A square and an equilateral triangle have the same perimeter. Let $A$ be the area of the circle circumscribed about the square and $B$ the area of the circle circumscribed around the triangle. Find $A/B$. $\mathrm{(A) \ } \frac{9}{16}\qquad$ $ \mathrm{(B) \ } \frac{3}{4}\qquad$ $ \mathrm{(C) \ } \frac{27}{32}\qquad$ $ \mathrm{(D) \ } \frac{3\sqrt{6}}{8}\qquad$ $ \mathrm{(E) \ } 1$
Solution
Let the common perimeter be $P$. The square has side $\frac P4$ and circumradius $\frac{P}{4}\cdot\frac{\sqrt2}2=\frac{P\sqrt2}8$, so $A=\pi\cdot\frac{2P^2}{64}=\frac{\pi P^2}{32}$.
The equilateral triangle has side $\frac P3$ and circumradius $\frac{P}{3\sqrt3}$, so $B=\pi\cdot\frac{P^2}{27}$.
A regular hexagon and an equilateral triangle have the same perimeter. Let $A$ be the area of the circle circumscribed about the hexagon and $B$ the area of the circle circumscribed about the triangle. Find $A/B$.
