2007 USAMO Problems/Problem 6
From AoPSWiki
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
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).
Let
and
(second intersection).
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: AoPS | ML) | ||
| Preceded by Problem 5 | 1 • 2 • 3 • 4 • 5 • 6 | Followed by Last Question |











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