Given an acute and scalene triangle $ ABC$, let $ H$ be its orthocenter, $ O$ its circumcenter, $ E$ and $ F$ the feet of the altitudes drawn from $ B$ and $ C$, respectively. Line $ AO$ intersects the circumcircle of the triangle again at point $ G$ and segments $ FE$ and $ BC$ at points $ X$ and $ Y$ respectively. Let $ Z$ be the point of intersection of line $ AH$ and the tangent line to the circumcircle at $ G$. Prove that $ HX$ is parallel to $ YZ$.
Problem
Source: Problem 5, Centroamerican Olympiad 2009
Tags: geometry, circumcircle, projective geometry, geometric transformation, homothety, geometry proposed
Luis González
08.10.2009 02:27
Rays $ AO,AH$ are isogonal WRT $ \angle BAC$ $ \Longrightarrow AX \perp EF.$ Since $ EF$ and $ BC$ are antiparallel WRT AB,AC, then $ HX$ and $ GD$ are antiparallel WRT $ AO,AH$ $ \Longrightarrow$ $ \angle AHX = \angle AGD.$ But the quadrilateral $ DYGZ$ is cyclic on account of the right angles $ \angle YDZ = \angle YGZ = 90^{\circ}.$ Therefore, $ \angle AHX = \angle AGD = \angle AZY$ $ \Longrightarrow$ $ HX \parallel YZ.$
ma 29
08.10.2009 11:30
Denote $ S = AH\cap EF; D = AH \cap BC ;A,V = AH \cap (O)$ We have :$ (AHSD)$ is harmonic ,so $ \frac {SA}{SH} = \frac {DA}{DH} = \frac {DA}{DV} = \frac {YA}{YG}$ $ = > SY//HG$ (1) On other hand,we see that $ SX//ZG$ (2) Since (1) ,(2) and the Pappus theorem then $ HX//YZ$
jayme
08.10.2009 17:28
Dear Mathlinkers, 1. like Luis observes the quadrilateral DYGZ is cyclic 2. after Luis's obsevation the quadrilateral HXGD is also cyclic 3. use now the Reim's theorem and we ae done Sincerely Jean-Louis
Dimitris X
08.10.2009 21:57
jayme wrote: 3. use now the Reim's theorem and we ae done Can you tell me what this theorem says dear jayme.
jayme
09.10.2009 13:51
Dear Dimitris and Mathlinkers, you will see what you want on http://perso.orange.fr/jl.ayme/ Sincerely Jean-Louis
Elementaryyy
10.10.2009 12:52
We have $ AX=sinC.cosA.AC$ $ AY=\frac {sinC}{cosKAX}.AC$ $ AH=cosA.2R$ $ AZ=\frac {2R}{cosKAX}$ then $ \frac {AX}{AY}=\frac {AH}{AZ}$ done.
thanhnam2902
11.10.2009 18:58
This is my solution: +) Easy we get ${ AG \perp EF}$ because $ \angle {AEX} + \angle{XAE} = \angle {AHF} + \angle{CBG} = \angle {ABC} + \angle{CBG} = 90^0$. +) Because $ XFBG$ and $ FBKH$ are cyclic $ \Longrightarrow AX.AG = AF.AB = AH.AK \Longrightarrow HKGX$ is cyclic $ \Longrightarrow \angle{HXY} = \angle{ZKG}$ +) But $ ZKXG$ is cyclic $ \Longrightarrow \angle{ZKG} = \angle{ZYG} \Longrightarrow XH // YZ$, Q.E.D.
Attachments:
Parallel Problem - Quy-CQT.pdf (9kb)
ElChapin
11.10.2009 21:53
Another solution.
Draw a line $ j$ through $ Y$ and perpendicular to $ AG$, that is parallel to $ EF$ and $ GZ$. Let $ V = AH\cap EF$ and $ T = j\cap AH$.
We'll prove that $ Z$ is the image of $ H$ under the homothety $ \sigma$, with center $ A$ and which leads $ X$ into $ Y$. The result follows from there.
$ V$ and $ Y$, and $ H$ and $ G$ are homologous in the similarity between $ \triangle AEF$ and $ \triangle ACB$, that implies $ \dfrac{AH}{AV} = \dfrac{AG}{AY}$. And by Thales' theorem $ \dfrac{AG}{AY} = \dfrac{AZ}{AT}$, so $ \dfrac{AH}{AV} = \dfrac{AZ}{AT}$. Since $ T$ is image of $ V$ under $ \sigma$ and, $ H$ and $ Z$ are on the same ray from $ A$ it follows that $ Z$ is image of $ H$ under $ \sigma$.
epsilonist
28.01.2010 17:23
Wow, I don't understand why but I think this problem is really nice. Thanks to the author.
eps0519
06.12.2011 01:30
Spelling mistake. thanhnam2902 wrote: This is my solution: +) Easy we get ${ AG \perp EF}$ because $ \angle {AEX} + \angle{XAE} = \angle {AHF} + \angle{CBG} = \angle {ABC} + \angle{CBG} = 90^0$. +) Because $ XFBG$ and $ FBKH$ are cyclic $ \Longrightarrow AX.AG = AF.AB = AH.AK \Longrightarrow HKGX$ is cyclic $ \Longrightarrow \angle{HXY} = \angle{ZKG}$ +) But $ ZKYG$ is cyclic $ \Longrightarrow \angle{ZKG} = \angle{ZYG} \Longrightarrow XH // YZ$, Q.E.D.
AlastorMoody
19.02.2019 06:44
Easy to see, $AX \perp EF$, hence, $\frac{AH}{AX}=\frac{\tfrac{AF}{\sin B}}{AF \cdot \sin C}=\frac{1}{\sin B \cdot \sin C}$, also notice, that if $AD \cap \odot (ABC)=K$, then, $DY||KG$ , hence, $\frac{AG}{AK}=\frac{AY}{AD}$ and since, $AG^2=AK \cdot AZ$, we have, $\frac{AZ}{AY}=\frac{AG}{AD}=\frac{\tfrac{c}{\sin C}}{c \sin B}=\frac{1}{\sin B \cdot \sin C}=\frac{AH}{AX} \implies HX||YZ$