Question

A square and an equilateral triangle have the same perimeter. Let $A$ be the area of the circle circumscribed about the square and $B$ be the area of the circle circumscribed about the triangle. Find $A/B.$

$\dfrac{9}{16}$

$\dfrac{3}{4}$

$\dfrac{27}{32}$

$\dfrac{3\sqrt{6}}{8}$

$1$

Solution:

Let the common perimeter be $12,$ so the square has side $3$ and the triangle has side $4.$

The square's circumradius is $\dfrac{3\sqrt2}{2},$ so $A=\pi\left(\dfrac{3\sqrt2}{2}\right)^2=\dfrac{9\pi}{2}.$

The triangle's circumradius is $\dfrac{4}{\sqrt3},$ so $B=\pi\left(\dfrac{4}{\sqrt3}\right)^2=\dfrac{16\pi}{3}.$

Then $\dfrac{A}{B}=\dfrac{9/2}{16/3}=\dfrac{27}{32}.$

Thus, the correct answer is C.

2003 AMC 12A Problem 11

Solution:

Problem 11 in Other Years