AIME I 2019/15 wrote:

Let $\overline{AB}$ be a chord of a circle $\omega$ , and let $P$ be a point on the chord $\overline{AB}$ . Circle $\omega_1$ passes through $A$ and $P$ and is internally tangent to $\omega$ . Circle $\omega_2$ passes through $B$ and $P$ and is internally tangent to $\omega$ . Circles $\omega_1$ and $\omega_2$ intersect at points $P$ and $Q$ . Line $PQ$ intersects $\omega$ at $X$ and $Y$ . Assume that $AP=5$ , $PB=3$ , $XY=11$ , and $PQ^2 = \tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

AoPS Link

Proof. Note by Radical Axis that the tangents from $A$ and $B$ and the line $XY$ concur at a point we will call $Z$ . Let $AQ$ meet $\omega$ again at $D$ .

Lemma: In $\triangle ABC$ with circumcircle $\Omega$ , let $D$ be a point on $\Omega$ such that $AD$ is the $A$ -symmedian. Let $T$ be the midpoint of $AD$ and let $BT$ meet the line through $C$ parallel to $AD$ at $E$ . Show that $E$ lies on $\Omega$ . (USAJMO 2011/5)

Proof. We now employ Barycentric Coordinates. Set $A=(1,0,0),B=(0,1,0),C=(0,0,1)$ . Since $D$ is on the $A$ -symmedian, it can be parameterized by $D=(t:b^2:c^2)$ but it is also on $\Omega$ so $\text{Pow}_\Omega (D)=0$ . Of course, $\Omega$ has equation $a^2yz+b^2zx+c^2xy=0$ , so plugging in $D$ gives $a^2b^2c^2+2tb^2c^2=0 \Rightarrow t=-\frac{a^2}2$ . So $D=\left(-\frac{a^2}2,b^2,c^2\right)=(-a^2:2b^2:2c^2).$ Thus, the point at infinity along the line $A$ -symmedian has coordinates $P_\infty=(b^2+c^2:-b^2:-c^2).$
Normalize $D$ by multiplying each component by $\frac{1}{-a^2+2b^2+2c^2}$ . The midpoint formula on normalized $D$ and $A$ gives $T=(-a^2+b^2+c^2:b^2:c^2)$ . Note that $E$ is the intersection of cevians $BT$ and $CP_\infty$ . Points on $BT$ can be parameterized by $(-a^2+b^2+c^2:s:c^2)$ while points on $CP_\infty$ can be parameterized by $(b^2+c^2:-b^2:t)$ . Hence, their intersection must have coordinates $E=[(b^2+c^2)(-a^2+b^2+c^2):-b^2(-a^2+b^2+c^2):c^2(b^2+c^2)].$ It remains to check that $\text{Pow}_\Omega (E)=0$ . Plugging $E$ into the circumcircle equation yields $a^2[-b^2(-a^2+b^2+c^2)c^2(b^2+c^2)]+b^2[(b^2+c^2)(-a^2+b^2+c^2)c^2(b^2+c^2)]+c^2[(b^2+c^2)(-a^2+b^2+c^2)(-b^2)(-a^2+b^2+c^2)]$ which is clearly true after factoring. $\square$

Consider the homothety centered at $A$ that maps $\triangle AQP\mapsto ADB$ . By construction, $QP\parallel DB$ so we can conclude that $Q$ is the midpoint of $\overline{XY}$ by our lemma.

Unfortunately, we cannot do the answer extraction with bary since we are not given the side lengths of $\triangle XAB$ , but Power of a Point on $P$ with respect to $\omega$ and minimal computations give away an answer of $PQ^2=\frac{61}{4}$ or $\boxed{65}.$