2008 AIME II Problem 13

Writing $w = \frac{1}{z},$ the half-plane $\mathrm{Re}\,z \gt \frac{1}{2}$ maps exactly onto the disk of radius $1$ centered at $1$

$S$ is the union of six unit disks centered at the sixth roots of unity; by symmetry find the area inside one $60^\circ$ wedge as a $120^\circ$ sector plus two triangles

The hexagon’s sides lie at distance $\frac{1}{2}$ from the origin, with one side on the line $\mathrm{Re}\,z = \frac{1}{2},$ so $R$ is the union of the six half-planes obtained by rotating $\mathrm{Re}\,z \gt \frac{1}{2}$ by multiples of $60^\circ.$ If $w = u + vi = \frac{1}{z},$ then $\mathrm{Re}\,z = \mathrm{Re}\,\frac{1}{w} = \frac{u}{u^2 + v^2} \gt \frac{1}{2}$ is equivalent to $u^2 + v^2 \lt 2u,$ i.e. $(u - 1)^2 + v^2 \lt 1.$ So each half-plane maps onto an open unit disk, and $S$ is the union of six unit disks centered at the sixth roots of unity.

Cut the plane into six $60^\circ$ wedges by the rays at angles $30^\circ + 60^\circ k;$ by symmetry, within each wedge $S$ coincides with the disk whose center lies in that wedge. The rays at $\pm 30^\circ$ meet the circle $|w - 1| = 1$ at $\left(\frac{3}{2}, \pm\frac{\sqrt{3}}{2}\right),$ so the piece of $S$ in that wedge consists of two triangles with vertices at $0,$ the center $1,$ and one of these points — each isosceles with two sides $1$ and apex angle $120^\circ,$ area $\frac{\sqrt{3}}{4}$ — together with the $120^\circ$ sector of the disk between them, area $\frac{\pi}{3}.$

Each wedge therefore contributes $\frac{\pi}{3} + \frac{\sqrt{3}}{2},$ and the total area is $$\begin{aligned} 6\left(\frac{\pi}{3} + \frac{\sqrt{3}}{2}\right) &= 2\pi + 3\sqrt{3} \\ &= 2\pi + \sqrt{27}. \end{aligned}$$ Thus $a = 2,$ $b = 27,$ and $a + b = 29.$