AMC 12 Mock Exam 2 Problem 15

Let the side length be $s$. The reflection of the hexagon across the perpendicular bisector of $\overline{BC}$ swaps $B$ with $C$ and $A$ with $D$, so it swaps the diagonals $\overline{AC}$ and $\overline{BD}$; hence their intersection $P$ lies on that perpendicular bisector, and $PB=PC$.

In triangle $BCD$, $BC=CD=s$ and $\angle BCD=120^\circ$, so $\angle DBC=30^\circ$. Thus $\triangle BCP$ is isosceles with base $BC=s$ and base angles $30^\circ$, so its height is $\frac s2\tan30^\circ=\frac{s}{2\sqrt3}$ and its area is $\frac12\cdot s\cdot\frac{s}{2\sqrt3}=\frac{s^2}{4\sqrt3}=\frac{\sqrt3}{12}s^2$.

The hexagon consists of six equilateral triangles of side $s$, with total area $\frac{3\sqrt3}{2}s^2$. The ratio is $\frac{\sqrt3/12}{3\sqrt3/2}=\boxed{\tfrac{1}{18}}$.