Circles $\omega$ and $\gamma$, both centered at $O$, have radii 20 and 17, respectively. Equilateral triangle $ABC$, whose interior lies in the interior of $\omega$ but in the exterior of $\gamma$, has vertex $A$ on $\omega$, and the line containing side $\overline{BC}$ is tangent to $\gamma$. Segments $\overline{AO}$ and $\overline{BC}$ intersect at $P$, and $\frac{BP}{CP} = 3$. Then $AB$ can be written in the form $\frac{m}{\sqrt{n}} - \frac{p}{\sqrt{q}}$ for positive integers $m, n, p, q$ with $\gcd(m, n) = \gcd(p, q) = 1$. What is $m + n + p + q$?
圆 $\omega$ 和 $\gamma$ 都以 $O$ 为圆心,半径分别为 20 和 17。正三角形 $ABC$ 的内部位于 $\omega$ 的内部但位于 $\gamma$ 的外部,顶点 $A$ 在 $\omega$ 上,边 $\overline{BC}$ 所在直线与 $\gamma$ 相切。线段 $\overline{AO}$ 和 $\overline{BC}$ 相交于 $P$,且 $\frac{BP}{CP} = 3$。则 $AB$ 可写成 $\frac{m}{\sqrt{n}} - \frac{p}{\sqrt{q}}$ 的形式,其中 $m, n, p, q$ 为正整数且 $\gcd(m, n) = \gcd(p, q) = 1$。求 $m + n + p + q$。