Let $P$ be an arbitrary point in the interior of triangle $\triangle ABC$. Lines$\overline{BP}$ and $\overline{CP}$ intersect $\overline{AC}$ and $\overline{AB}$ at $E$ and $F$, respectively. Let $K$ and $L$ be the midpoints of the segments $BF$ and $CE$, respectively. Let the lines through $L$ and $K$ parallel to $\overline{CF}$ and $\overline{BE}$ intersect $\overline{BC}$ at $S$ and $T$, respectively; moreover, denote by $M$ and $N$ the reflection of $S$ and $T$ over the points $L$ and $K$, respectively. Prove that as $P$ moves in the interior of triangle $\triangle ABC$, line $\overline{MN}$ passes through a fixed point. Proposed by Ali Zamani
Problem
Source: 7th Iranian Geometry Olympiad (Elementary) P4
Tags: geometry, IGO
aops-g5-gethsemanea2
24.04.2022 15:02
hi @dungnguyentien and @DrGeometry, not sure why you guys applied Thales theorem in the solutions... the theorem involves circles and right angles but I can't find any of those here... please clarify, thanks
Tafi_ak
07.08.2022 18:21
Notice that $MESC$ is a parallelogram. So $ME\parallel BC$. Let $D$ be the midpoint of $BC$ and $J=ME\cap CP$. The point $M$ is the midpoint of $JE$ because $ML\parallel JC$ and $L$ is the midpoint of $EC$. So notice that trapezoid $JBCE$ and $D,M$ are the midpoints of $JE,BC$ respectively and $P$ is the intersection of the diagonal. Therefore $D,P,M$ are collinear. Similarly $D, P, N$ are collinear. So $D$ is the desired fixed point.