Saudi Arabian Mathematical Competitions, 2012

Let $O_1$, $O_2$, $O_3$ be the centers of the three circles and $S$ the area of the common region. The three sectors with centers $O_1$, $O_2$, $O_3$ which subtend the arcs $O_2O_3$, $O_1O_3$, $O_2O_1$, respectively, cover the surface of area $S$ and twice more the surface of triangle $O_1O_2O_3$.

The area of triangle $O_1O_2O_3$ is $\frac{R^2\sqrt{3}}{4}$.

On the other hand, the area of each of these three circular sectors equals $\frac{1}{3}$ the area of a semicircle of radius $R$, hence it is $\frac{1}{6}\pi R^2$.

Hence
$$\frac{1}{2}\pi R^2 = S + 2 \cdot \frac{R^2\sqrt{3}}{4}.$$
$$\text{Therefore } S = \frac{1}{2}(\pi - \sqrt{3})R^2.$$