A Beauty from Evan Chen

For discussing Olympiad level Geometry Problems
tanmoy
Posts:312
Joined:Fri Oct 18, 2013 11:56 pm
Location:Rangpur,Bangladesh
A Beauty from Evan Chen

Unread post by tanmoy » Sat Nov 12, 2016 3:46 pm

Let $ABC$ be a scalene triangle with orthocenter $H$ and circumcenter $O$. Denote by $M$, $N$ the midpoints of $\overline{AH}$, $\overline{BC}$. Suppose the circle $\gamma$ with diameter $\overline{AH}$ meets the circumcircle of $ABC$ at $G \neq A$,and meets line $AN$ at a point $Q \neq A$. The tangent to $\gamma$ at $G$ meets line $OM$ at $P$. Show that the circumcircles of $\triangle GNQ$ and $\triangle MBC$ intersect at a point $T$ on $\overline{PN}$

Source:
USA TSTST $2016$,Problem $2$;Proposed by Evan Chen
"Questions we can't answer are far better than answers we can't question"

joydip
Posts:49
Joined:Tue May 17, 2016 11:52 am

Re: A Beauty from Evan Chen

Unread post by joydip » Tue Nov 22, 2016 12:54 am

Let $AG$ and $MO$ meet $BC$ at $K,S$ respectively .$KM$ meets $PN$ at $R$.
As $M$ is the center of $\gamma$,so $MO$ is the perpendicular bisector of $AG$ . $ \angle PAM =\angle PGM =90^o \Rightarrow AP\| BC \Rightarrow \angle APM =\angle OSN.$
$ON=\frac {1}{2} AH =AM $ and $\angle ONS =90^o =\angle PAM $ gives $\triangle PAM \cong \triangle SNO$ .So $AP=SN \Rightarrow APNS$ is a parallelogram .
$SM\perp AK ,AM\perp KS \Rightarrow M$ is the orthocenter of $\triangle AKS$ .So $KM\perp AS \Rightarrow KM\perp PN \Rightarrow APGRM $ is cyclic. So.$KR.KM=KG.KA=KB.KC$ , So $BRMC$ is cyclic.
$\angle PRG= \angle PMG =\frac {1}{2} \angle AMG=\angle AQG \Rightarrow \angle GRN =\angle GQN $ .So $GRQN$ is cyclic.$[T=R]$
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