2002 IMO Shortlist Problems/G4
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.