10.313. Two circles of radius $R$ intersect in such a way that each passes through the center of the other. Two other circles of the same radius have their centers at the points of intersection of the first two circles. Find the area common to all four circles.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0787.jpg?height=408&width=341&top_left_y=916&top_left_x=96)

Fig. 10.102

Each of the last two circles passes through the centers of the first two (Fig. 10.102), so the length of their common chord \(O_{1} O_{2}=R\). The desired area is twice the area of the segment with a central angle of \(60^{\circ}\), i.e.,

\[ S=2\left(\frac{\pi R^{2}}{6}-\frac{R^{2} \sqrt{3}}{4}\right)=\frac{R^{2}(2 \pi-3 \sqrt{3})}{6} \].

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