10.397 Six equal circles are inscribed in a circle of radius $R$, each touching the given circle and two adjacent ones. Calculate the area of the figure bounded by these six circles.
Solution. Let $O_{1}, o_{2}, o_{3}, O_{4}, o_{5}, o_{6}$ be the centers of six circles of radius $r$, each touching two adjacent circles and inscribed in a given circle of radius $R$ with center $P$. Clearly, $P$ coincides with the center of the regular hexagon $T$ with vertices $O_{i}, i=1,2, \ldots 6$, and side length $2 r$, and $P O_{i}=R-r$ (the line $P O_{i}$ passes through the point of tangency of the circles $P$ and $O_{i}$).

From this, $2 r=O_{1} O_{2}=P O_{1}=P O_{2}=R-r \Rightarrow r=R / 3$.

The area of the figure $F$, bounded by the circles $O_{i}, i=1,2, \ldots 6$, is equal to the area of $T$ minus the area of six equal sectors of $120^{\circ}$ and radius $r$, i.e.,

$$
S=6 S_{\Delta O_{1} O_{2} P}-6 \cdot \frac{\pi r^{2}}{3}=6 \cdot \frac{(2 r)^{2} \sqrt{3}}{4}-2 \pi r^{2}=\frac{2}{9} R^{2}(3 \sqrt{3}-\pi)
$$

Answer: $2 R^{2}(3 \sqrt{3}-\pi) / 9$.