Problem: A circle with a diameter $\overline{PQ}\,$ of length 10 is internally tangent at $P^{}_{}$ to a circle of radius 20. Square $ABCD\,$ is constructed with $A\,$ and $B\,$ on the larger circle, $\overline{CD}\,$ tangent at $Q\,$ to the smaller circle, and the smaller circle outside $ABCD\,$. The length of $\overline{AB}\,$ can be written in the form $m + \sqrt{n}\,$, where $m\,$ and $n\,$ are integers. Find $m + n\,$. Solution: Within our square, draw a line from the center of the square $E$ through the center of the smaller circle extending in both directions of the greater circle. Call the point of intersection of the square that is not tangent to the small circle $D$. Draw a line from the center of the big circle $O$ through the top or bottom diagonal facing away from the circle. Now, we are in good shape to use the problem conditions. We want to find the side length $s$. We know that $OA = 20$. We also know that $O Q=O P-P Q=20-10=10$ and $O D=A B-O Q=A B-10$. The other leg, $AD = \frac{1}{2} A B$. Let $x = AB$. Then, $ \begin{aligned} & (x10)^2+\left(\frac{1}{2} x\right)^2=20^2 \\ & x^2-20 x+100+\frac{1}{4} x^2-400=0 \\ & x^2-16 x-240=0\end{aligned} $ So $x = AE=8+\sqrt{304}$ meaning $m+n = 312$.