1963 AHSME Problems/Problem 15

Problem

A circle is inscribed in an equilateral triangle, and a square is inscribed in the circle. The ratio of the area of the triangle to the area of the square is:

$\textbf{(A)}\ \sqrt{3}:1\qquad \textbf{(B)}\ \sqrt{3}:\sqrt{2}\qquad \textbf{(C)}\ 3\sqrt{3}:2\qquad \textbf{(D)}\ 3:\sqrt{2}\qquad \textbf{(E)}\ 3:2\sqrt{2}$

Solution

$[asy] draw(circle((0,0),100)); draw((173.205,-100)--(-173.205,-100)--(0,200)--(173.205,-100)); draw((-100,0)--(0,100)--(100,0)--(0,-100)--(-100,0)); draw((0,0)--(0,-100),dotted); draw((0,0)--(-173.205,-100),dotted); [/asy]$ Let the radius of the circle be $r$. That means the diameter of the circle is $2r$, so the side length of the square is $r\sqrt{2}$. Therefore, the area of the square is $2r^2$.

By using 30-60-90 triangles, half of the side length of an equilateral triangle is $r\sqrt{3}$, so each side is $2r\sqrt{3}$ units long. Thus, the area of the equilateral triangle is $3r^2\sqrt{3}$.

The ratio of the area of the equilateral triangle to the area of the square is $\frac{3r^2 \sqrt{3}}{2r^2} = \frac{3 \sqrt{3}}{2}$, so the answer is $\boxed{\textbf{(C)}}$.