National Olympiad Final Round

The circles drawn around two neighbouring vertices of the hexagon intersect, since $2 \cdot 1 > \sqrt{3}$. Hence every two neighbouring circles have a common region of the shape of a lens. Since a regular hexagon can be put together from six equilateral triangles, the circumradius of the hexagon equals $\sqrt{3}$. The altitude of one equilateral triangle is $\sqrt{(\sqrt{3})^2 - (\frac{\sqrt{3}}{2})^2} = \frac{3}{2}$.

The distance between a vertex and the second one counting from that vertex along the circumference is $3$, since it equals twice the altitude of the equilateral triangle (Fig. 17). As $2 \cdot 1 < 3$, circles drawn around such two vertices do not intersect. Thus the circles drawn around opposite vertices do not intersect either.

Fig. 17

Fig. 18

$(\frac{AB}{2})^2 = AC^2$, i.e., $(\frac{CD}{2})^2 + (\frac{\sqrt{3}}{2})^2 = 1^2$, whence $CD = 1$, implying that the triangle $ACD$ is equilateral. Thus the area of sector $ACD$ and triangle $ACD$ equal $\frac{1}{6}\pi$ and $\frac{\sqrt{3}}{4}$, respectively. The area of the region painted blue is $6\pi - 12(\frac{1}{6}\pi - \frac{\sqrt{3}}{4})$, which equals $4\pi + 3\sqrt{3}$.