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 the center of its circumscribed circle, and let <math>A_1, \, B_1,</math> and <math>C_1</math> be the midpoints of sides <math>BC, \, CA,</math> and <math>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.
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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 ==
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== Solution 1 ==
 
'''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
*[[Isogonal conjugate]]
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== Solution 2 ==
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Let <math>AH</math> be the altitude of <math>\triangle ABC \implies</math>
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<cmath>\angle BAH = 90^\circ - \angle ABC, \angle OAC  = \angle OCA =</cmath>
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<cmath>=  \frac{180^\circ - \angle AOC}{2} = \frac{180^\circ - 2\angle ABC}{2} = \angle BAH.</cmath>
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Hence <math>AH</math> and <math>AO</math> are isogonals with respect to the angle <math>\angle BAC.</math>
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<cmath>\triangle OAA_1 \sim \triangle OA_2A, AH || A_1O \implies \angle AA_1O = \angle A_2AO = \angle A_1AH.</cmath>
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<math>AA_2</math> and <math>AA_1</math> are isogonals with respect to the angle <math>\angle BAC.</math>
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Similarly <math>BB_2</math> and <math>BB_1</math> are isogonals with respect to <math>\angle ABC.</math>
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Similarly <math>CC_2</math> and <math>CC_1</math> are isogonals with respect to <math>\angle ACB.</math>
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Let <math>G = AA_1 \cap BB_1</math> be the centroid of <math>\triangle ABC, L = AA_2 \cap BB_2.</math>
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<math>L</math> is the isogonal conjugate of a point <math>G</math> with respect to a triangle <math>\triangle ABC.</math>
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<cmath>G \in CC_1 \implies L \in CC_2.</cmath>
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<i><b>Corollary</b></i>
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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.
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<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.
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'''vladimir.shelomovskii@gmail.com, vvsss'''
  
 
==See Also==
 
==See Also==

Latest revision as of 17:19, 15 February 2025

Problem

Given a nonisosceles, nonright triangle $ABC,$ let $O$ denote its circumcenter, and let $A_1, \, B_1,$ and $C_1$ be the midpoints of sides $\overline{BC}$, $\overline{CA}$and $\overline{AB}$ respectively. Point $A_2$ is located on the ray $OA_1$ so that $\triangle OAA_1$ is similar to $\triangle OA_2A$. Points $B_2$ and $C_2$ on rays $OB_1$ and $OC_1,$ respectively, are defined similarly. Prove that lines $AA_2, \, BB_2,$ and $CC_2$ are concurrent.

Solution 1

LEMMA 1: In $\triangle ABC$ with circumcenter $O$, $\angle OAC = 90 - \angle B$.

PROOF of Lemma 1: The arc $AC$ equals $2\angle B$ which equals $\angle AOC$. Since $\triangle AOC$ is isosceles we have that $\angle OAC = \angle OCA = 90 - \angle B$. QED


Define $H \in BC$ s.t. $AH \perp BC$. Since $OA_1 \perp BC$, $AH \parallel OA_1$. Let $\angle AA_2O = \angle A_1AO = x$ and $\angle AA_1O = \angle A_2AO = y$. Since we have $AH \parallel OA_1$, we have that $\angle HAA_2 = x$. Also, we have that $\angle A_2AA_1 = y-x$. Furthermore, $\angle BAH = 90 - \angle B = \angle OAC$, by lemma 1. Therefore, $\angle A_1AC = 90 - \angle B + x = \angle BAA_2$. Since $A_1$ is the midpoint of $BC$, $AA_1$ is the median. However $\angle A_1AC =  \angle BAA_2$ tells us that $AA_2$ is just $AA_1$ reflected across the internal angle bisector of $A$. By definition, $AA_2$ is the $A$-symmedian. Likewise, $BB_2$ is the $B$-symmedian and $CC_2$ is the $C$-symmedian. Since the symmedians concur at the symmedian point, we are done.

QED

Solution 2

Let $AH$ be the altitude of $\triangle ABC \implies$ \[\angle BAH = 90^\circ - \angle ABC, \angle OAC  = \angle OCA =\] \[=  \frac{180^\circ - \angle AOC}{2} = \frac{180^\circ - 2\angle ABC}{2} = \angle BAH.\] Hence $AH$ and $AO$ are isogonals with respect to the angle $\angle BAC.$ \[\triangle OAA_1 \sim \triangle OA_2A, AH || A_1O \implies \angle AA_1O = \angle A_2AO = \angle A_1AH.\] $AA_2$ and $AA_1$ are isogonals with respect to the angle $\angle BAC.$

Similarly $BB_2$ and $BB_1$ are isogonals with respect to $\angle ABC.$

Similarly $CC_2$ and $CC_1$ are isogonals with respect to $\angle ACB.$

Let $G = AA_1 \cap BB_1$ be the centroid of $\triangle ABC, L = AA_2 \cap BB_2.$

$L$ is the isogonal conjugate of a point $G$ with respect to a triangle $\triangle ABC.$

\[G \in CC_1 \implies L \in CC_2.\]

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.

$AA_1, BB_1, CC_1$ are medians, therefore $AA_2, BB_2, CC_2$ 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 (ProblemsResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5
All USAMO Problems and Solutions

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