1995 USAMO Problems/Problem 3
Problem
Given a nonisosceles, nonright triangle let
denote its circumcenter, and let
and
be the midpoints of sides
and
respectively. Point
is located on the ray
so that
is similar to
. Points
and
on rays
and
respectively, are defined similarly. Prove that lines
and
are concurrent.
Solution
LEMMA 1: In with circumcenter
,
.
PROOF of Lemma 1: The arc equals
which equals
. Since
is isosceles we have that
.
QED
Define s.t.
. Since
,
. Let
and
. Since we have
, we have that
. Also, we have that
. Furthermore,
, by lemma 1. Therefore,
. Since
is the midpoint of
,
is the median. However
tells us that
is just
reflected across the internal angle bisector of
. By definition,
is the
-symmedian. Likewise,
is the
-symmedian and
is the
-symmedian. Since the symmedians concur at the symmedian point, we are done.
QED
See Also
1995 USAMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.