, 2014

Solution:

$\boxed{\frac{\pi \sqrt{3}}{12}}$ Let the side length of the hexagon be $s$. The length of $BD$ is $s \sqrt{3}$, so the area of rectangle $ABDE$ is $s^{2} \sqrt{3}$. Equilateral triangle $BDF$ has side length $s \sqrt{3}$. The inradius of an equilateral triangle is $\sqrt{3} / 6$ times the length of its side, and so has length $\frac{s}{2}$. Thus, the area of circle $P$ is $\frac{\pi s^{2}}{4}$, so the ratio is $\frac{\pi s^{2} / 4}{s^{2} \sqrt{3}}=\frac{\pi \sqrt{3}}{12}$.