Difference between revisions of "2014 USAMO Problems/Problem 5"
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==Solution== | ==Solution== | ||
+ | Let <math>O_1</math> be the center of <math>(AHPC)</math>, <math>O</math> be the center of <math>(ABC)</math>. Note that <math>(O_1)</math> is the reflection of <math>(O)</math> across <math>AC</math>, so <math>AO=AO_1</math>. Additionally | ||
+ | <cmath> | ||
+ | \angle AYC=180-\angle APC=180-\angle AHC=\angle B | ||
+ | </cmath> | ||
+ | so <math>Y</math> lies on <math>(O)</math>. Now since <math>XO,OO_1,XO_1</math> are perpendicular to <math>AB,AC,</math> and their bisector, <math>XOO_1</math> is isosceles with <math>XO=OO_1</math>, and <math>\angle XOO_1=180-\angle A</math>. Also | ||
+ | <cmath> | ||
+ | \angle AOY=2\angle ACY=2(90-\angle PAC)=2(90-\frac{A}{2})=180-\angle A = \angle XOO_1 | ||
+ | </cmath> | ||
+ | But <math>YO=OA</math> as well, and <math>\angle YOX=\angle AOO_1</math>, so <math>\triangle OYX\cong \triangle OAO_1</math>. Thus <math>XY=AO_1=AO</math>. |
Revision as of 17:29, 30 December 2014
Problem
Let be a triangle with orthocenter
and let
be the second intersection of the circumcircle of triangle
with the internal bisector of the angle
. Let
be the circumcenter of triangle
and
the orthocenter of triangle
. Prove that the length of segment
is equal to the circumradius of triangle
.
Solution
Let be the center of
,
be the center of
. Note that
is the reflection of
across
, so
. Additionally
so
lies on
. Now since
are perpendicular to
and their bisector,
is isosceles with
, and
. Also
But
as well, and
, so
. Thus
.