Difference between revisions of "1995 USAMO Problems/Problem 3"
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==Problem== | ==Problem== | ||
− | Given a nonisosceles, nonright triangle <math>ABC,</math> let <math>O</math> denote | + | Given a nonisosceles, nonright triangle <math>ABC,</math> let <math>O</math> denote its circumcenter, and let <math>A_1, \, B_1,</math> and <math>C_1</math> be the midpoints of sides <math>\overline{BC}</math>, <math>\overline{CA}</math>and <math>\overline{AB}</math> respectively. Point <math>A_2</math> is located on the ray <math>OA_1</math> so that <math>\triangle OAA_1</math> is similar to <math>\triangle OA_2A</math>. Points <math>B_2</math> and <math>C_2</math> on rays <math>OB_1</math> and <math>OC_1,</math> respectively, are defined similarly. Prove that lines <math>AA_2, \, BB_2,</math> and <math>CC_2</math> are concurrent. |
− | |||
== Solution == | == Solution == | ||
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QED | QED | ||
+ | *[[Isogonal conjugate]] | ||
==See Also== | ==See Also== |
Latest revision as of 01:42, 19 April 2024
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.