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 $EF$.
Iran TST Incircle configuration
Re: Iran TST Incircle configuration
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
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"