2001 USAMO Problems/Problem 2
Problem
Let be a triangle and let
be its incircle. Denote by
and
the points where
is tangent to sides
and
, respectively. Denote by
and
the points on sides
and
, respectively, such that
and
, and denote by
the point of intersection of segments
and
. Circle
intersects segment
at two points, the closer of which to the vertex
is denoted by
. Prove that
.
Solution
It is well known that the excircle opposite is tangent to
at the point
. (Proof: let the points of tangency of the excircle with the lines
be
respectively. Then
. It follows that
, and
, so
.)
Now consider the homothety that carries the incircle of to its excircle. The homothety also carries
to
(since
are collinear), and carries the tangency points
to
. It follows that
.
![[asy][/asy]](http://latex.artofproblemsolving.com/0/6/1/061f07da08b2b0c2554b2543f4a362d2aedbcc4b.png)
By Menelaus' Theorem it follows that . It easily follows that
.
See also
2001 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |