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2014 USAMO Problems/Problem 5

Revision as of 11:06, 27 March 2022 by Scarface (talk | contribs) (→‎Solution 2)

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 1

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 $XY=AO_1=AO$.

Solution 2

Since $AHPC$ is a cyclic quadrilateral, $\angle AHC = \angle APC$. $\angle AHC = 90^\circ + \angle ABC$ and $\angle APC = 90^\circ + \angle AYC$, we find $\angle ABC = \angle AYC$. That is, $ABYC$ is a cyclic quadrilateral. Let $D$ be mid-point of $\overline{AB}$. $O, X, D$ are collinear and $OX \perp AB$. Let $M$ be second intersection of $AP$ with circumcircle of the triangle $ABC$. Let $YP \cap AC = E$, $YM \cap AB = F$. Since $M$ is mid-point of the arc $BC$, $OM\perp BC$. Since $AYMC$ is a cyclic quadrilateral, $\angle CYM = \angle CAM = \angle BAC /2$. Since $Y$ is the orthocenter of triangle $APC$, $\angle PYC = \angle CAP = \angle BAC /2$. Thus, $\angle PYM = \angle BAC$ and $AEYF$ is a cyclic quadrilateral. So, $YF \perp AB$ and $OX \parallel MY$. We will prove that $XYMO$ is a parallelogram.

https://wiki-images.artofproblemsolving.com//7/7b/Usamo2014-5.png

We see that $YPM$ is an isosceles triangle and $YM=YP$. Also $XB=XP$ and $\angle BXP = 2\angle BAP = \angle BAC = \angle PYM$. Then, $BXP \sim MYP$. By spiral similarity, $BPM \sim XPY$ and $\angle XYP = \angle BMP = \angle BCA$. Hence, $\angle XYP = \angle BCA$, $XY \perp BC$. Since $OM \perp BC$, we get $XYMO$ is a parallelogram. As a result, $OM = XY$. (Lokman GÖKÇE)

See also

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