2003 AMC 12A Problem 11

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

\frac{A}{B}.

\dfrac{9}{16}

\dfrac{3}{4}

\dfrac{27}{32}

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

1

Choose a convenient common perimeter, such as

12

A square’s circumradius is half its diagonal; an equilateral triangle’s is

\dfrac{\text{side}}{\sqrt3}

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

Thus, the correct answer is C.

Problem 11 in other years