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. It seems like at least.

~notverysmart

2009 USAMO (Problems • Resources)
Preceded by First question	Followed by Problem 2
1 • 2 • 3 • 4 • 5 • 6
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2009 USAMO Problems/Problem 1

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Problem

Solution 1

Solution 2

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