[Square](https://artofproblemsolving.com/wiki/index.php/Square) $ABCD$ is inscribed in a [circle](https://artofproblemsolving.com/wiki/index.php/Circle). Square $EFGH$ has vertices $E$ and $F$ on $\overline{CD}$ and vertices $G$ and $H$ on the circle. If the area of square $ABCD$ is $1$, then the area of square $EFGH$ 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](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem), the radius of $\odot O = OC = a\sqrt{2}$.


Now consider [right triangle](https://artofproblemsolving.com/wiki/index.php/Right_triangle) $OGI$, where $I$ is the midpoint of $\overline{GH}$. Then, by the Pythagorean Theorem,
\begin{align*} OG^2 = 2a^2 &= OI^2 + GI^2 = (a+2b)^2 + b^2 \\ 0 &= a^2 - 4ab - 5b^2 = (a - 5b)(a + b) \end{align*}
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
\[0 = 1 - 4\left(\frac{b}{a}\right) - 5\left(\frac{b}{a}\right)^2\]
This is a quadratic in $\frac{b}{a}$, and solving it gives $\frac{b}{a} = \frac{1}{5},-1$. The negative solution is extraneous, and so the ratio of the areas is $\left(\frac{1}{5}\right)^2 = \frac{1}{25}$ and the answer is $10\cdot 25 + 1 = \boxed{251}$.
Remark: The division by $a^2$ is equivalent to simply setting the original area of square $ABCD$ to 1.