\sqrt{3}}=\sqrt{3} R$. Then the side of the regular hexagon is $a_{6}=\frac{1}{3} a_{3}=\frac{\sqrt{3}}{3} R$. The radius of the inscribed circle in it $r=\frac{a_{6}}{2 \operatorname{tg} \frac{180^{\circ}}{6}}=\frac{a_{6} \sqrt{3}}{2}=\frac{\sqrt{3} \cdot \sqrt{3} R}{3 \cdot 2}=\frac{R}{2}$. The area of the hexagon: $S_{6}=\frac{6 a_{6} r}{2}=\frac{6 R \sqrt{3} \cdot R}{2 \cdot 2 \cdot 3}=\frac{\sqrt{3} R^{2}}{2}$. Answer: $\frac{\sqrt{3} R^{2}}{2}$.