2019 AIME I Problem 15

$A$ lies on both $\omega$ and $\omega_1,$ but tangent circles meet only once, so $\omega_1$ is tangent to $\omega$ at $A$ itself (and $\omega_2$ at $B$)

Let $Z$ be the intersection of the tangents to $\omega$ at $A$ and $B:$ then $Z$ lies on line $PQ$ with $ZP \cdot ZQ = ZX \cdot ZY = ZA^2,$ and $ZA^2 - ZP^2 = AP \cdot PB$

Since $A$ lies on both $\omega$ and $\omega_1$ and internally tangent circles meet only at their point of tangency, $\omega_1$ is tangent to $\omega$ at $A;$ likewise $\omega_2$ is tangent at $B.$ Let $Z$ be the intersection of the tangent lines to $\omega$ at $A$ and $B.$ Each tangent line is also tangent to the corresponding inner circle, so the powers of $Z$ with respect to $\omega_1$ and $\omega_2$ are $ZA^2$ and $ZB^2,$ which are equal. Hence $Z$ lies on the radical axis $PQ,$ and along the line through $Z,$ $X,$ $P,$ $Q,$ $Y:$ $$ZP \cdot ZQ = ZA^2 = ZX \cdot ZY,$$ the last equality because $ZA$ is tangent to $\omega.$

Because $ZA = ZB,$ the point $Z$ lies on the perpendicular bisector of $\overline{AB};$ if $M$ is the midpoint of $\overline{AB},$ then $ZA^2 - ZP^2$ $= MA^2 - MP^2$ $= 4^2 - 1^2 = 15.$ Also the power of $P$ in $\omega$ gives $XP \cdot PY = AP \cdot PB = 15.$ Set $s = ZP$ and $u = ZX,$ so $ZY = u + 11.$ The relations become $$\begin{aligned} u(u + 11) &= s^2 + 15, \\ (s - u)(u + 11 - s) &= 15. \end{aligned}$$ Expanding the second and substituting the first yields $u = s - \frac{11}{2} + \frac{15}{s},$ and substituting back gives $\left(s + \frac{15}{s}\right)^2 - \frac{121}{4} = s^2 + 15,$ so $\frac{225}{s^2} = \frac{61}{4}.$

Finally $ZQ = \frac{ZA^2}{ZP} = s + \frac{15}{s},$ so $PQ = ZQ - ZP = \frac{15}{s}$ and $PQ^2 = \frac{225}{s^2} = \frac{61}{4}.$ Therefore $m + n = 61 + 4 = 65.$