jmc

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, $$\begin{array}{rl}O{G}^2=2a^2& =O{I}^2+G{I}^2=(a+2b{)}^2+b^2\\ 0& =a^2-4ab-5b^2=(a-5b)(a+b)\end{array}$$ 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 $\frac{[EFGH]}{[ABCD]}={\left(\frac{1}{5}\right)}^2=\frac{1}{25}$, 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 $10\cdot 25+1=\boxed{251}$.