2003 AMC 12A Problem 11

Problem 11 of 25IntermediateGeometry

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

\frac{A}{B}.

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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},

A=\pi\left(\dfrac{3\sqrt2}{2}\right)^2=\dfrac{9\pi}{2}.

The triangle’s circumradius is

\dfrac{4}{\sqrt3},

B=\pi\left(\dfrac{4}{\sqrt3}\right)^2=\dfrac{16\pi}{3}.

Then

\dfrac{A}{B}=\dfrac{\frac{9}{2}}{\frac{16}{3}}=\dfrac{27}{32}.

Thus, the correct answer is C.