AIME 2016 I, Problem 15

Circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y$. Line $\ell$ is tangent to $\omega_1$ and $\omega_2$ at $A$ and $B$, respectively, with line $AB$ closer to point $X$ than to $Y$. Circle $\omega$ passes through $A$ and $B$ intersecting $\omega_1$ again at $D \neq A$ and intersecting $\omega_2$ again at $C \neq B$. The three points $C$, $Y$, $D$ are collinear, $XC = 67$, $XY = 47$, and $XD = 37$. Find $AB^2$.

Solution: Let $Z = XY \cap AB$. By the radical axis theorem $AD, XY, BC$ are concurrent, say at $P$. Moreover, $\triangle DXP \sim \triangle PXC$ by simple angle chasing. Let $y = PX, x = XZ$. Then \[\frac{y}{37} = \frac{67}{y} \qquad \implies \qquad y^2 = 37 \cdot 67.\] Now, $AZ^2 = \tfrac 14 AB^2$, and by power of a point, \begin{align}         x(y-x) &= \tfrac 14 AB^2, \quad \text{and} \\         x(47+x) &= \tfrac 14 AB^2     \end{align} Solving, we get \[\tfrac 14 AB^2 = \tfrac 12 (y-47)\cdot \tfrac 12 (y+47) \qquad \implies\] \[\qquad AB ^ 2 = 37\cdot 67 - 47^2 = \fbox{270}\]

Answer: 270