# 2009 USAMO Problems/Problem 1

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## Problem

Given circles $\omega_1$ and $\omega_2$ intersecting at points $X$ and $Y$, let $\ell_1$ be a line through the center of $\omega_1$ intersecting $\omega_2$ at points $P$ and $Q$ and let $\ell_2$ be a line through the center of $\omega_2$ intersecting $\omega_1$ at points $R$ and $S$. Prove that if $P, Q, R$ and $S$ lie on a circle then the center of this circle lies on line $XY$.

## Solution

Let $\omega_3$ be the circumcircle of $PQRS$. Define $r_i$ to be the radius and $O_i$ to be the center of the circle $\omega_i, i = 1,2,3$. Then $O_1$ lies on the line passing through the intersections of $\omega_2, \omega_3$, or their radical axis, and similarly $O_2$ lies on the radical axis of $\omega_1, \omega_3$. Then, the power of $O_1$ with respect to $\omega_2,\omega_3$ are the same, and similarly for $O_2$: \begin{align*}O_1O_2^2 - r_2^2 &= O_1O_3^2 - r_3^2 \\ O_2O_1^2 - r_1^2 &= O_2O_3^2 - r_3^2 \end{align*} Subtracting gives $O_1O_3^2 - r_1^2 = O_2O_3^2 - r_2^2$, so $O_3$ lies on the radical axis of $\omega_1,\omega_2$. Thus $X,Y,O_3$ are collinear.