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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== Solution == | == 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>. | |
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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 | QED | ||
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QED | QED | ||
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+ | ==See Also== | ||
+ | {{USAMO box|year=1995|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | [[Category:Olympiad Geometry Problems]] |
Revision as of 08:06, 19 July 2016
Problem
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.
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.