2014 USAMO Problems/Problem 5

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 $XY=AO_1=AO$ .

2014 USAMO (Problems • Resources)
Preceded by Problem 4	Followed by Problem 6
1 • 2 • 3 • 4 • 5 • 6
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2014 USAMO Problems/Problem 5

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