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. | ||
+ | |||
+ | |||
+ | ''' | ||
+ | == Bold text == | ||
+ | Solution''' | ||
+ | |||
+ | |||
+ | LEMMA 1: | ||
+ | |||
+ | In <math>\triangle ABC</math> with circumcenter <math>O</math>, <math>\angle OAC = 90 - \angle B</math>. | ||
+ | |||
+ | PROOF of Lemma 1: | ||
+ | |||
+ | 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 | ||
+ | |||
+ | -------------------------------------------------------------- | ||
+ | |||
+ | 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. | ||
+ | |||
+ | QED |
Revision as of 20:58, 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