2014 USAMO Problems/Problem 5


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 $ABC$.


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 $XY=AO_1=AO$.

See also

2014 USAMO (ProblemsResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6
All USAMO Problems and Solutions
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