2019 AMC 12A Problem 22

Let $s = AB.$ Since $\dfrac{BP}{CP} = 3,$ we have $BP = \dfrac{3s}{4}$ and $CP = \dfrac{s}{4}.$ Put $P$ at the origin with $BC$ on the $x$-axis, $B = \left(-\tfrac{3s}{4}, 0\right),$ $C = \left(\tfrac{s}{4}, 0\right),$ and apex $A = \left(-\tfrac{s}{4}, \tfrac{s\sqrt{3}}{2}\right).$ Points $P, O, A$ are collinear, so $O = t \cdot A$ for some scalar $t.$ Two conditions pin it down: $O$ is at distance $17$ from line $BC,$ giving $|t| \cdot \dfrac{s\sqrt{3}}{2} = 17,$ and $A$ is on $\omega,$ giving $|t - 1| \cdot \dfrac{s\sqrt{13}}{4} = 20$ since $|A| = \dfrac{s\sqrt{13}}{4}.$ Solving, $|t| s = \dfrac{34}{\sqrt{3}}$ and $|t - 1| s = \dfrac{80}{\sqrt{13}}.$ The valid configuration has $O$ and $A$ on opposite sides of $P,$ so $|t-1|s=|t|s+s.$ Therefore $$AB = s = \dfrac{80}{\sqrt{13}} - \dfrac{34}{\sqrt{3}}.$$ Then $m + n + p + q = 80 + 13$ $+ 34 + 3 = 130.$ Thus, the correct answer is E.