Saudi Arabia 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 $O_2{O}_3$, $O_1{O}_3$, $O_2{O}_1$, respectively, cover the surface of area $S$ and twice more the surface of triangle $O_1{O}_2{O}_3$.

The area of triangle $O_1{O}_2{O}_3$ 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 $$\frac{1}{2}\pi R^2=S+2\cdot \frac{R^2\sqrt{3}}{4}.$$ $$\text{Therefore }S=\frac{1}{2}(\pi -\sqrt{3})R^2.$$