Problem (Number Theory):
Let $T=688$. Let $a$ be the least nonzero digit in $T$, and let $b$ be the greatest digit in $T$. In square $N O R M, N O=b$, and points $P_{1}$ and $P_{2}$ lie on $\overline{N O}$ and $\overline{O R}$, respectively, so that $O P_{1}=O P_{2}=a$. A circle centered at $O$ has radius $a$, and quarter-circular arc $\widehat{P_{1} P_{2}}$ is drawn. There is a circle that is tangent to $\widehat{P_{1} P_{2}}$ and to sides $\overline{M N}$ and $\overline{M R}$. The radius of this circle can be written in the form $x-y \sqrt{2}$, where $x$ and $y$ are positive integers. Compute $x+y$.

Solution:
Let $r$ and $Q$ denote the respective radius and center of the circle whose radius is concerned. Let this circle be tangent to arc $\widehat{P_{1} P_{2}}$ at point $P$, and let it be tangent to sides $\overline{M N}$ and $\overline{M R}$ at points $T_{1}$ and $T_{2}$, respectively.



Note that $Q$ lies on diagonal $\overline{M O}$ because it is equidistant to $\overline{M N}$ and $\overline{M R}$. Points $Q, P$, and $O$ must be collinear because the circles centered at $Q$ and $O$ are mutually tangent at point $P$. It therefore follows that $P$ also lies on diagonal $\overline{M O}$. Because triangles $Q T_{1} M$ and $Q T_{2} M$ are isosceles right triangles, it follows that $M Q=r \sqrt{2}$. Thus

$$
b \sqrt{2}=M O=M Q+Q P+P O=r \sqrt{2}+r+a
$$

Solving this equation yields $r=a+2 b-(a+b) \sqrt{2}$. With $T=688, a=6$ and $b=8$, so $r=22-14 \sqrt{2}$, hence $x+y=22+14=\mathbf{3 6}$.

Final answer: 36