# Difference between revisions of "1966 AHSME Problems/Problem 4"

## Problem

Circle I is circumscribed about a given square and circle II is inscribed in the given square. If $r$ is the ratio of the area of circle I to that of circle II, then $r$ equals: $\text{(A) } \sqrt{2} \quad \text{(B) } 2 \quad \text{(C) } \sqrt{3} \quad \text{(D) } 2\sqrt{2} \quad \text{(E) } 2\sqrt{3}$

## Solution

Make half of the square's side $x$. Now the radius of the smaller circle is $x$, so it's area is $\pi x^2$. Now find the diameter of the bigger circle. Since half of the square's side is $x$, the full side is $2x$. Using the Pythagorean theorem, you get the diagonal to be $2\sqrt{2}x$. Half of that is the radius, or $x\sqrt{2}$. Using the same equation as before, you get the area of the larger circle to be $2x^2 \pi$. Putting one over the other and dividing, you get two as the answer: or $\boxed{(B)}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 