smc

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 $△BCD$ and $△BAD$ are equilateral, ${\ell }_{BC}$ contains $D$, ${\ell }_{BD}$ contains $A$ and $C$, and ${\ell }_{BA}$ contains $D$. Then $△BEF\cong △BGF\cong △BGH\cong △BIH$ 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 $\boxed{\frac{2\sqrt{3}}{3}}$.