2007 USAMO Problems/Problem 6
Problem
Let be an acute triangle with
,
, and
being its incircle, circumcircle, and circumradius, respectively. Circle
is tangent internally to
at
and tangent externally to
. Circle
is tangent internally to
at
and tangent internally to
. Let
and
denote the centers of
and
, respectively. Define points
,
,
,
analogously. Prove that

with equality if and only if triangle is equilateral.
Solution
An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.
Lemma:
Proof:
Note and
lie on
since for a pair of tangent circles, the point of tangency and the two centers are collinear.
Let touch
,
, and
at
,
, and
, respectively. Note
. Consider an inversion,
, centered at
, passing through
,
. Since
,
is orthogonal to the inversion circle, so
. Consider
. Note that
passes through
and is tangent to
, hence
is a line that is tangent to
. Furthermore,
because
is symmetric about
, so the inversion preserves that reflective symmetry. Since it is a line that is symmetric about
, it must be perpendicular to
. Likewise,
is the other line tangent to
and perpendicular to
.
Let and
(second intersection).
Evidently, and
. We want:

by inversion. Note that , and they are tangent to
, so the distance between those lines is
. Drop a perpendicular from
to
, touching at
. Then
. Then
,
=
. So

Note that . Applying the double angle formulas and
, we get


End Lemma
The problem becomes:


which is true because , equality is when the circumcenter and incenter coincide. As before,
, so, by symmetry,
. Hence the inequality is true iff
is equilateral.
Comment: It is much easier to determine by considering
. We have
,
,
, and
. However, the inversion is always nice to use. This also gives an easy construction for
because the tangency point is collinear with the intersection of
and
.
See also
2007 USAMO (Problems • Resources) | ||
Preceded by Problem 5 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |