10.396 Three equal circles are inscribed in a circle of radius $R$, touching the outer circle and each other. Calculate the area of the figure bounded by these three circles.
Solution. Let $O_{1}, O_{2}, O_{3}$ be the centers of three mutually tangent circles of radius $r$, inscribed in a given circle of radius $R$ with center $O_{4}$.

Obviously, $O_{4}$ coincides with the center of the equilateral $\triangle O_{1} O_{2} O_{3}$ with side $2 r$, and $O_{4} O_{1}=R-r$ (the line $O_{4} O_{1}$ passes through the point of tangency of the circles $O_{4}$ and $O_{1}$). Hence, $2 r=O_{1} O_{2}=\sqrt{3} \cdot O_{4} O_{1}=\sqrt{3}(R-r) \Rightarrow r=\sqrt{3}(2-\sqrt{3}) R$.

The area of the figure $\Phi$, bounded by the circles $O_{1}, O_{2}, O_{3}$, is equal to $S_{\triangle O_{1} O_{2} O_{3}}$ minus the areas of three equal sectors of $60^{\circ}$ and radius $r$, i.e.,

$S_{\Phi}=\frac{(2 r)^{2} \sqrt{3}}{4}-3 \cdot \frac{\pi r^{2}}{6}$.

Answer: $\frac{1}{2} R^{2}(7-4 \sqrt{3})(6 \sqrt{3}-3 \pi)$.