Acute triangle $A B C$ has circumcircle $\Gamma$. Let $M$ be the midpoint of $B C$. Points $P$ and $Q$ lie on $\Gamma$ so that $\angle A P M=90^{\circ}$ and $Q \neq A$ lies on line $A M$. Segments $P Q$ and $B C$ intersect at $S$. Suppose that $B S=1, C S=3, P Q=8 \sqrt{\frac{7}{37}}$, and the radius of $\Gamma$ is $r$. If the sum of all possible values of $r^{2}$ can be expressed as $\frac{a}{b}$ for relatively prime positive integers $a$ and $b$, compute $100 a+b$.
Solution: Let $A^{\prime}$ be the $A$-antipode in $\Gamma$, let $O$ be the center of $\Gamma$, and let $T:=A A^{\prime} \cap B C$. Note that $A^{\prime}$ lies on line $P M$. The key observation is that $T$ is the reflection of $S$ about $M$; this follows by the Butterfly Theorem on chords $\overline{P A^{\prime}}$ and $\overline{A Q}$. Let $\theta:=\angle A M P$ and $x=O T=O S$. Observe that $\cos \theta=\frac{P M}{A M}=\frac{P Q}{A A^{\prime}}=\frac{P Q}{2 r}$. We find the area of $\triangle A M A^{\prime}$ in two ways. First, we have \[ \begin{aligned} 2\left[A M A^{\prime}\right] & =A M \cdot M A \cdot \sin \theta \\ & =A M \cdot \frac{M B \cdot M C}{P M} \cdot \sin \theta \\ & =4 \tan \theta \\ & =8 r \sqrt{\frac{37}{448}-\frac{1}{4 r^{2}}} \end{aligned} \] On the other hand, \[ \begin{aligned} 2\left[A M A^{\prime}\right] & =M T \cdot A A^{\prime} \cdot \sin \angle O T M \\ & =2 r \sqrt{1-\frac{1}{x^{2}}} \end{aligned} \] Setting the two expressions equal and squaring yields $\frac{37}{28}-\frac{4}{r^{2}}=1-\frac{1}{x^{2}}$. By Power of a Point, $3=B S \cdot S C=r^{2}-x^{2}$, so $x^{2}=r^{2}-3$. Substituting and solving the resulting quadratic in $r^{2}$ gives $r^{2}=\frac{16}{3}$ and $r^{2}=7$. Thus $\frac{a}{b}=\frac{37}{3}$, so $100 a+b=3703$. $\fbox{3703}$.