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 1 == | |
− | == Solution == | ||
'''LEMMA 1: ''' In <math>\triangle ABC</math> with circumcenter <math>O</math>, <math>\angle OAC = 90 - \angle B</math>. | '''LEMMA 1: ''' In <math>\triangle ABC</math> with circumcenter <math>O</math>, <math>\angle OAC = 90 - \angle B</math>. | ||
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QED | QED | ||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | Let <math>AH</math> be the altitude of <math>\triangle ABC \implies</math> | ||
+ | <cmath>\angle BAH = 90^\circ - \angle ABC, \angle OAC = \angle OCA =</cmath> | ||
+ | <cmath>= \frac{180^\circ - \angle AOC}{2} = \frac{180^\circ - 2\angle ABC}{2} = \angle BAH.</cmath> | ||
+ | Hence <math>AH</math> and <math>AO</math> are isogonals with respect to the angle <math>\angle BAC.</math> | ||
+ | <cmath>\triangle OAA_1 \sim \triangle OA_2A, AH || A_1O \implies \angle AA_1O = \angle A_2AO = \angle A_1AH.</cmath> | ||
+ | <math>AA_2</math> and <math>AA_1</math> are isogonals with respect to the angle <math>\angle BAC.</math> | ||
+ | |||
+ | Similarly <math>BB_2</math> and <math>BB_1</math> are isogonals with respect to <math>\angle ABC.</math> | ||
+ | |||
+ | Similarly <math>CC_2</math> and <math>CC_1</math> are isogonals with respect to <math>\angle ACB.</math> | ||
+ | |||
+ | Let <math>G = AA_1 \cap BB_1</math> be the centroid of <math>\triangle ABC, L = AA_2 \cap BB_2.</math> | ||
+ | |||
+ | <math>L</math> is the isogonal conjugate of a point <math>G</math> with respect to a triangle <math>\triangle ABC.</math> | ||
+ | |||
+ | <cmath>G \in CC_1 \implies L \in CC_2.</cmath> | ||
+ | |||
+ | <i><b>Corollary</b></i> | ||
+ | |||
+ | If median and symmedian start from any vertex of the triangle, then the angle formed by the symmedian and the angle side has the same measure as the angle between the median and the other side of the angle. | ||
+ | |||
+ | <math>AA_1, BB_1, CC_1</math> are medians, therefore <math>AA_2, BB_2, CC_2</math> are symmedians, so the three symmedians meet at a point which is triangle center called the Lemoine point. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
==See Also== | ==See Also== |
Latest revision as of 17:19, 15 February 2025
Contents
[hide]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 1
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
Solution 2
Let be the altitude of
Hence
and
are isogonals with respect to the angle
and
are isogonals with respect to the angle
Similarly and
are isogonals with respect to
Similarly and
are isogonals with respect to
Let be the centroid of
is the isogonal conjugate of a point
with respect to a triangle
Corollary
If median and symmedian start from any vertex of the triangle, then the angle formed by the symmedian and the angle side has the same measure as the angle between the median and the other side of the angle.
are medians, therefore
are symmedians, so the three symmedians meet at a point which is triangle center called the Lemoine point.
vladimir.shelomovskii@gmail.com, vvsss
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.