Iran TST Incircle configuration

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rah4927
Posts:110
Joined:Sat Feb 07, 2015 9:47 pm
Iran TST Incircle configuration

Unread post by rah4927 » Fri Aug 05, 2016 12:23 am

Let $ABC$ be a triangle with incenter $I$ and contact
triangle $DEF$. Let $M$ be the foot of the perpendicular from $D$ to $EF$ and let $P$ be the
midpoint of $DM$. If $H$ is the orthocenter of triangle $BIC$, prove that $PH$ bisects $EF$.

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

Re: Iran TST Incircle configuration

Unread post by tanmoy » Fri Aug 05, 2016 1:10 am

Oh,that famous problem. :)
Let $K,L$ lie on $BI$ and $CI$,respectively so that $BK \perp CK$ and $CL \perp BL$.From this: viewtopic.php?f=25&p=16649#p16649 ,$K,L$ lie on $EF$.$I$ is the orthocenter of $\Delta HBC$.Let $N$ be the midpoint of $EF$ and $HD \cap EF=Q$.So,$-1=(H,I,Q,D)=N(H,I,Q,D)=(HN \cap DN,\infty,M,D)=(P,\infty,M,D)$.So,$P,N,H$ are collinear.Done
"Questions we can't answer are far better than answers we can't question"

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