Problem 159

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 $m + n + p + q = 80 + 13 + 34 + 3 = \boxed{\textbf{(E) } 130}$.