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Difference between revisions of "2014 USAMO Problems/Problem 5"

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==Solution==
 
==Solution==
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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>
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\angle AYC=180-\angle APC=180-\angle AHC=\angle B
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</cmath>
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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
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<cmath>
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\angle AOY=2\angle ACY=2(90-\angle PAC)=2(90-\frac{A}{2})=180-\angle A = \angle XOO_1
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</cmath>
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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 16:29, 30 December 2014

Problem

Let $ABC$ be a triangle with orthocenter $H$ and let $P$ be the second intersection of the circumcircle of triangle $AHC$ with the internal bisector of the angle $\angle BAC$. Let $X$ be the circumcenter of triangle $APB$ and $Y$ the orthocenter of triangle $APC$. Prove that the length of segment $XY$ is equal to the circumradius of triangle $ABC$.

Solution

Let $O_1$ be the center of $(AHPC)$, $O$ be the center of $(ABC)$. Note that $(O_1)$ is the reflection of $(O)$ across $AC$, so $AO=AO_1$. Additionally \[\angle AYC=180-\angle APC=180-\angle AHC=\angle B\] so $Y$ lies on $(O)$. Now since $XO,OO_1,XO_1$ are perpendicular to $AB,AC,$ and their bisector, $XOO_1$ is isosceles with $XO=OO_1$, and $\angle XOO_1=180-\angle A$. Also \[\angle AOY=2\angle ACY=2(90-\angle PAC)=2(90-\frac{A}{2})=180-\angle A = \angle XOO_1\] But $YO=OA$ as well, and $\angle YOX=\angle AOO_1$, so $\triangle OYX\cong \triangle OAO_1$. Thus $XY=AO_1=AO$.

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