2002 IMO Shortlist Problems/G4
From AoPSWiki
Problem
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.
Solution
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.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
