# 2002 IMO Shortlist Problems/G4

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## Problem

Circles $\displaystyle S_1$ and $\displaystyle S_2$ intersect at points $\displaystyle P$ and $\displaystyle Q$. Distinct points $\displaystyle A_1$ and $\displaystyle B_1$ (not at $\displaystyle P$ or $\displaystyle Q$) are selected on $\displaystyle S_1$. The lines $\displaystyle A_1P$ and $\displaystyle B_1P$ meet $\displaystyle S_2$ again at $\displaystyle A_2$ and $\displaystyle B_2$ respectively, and the lines $\displaystyle A_1B_1$ and $\displaystyle A_2B_2$ meet at $\displaystyle{} C$. Prove that, as $\displaystyle A_1$ and $\displaystyle B_1$ vary, the circumcenters of triangles $\displaystyle A_1A_2C$ all lie on one fixed circle.

## Solution

We will use directed angles mod $\displaystyle \pi$.

Since $\displaystyle A_1, B_1, C$ are collinear, $\angle CA_1Q = \angle B_1A_1Q$. Since $\displaystyle A_1, B_1, P, Q$ all lie on $\displaystyle S_1$, $\angle B_1A_1Q = \angle B_1PQ$. Hence, $\angle CA_1Q = \angle B_1PQ$. Similarly, $\angle CA_2Q = \angle B_2PQ$. But since $\displaystyle B_1, B_2, P$ are collinear, $\angle B_1PQ = \angle B_2PQ$. This means that $\angle CA_1Q = \angle CA_2Q$, so $\displaystyle C, A_1, A_2, Q$ are concyclic. This means that, regardless of the location of $\displaystyle B_1$, the circumcenter of $\displaystyle A_1A_2C$ is the circumcenter of $\displaystyle A_1A_2Q$.

Note that as $\displaystyle A_1$ varies, the values of $\angle QA_1A_2 = \angle QA_1P$ and $\angle QA_2A_1 = \angle QA_2P$ stay fixed, at half the measure of arc $\displaystyle QP$ on circles $\displaystyle S_1$ and $\displaystyle S_2$, respectively. Therefore all triangles $\displaystyle QA_1A_2$, are similar. If $\displaystyle X$ denotes the circumcenter of triangle $\displaystyle QA_1A_2$, then we must also have all triangles $\displaystyle QA_1X$ are similar. Since $\displaystyle Q$ is fixed, this means that there exists a spiral similarity that maps every point $\displaystyle A_1$ to its corresponding point $\displaystyle X$. This means that the locus of $\displaystyle X$ must be the image of the locus of $\displaystyle A_1$ under the spiral similarity. But the locus of $\displaystyle A_1$ is a circle, and the image of a circle under a spiral similarity is another circle. Q.E.D.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.