AMC12 2011 B

Suppose that $P$ is a point in the rhombus $ABCD$ and let $\ell_{BC}$ be the perpendicular bisector of $\overline{BC}$. Then $PB < PC$ if and only if $P$ is on the same side of $\ell_{BC}$ as $B$. The line $\ell_{BC}$ divides the plane into two half-planes; let $S_{BC}$ be the half-plane containing $B$. Let us define similarly $\ell_{BD},S_{BD}$ and $\ell_{BA},S_{BA}$. Then $R$ is equal to $ABCD \cap S_{BC} \cap S_{BD} \cap S_{BA}$. The region turns out to be an irregular pentagon. We can make it easier to find the area of this region by dividing it into four triangles: Since $\triangle BCD$ and $\triangle BAD$ are equilateral, $\ell_{BC}$ contains $D$, $\ell_{BD}$ contains $A$ and $C$, and $\ell_{BA}$ contains $D$. Then $\triangle BEF \cong \triangle BGF \cong \triangle BGH \cong \triangle BIH$ with $BE = 1$ and $EF = \frac{1}{\sqrt{3}}$ so $[BEF] = \frac{1}{2}\cdot 1 \cdot \frac{\sqrt{3}}{3}$. Multiply this by 4 and it turns out that the pentagon has area $\boxed{(C)\frac{2\sqrt{3}}{3}}$.