3. Let $A, B, C$, and $D$ be points on a circle such that $A B=11$ and $C D=19$. Point $P$ is on segment $A B$ with $A P=6$, and $Q$ is on segment $C D$ with $C Q=7$. The line through $P$ and $Q$ intersects the circle at $X$ and $Y$. If $P Q=27$, find $X Y$.
Answer: 31
Solution: Suppose $X, P, Q, Y$ lie in that order. Let $P X=x$ and $Q Y=y$. By power of a point from $P, x \cdot(27+y)=30$, and by power of a point from $Q, y \cdot(27+x)=$ 84. Subtracting the first from the second, $27 \cdot(y-x)=54$, so $y=x+2$. Now, $x \cdot(29+x)=30$, and we find $x=1,-30$. Since -30 makes no sense, we take $x=1$ and obtain $X Y=1+27+3=31$