smc

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, $△PMA\sim △PSO$. If we let $s$ be the side length of $△ABC$, 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 $△PMA\sim △PSO$) 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 $m+n+p+q=80+13+34+3=\boxed{130}$.