2002 IMO Shortlist Problems/G4
Circles and intersect at points and . Distinct points and (not at or ) are selected on . The lines and meet again at and respectively, and the lines and meet at . Prove that, as and vary, the circumcenters of triangles all lie on one fixed circle.
We will use directed angles mod .
Since are collinear, . Since all lie on , . Hence, . Similarly, . But since are collinear, . This means that , so are concyclic. This means that, regardless of the location of , the circumcenter of is the circumcenter of .
Note that as varies, the values of and stay fixed, at half the measure of arc on circles and , respectively. Therefore all triangles , are similar. If denotes the circumcenter of triangle , then we must also have all triangles are similar. Since is fixed, this means that there exists a spiral similarity that maps every point to its corresponding point . This means that the locus of must be the image of the locus of under the spiral similarity. But the locus of is a circle, and the image of a circle under a spiral similarity is another circle. Q.E.D.
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