China TST $2011$ Day 2

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rah4927
Posts:110
Joined:Sat Feb 07, 2015 9:47 pm
China TST $2011$ Day 2

Unread post by rah4927 » Mon Aug 15, 2016 5:55 pm

Let $H$ be the orthocenter of an acute trangle $ABC$ with circumcircle $\Gamma$. Let $P$ be a point on the arc $BC$ (not containing $A$) of $\Gamma$, and let $M$ be a point on the arc $CA$ (not containing $B$) of $\Gamma$ such that $H$ lies on the segment $PM$. Let $K$ be another point on $\Gamma$ such that $KM$ is parallel to the Simson line of $P$ with respect to triangle $ABC$. Let $Q$ be another point on $\Gamma$ such that $PQ \parallel BC$. Segments $BC$ and $KQ$ intersect at a point $J$. Prove that $\triangle KJM$ is an isosceles triangle.

rah4927
Posts:110
Joined:Sat Feb 07, 2015 9:47 pm

Re: China TST $2011$ Day 2

Unread post by rah4927 » Mon Aug 15, 2016 7:15 pm

$AQ$ is perpendicular to the simson line of $P$ .

tanmoy
Posts:312
Joined:Fri Oct 18, 2013 11:56 pm
Location:Rangpur,Bangladesh

Re: China TST $2011$ Day 2

Unread post by tanmoy » Mon Aug 15, 2016 9:20 pm

Very nice problem.
Lemma: Let $l$ be the simson line of a point $P$ lying on the minor arc $BC$ of the circumcircle of a triangle $\triangle ABC$.And let $PA_{1} \perp BC$ and it cuts $(ABC)$ again at $P_{1}$.Then $AP_{1} \parallel l$.
Proof of the Lemma: Let $PB_{1} \perp AC$.Then $AP_{1}PC$ and $A_{1}PCB_{1}$ are concyclic.Then by Reim's Theorem, $AP_{1} \parallel l$.

Back to the main problem.Let $PM \cap BC=R$ and the reflection of $H$ in $BC$ be $H_{1}$.$MKPQ$ is cyclic and $RJ \parallel PQ$.So, by Reim's Theorem, $KMJR$ is cyclic.

Now,we have to prove that $\measuredangle JRM=\measuredangle BRP=\measuredangle BRK$ i.e. $BC$ is the bisector of $\angle KRP$.But it is well known that $BC$ bisects $\angle H_{1}RH$.So,we have to prove that points $K,R,H_{1}$ are collinear i.e. we have to prove that $\measuredangle RH_{1}H=\measuredangle KH_{1}H$.

From the lemma,we know that $P_{1}M=AK$.So,$\measuredangle RH_{1}H=\measuredangle HPP_{1}=\measuredangle MPP_{1}=\measuredangle KH_{1}A=\measuredangle KH_{1}H$.So,from the last line of the 3rd paragraph,we are done.

NOTE: $K$ is the Anti-Steiner Point of the line $PHM$
"Questions we can't answer are far better than answers we can't question"

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