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

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\angle AOY=2\angle ACY=2(90-\angle PAC)=2(90-\frac{A}{2})=180-\angle A = \angle XOO_1
 
\angle AOY=2\angle ACY=2(90-\angle PAC)=2(90-\frac{A}{2})=180-\angle A = \angle XOO_1
 
</cmath>
 
</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>.
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But <math>YO=OA</math> as well, and <math>\angle YOX=\angle AOO_1</math>, so <math>\triangle OYX\sim \triangle OAO_1</math>. Thus <math>XY=AO_1=AO</math>.

Revision as of 10:59, 10 July 2019

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 $\triangle OYX\sim \triangle OAO_1$. Thus $XY=AO_1=AO$.

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