Difference between revisions of "1995 USAMO Problems/Problem 3"
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+ | '''Inversive Solution''' | ||
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+ | Consider the inversion <math>\Omega</math> centered at <math>O</math> with radius <math>OA</math>. From the given similarity, we obtain <math>OA^2=OA_1\cdot OA_2</math>, i.e. <math>A_1</math> and <math>A_2</math> are inverses wrt to <math>(ABC)</math>. Hence <cmath>\angle OBA_2=\angle OA_1C=90^\circ.</cmath>Similary <math>A_2C</math> is tangent to <math>(ABC).</math> Thus, <math>AA_2</math> is a symmedian. Since symmedians concur at the Lemoine point, we are done. | ||
==See Also== | ==See Also== |
Revision as of 23:14, 22 April 2017
Problem
Given a nonisosceles, nonright triangle let
denote the center of its circumscribed circle, 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, i.e. these three lines intersect at a point.
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
Inversive Solution
Consider the inversion centered at
with radius
. From the given similarity, we obtain
, i.e.
and
are inverses wrt to
. Hence
Similary
is tangent to
Thus,
is a symmedian. Since symmedians concur at the Lemoine point, we are done.
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.