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

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Since <math>AHPC</math> is a cyclic quadrilateral, <math>\angle AHC = \angle APC</math>. <math>\angle AHC = 90^\circ + \angle ABC</math> and <math>\angle APC = 90^\circ + \angle AYC</math>, we find <math>\angle ABC = \angle AYC</math>. That is, <math>ABYC</math> is a cyclic quadrilateral. Let <math>D</math> be mid-point of <math>\overline{AB}</math>. <math>O, X, D</math> are collinear and <math>OX \perp AB</math>. Let <math>M</math> be second intersection of <math>AP</math> with circumcircle of the triangle <math>ABC</math>. Let <math>YP \cap AC = E</math>, <math>YM \cap AB = F</math>. Since <math>M</math> is mid-point of the arc <math>BC</math>, <math>OM\perp BC</math>. Since <math>AYMC</math> is a cyclic quadrilateral, <math>\angle CYM = \angle CAM = \angle BAC /2</math>. Since <math>Y</math> is the orthocenter of triangle <math>APC</math>, <math>\angle PYC = \angle CAP = \angle BAC /2</math>. Thus, <math>\angle PYM = \angle BAC</math> and <math>AEYF</math> is a cyclic quadrilateral. So, <math>YF \perp AB</math> and <math>OX \parallel MY</math>. We will prove that <math>XYMO</math> is a parallelogram.
 
Since <math>AHPC</math> is a cyclic quadrilateral, <math>\angle AHC = \angle APC</math>. <math>\angle AHC = 90^\circ + \angle ABC</math> and <math>\angle APC = 90^\circ + \angle AYC</math>, we find <math>\angle ABC = \angle AYC</math>. That is, <math>ABYC</math> is a cyclic quadrilateral. Let <math>D</math> be mid-point of <math>\overline{AB}</math>. <math>O, X, D</math> are collinear and <math>OX \perp AB</math>. Let <math>M</math> be second intersection of <math>AP</math> with circumcircle of the triangle <math>ABC</math>. Let <math>YP \cap AC = E</math>, <math>YM \cap AB = F</math>. Since <math>M</math> is mid-point of the arc <math>BC</math>, <math>OM\perp BC</math>. Since <math>AYMC</math> is a cyclic quadrilateral, <math>\angle CYM = \angle CAM = \angle BAC /2</math>. Since <math>Y</math> is the orthocenter of triangle <math>APC</math>, <math>\angle PYC = \angle CAP = \angle BAC /2</math>. Thus, <math>\angle PYM = \angle BAC</math> and <math>AEYF</math> is a cyclic quadrilateral. So, <math>YF \perp AB</math> and <math>OX \parallel MY</math>. We will prove that <math>XYMO</math> is a parallelogram.
  
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https://wiki-images.artofproblemsolving.com//7/7b/Usamo2014-5.png
  
 
We see that <math>YPM</math> is an isosceles triangle and <math>YM=YP</math>. Also <math>XB=XP</math> and <math>\angle BXP = 2\angle BAP = \angle BAC = \angle PYM</math>. Then, <math> BXP \sim MYP </math>. By spiral similarity, <math> BPM \sim XPY </math> and <math>\angle XYP = \angle BMP = \angle BCA</math>. Hence, <math>\angle XYP = \angle BCA</math>, <math>XY \perp BC</math>. Since <math>OM \perp BC</math>, we get <math>XYMO</math> is a parallelogram. As a result, <math>OM = XY</math>.
 
We see that <math>YPM</math> is an isosceles triangle and <math>YM=YP</math>. Also <math>XB=XP</math> and <math>\angle BXP = 2\angle BAP = \angle BAC = \angle PYM</math>. Then, <math> BXP \sim MYP </math>. By spiral similarity, <math> BPM \sim XPY </math> and <math>\angle XYP = \angle BMP = \angle BCA</math>. Hence, <math>\angle XYP = \angle BCA</math>, <math>XY \perp BC</math>. Since <math>OM \perp BC</math>, we get <math>XYMO</math> is a parallelogram. As a result, <math>OM = XY</math>.

Revision as of 11:06, 27 March 2022

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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