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 $\dfrac{BP}{CP} = 3$. Then $AB$ can be written in the form $\dfrac{m}{\sqrt{n}} - \dfrac{p}{\sqrt{q}}$ for positive integers $m$, $n$, $p$, $q$ with $\text{gcd}(m,n) = \text{gcd}(p,q) = 1$. What is $m+n+p+q$?
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$\textbf{(A) } 42 \qquad \textbf{(B) }86 \qquad \textbf{(C) } 92 \qquad \textbf{(D) } 114 \qquad \textbf{(E) } 130$
Let $S$ be the point of tangency between $\overline{BC}$ and $\gamma$, and $M$ be the midpoint of $\overline{BC}$. Note that $AM \perp BS$ and $OS \perp BS$. This implies that $\angle OAM \cong \angle AOS$, and $\angle AMP \cong \angle OSP$. Thus, $\triangle PMA \sim \triangle PSO$.
If we let $s$ be the side length of $\triangle ABC$, then it follows that $AM = \frac{\sqrt{3}}{2}s$ and $PM = \frac{s}{4}$. This implies that $AP = \frac{\sqrt{13}}{4}s$, so $\frac{AM}{AP} = \frac{2\sqrt{3}}{\sqrt{13}}$. Furthermore, $\frac{AM + SO}{AO} = \frac{AM}{AP}$ (because $\triangle PMA \sim \triangle PSO$) so this gives us the equation
\[\frac{\frac{\sqrt{3}}{2}s + 17}{20} = \frac{2\sqrt{3}}{\sqrt{13}}\]
to solve for the side length $s$, or $AB$. Thus,
\[\frac{\sqrt{39}}{2}s + 17\sqrt{13} = 40\sqrt{3}\]
\[\frac{\sqrt{39}}{2}s = 40\sqrt{3} - 17\sqrt{13}\]
\[s = \frac{80}{\sqrt{13}} - \frac{34}{\sqrt{3}} = AB\]
The problem asks for $m + n + p + q = 80 + 13 + 34 + 3 = \boxed{\textbf{(E) } 130}$.