2002 IMO Shortlist Problems/G4

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.