3. [4] Let $A B C D E F$ be a regular hexagon. Let $P$ be the circle inscribed in $\triangle B D F$. Find the ratio of the area of circle $P$ to the area of rectangle $A B D E$.
Answer: $\frac{\pi \sqrt{3}}{12}$ Let the side length of the hexagon be $s$. The length of $B D$ is $s \sqrt{3}$, so the area of rectangle $A B D E$ is $s^{2} \sqrt{3}$. Equilateral triangle $B D F$ 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 $\frac{\pi s^{2} / 4}{s^{2} \sqrt{3}}=\frac{\pi \sqrt{3}}{12}$.