2001 AIME II Problems/Problem 6

Problem

Square $ABCD$ is inscribed in a circle. Square $EFGH$ has vertices $E$ and $F$ on $\overline{CD}$ and vertices $G$ and $H$ on the circle. The ratio of the area of square $EFGH$ to the area of square $ABCD$ can be expressed as $\frac {m}{n}$ where $m$ and $n$ are relatively prime positive integers and $m < n$ . Find $10n + m$ .

Let $O$ be the center of the circle, and $2a$ be the side length of $ABCD$ , $2b$ be the side length of $EFGH$ . By the Pythagorean Theorem, the radius of $\odot O = OC = a\sqrt{2}$ .

Now consider right triangle $OGI$ , where $I$ is the midpoint of $\overline{GH}$ . Then, by the Pythagorean Theorem,

Thus $a = 5b$ (since lengths are positive, we discard the other root). The ratio of the areas of two similar figures is the square of the ratio of their corresponding side lengths, so $\frac{[EFGH]}{[ABCD]} = \left(\frac 15\right)^2 = \frac{1}{25}$ , and the answer is $10n + m = \boxed{251}$ .

Another way to proceed from $0 = a^2 - 4ab - 5b^2$ is to note that $\frac{b}{a}$ is the quantity we need; thus, we divide by $a^2$ to get