10.312. A regular triangle is inscribed in a circle of radius $R$, and its heights are extended to intersect the circle. These intersection points are then connected to form a new triangle. Calculate the part of the circle's area that lies outside these triangles.
## Solution.

The desired area is equal to $6 S_{1}$, where $S_{1}$ is the area of the curvilinear triangle $A P F$ (Fig. 10.101), $S_{1}$, in turn, is the difference between the sector $A O F$ and twice the area of the triangle $A O P . \angle A O F=60^{\circ}$, therefore,

$$
S_{A O F}=\frac{1}{6} \pi R^{2} . S_{\triangle A O P}=\frac{1}{2} A P \cdot O K=\frac{1}{2} \cdot \frac{1}{3} A C \cdot O K=\frac{1}{6} R \sqrt{3} \cdot \frac{1}{2} R=\frac{R^{2} \sqrt{3}}{12}
$$

Therefore, the desired area

$$
S=6 S_{1}=6\left(S_{A O F}-2 S_{\triangle A O P}\right)=6\left(\frac{\pi R^{2}}{6}-\frac{R^{2} \sqrt{3}}{6}\right)=R^{2}(\pi-\sqrt{3})
$$

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