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

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

Remarks

Conveniently, this analytic solution takes care of all configuration issues we may have encountered had we used a more traditional solution, such as angle-chasing(which would indeed work).

~AopsUser101

2009 USAMO (Problems • Resources)
Preceded by First question	Followed by Problem 2
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2009 USAMO Problems/Problem 1

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Solution

Remarks

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