2011 AMC 12B Problem 16

Let $E$ and $H$ be the midpoints of $AB$ and $BC.$ The perpendicular bisector of $AB$ through $E$ meets diagonal $AC$ at $F,$ and the perpendicular bisector of $BC$ through $H$ meets $AC$ at $G.$ The region $R$ is the pentagon $BEFGH.$ Triangle $AFE$ is a $30$-$60$-$90^\circ$ triangle with $AE=1,$ so its area is $\dfrac12\cdot1\cdot\dfrac{1}{\sqrt3}=\dfrac{\sqrt3}{6}.$ Triangles $BFE$ and $BGH$ are congruent to it, and $\triangle FBG$ is equilateral, splitting into two more copies. Hence $R$ consists of four congruent triangles, giving area $4\cdot\dfrac{\sqrt3}{6}=\dfrac{2\sqrt3}{3}.$ Thus, the correct answer is C.