# 2009 USAMO Problems/Problem 1

## 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 1

Let $\omega_3$ be the circumcircle of $PQRS$, $r_i$ to be the radius of $\omega_i$, and $O_i$ to be the center of the circle $\omega_i$, where $i \in \{1,2,3\}$. Note that $SR$ and $PQ$ are the radical axises of $O_1$ , $O_3$ and $O_2$ , $O_3$ respectively. Hence, by power of a point(the power of $O_1$ can be expressed using circle $\omega_2$ and $\omega_3$ and the power of $O_2$ can be expressed using circle $\omega_1$ and $\omega_3$), $$O_1O_2^2 - r_2^2 = O_1O_3^2 - r_3^2$$ $$O_2O_1^2 - r_1^2 = O_2O_3^2 - r_3^2$$ Subtracting these two equations yields that $O_1O_3^2 - r_1^2 = O_2O_3^2 - r_2^2$, so $O_3$ must lie on the radical axis of $\omega_1$ , $\omega_2$.

~AopsUser101

## Solution 2

Define $\omega_i$ and $O_i$ similarly to above. Note that $O_1O_3$ is perpendicular to RS and $O_2 O_3$ is perpendicular to PQ. Thus, the intersection of PQ and RS must be the orthocenter of triangle $O_1O_2O_3$. Define this as point $H$. Extending line $O_3H$ to meet $O_1O_2$, we note that $O_3H$ is perpendicular to $O_1O_2$.

In addition, note that by the radical axis theorem, the intersection of $PQ$ and $RS$ must also lie on the radical axis of $\omega_1$ and $\omega_2$. Because the radical axis of $\omega_1$ and $\omega_2$ is perpendicular to $O_1O_2$ and contains $H$, it must also contain $O_3$, and we are done.