# Difference between revisions of "1995 USAMO Problems/Problem 3"

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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, i.e. these three lines intersect at a point. | 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, i.e. these three lines intersect at a point. | ||

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+ | ''' | ||

+ | == Bold text == | ||

+ | Solution''' | ||

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+ | LEMMA 1: | ||

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+ | In <math>\triangle ABC</math> with circumcenter <math>O</math>, <math>\angle OAC = 90 - \angle B</math>. | ||

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+ | PROOF of Lemma 1: | ||

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+ | The arc <math>AC</math> equals <math>2\angle B</math> which equals <math>\angle AOC</math>. Since <math>\triangle AOC</math> is isosceles we have that <math>\angle OAC = \angle OCA = 90 - \angle B</math>. | ||

+ | QED | ||

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+ | -------------------------------------------------------------- | ||

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

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+ | QED |

## Revision as of 19:58, 4 July 2012

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.

## Bold text

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