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
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 inversions map a circle's center collinear with the center of inversion. Likewise,
is the other line tangent to
and perpendicular to
.
Let w_{A} AO=X and w_{A}'
AO=X' [second intersection].
Let \Omega_{A} AO=Y and \Omega_{A}'
AO=Y' [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
.
Solution by AoPS user Altheman
2007 USAMO (Problems • Resources) | ||
Preceded by Problem 5 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |