Iranian Geometry Olympiad 2020 (Elementary) P4

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IftakharTausifFarhan
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Iranian Geometry Olympiad 2020 (Elementary) P4

Unread post by IftakharTausifFarhan » Sat Dec 12, 2020 5:00 pm

Let $P$ be an arbitrary point in the interior of triangle $ABC$. Lines $BP$ and $CP$ intersect $AC$ and $AB$ 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 $CF$ and $BE$ intersect $BC$ 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 $ABC$, line $MN$ passes through a fixed point.

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Anindya Biswas
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Re: Iranian Geometry Olympiad 2020 (Elementary) P4

Unread post by Anindya Biswas » Sun Jan 31, 2021 8:22 pm

Here's a simple argument using congruence and similarity of triangles and properties of parallel lines:
In $\triangle KBT$ and $\triangle KFN$,
$KB=KF$, $KT=KN$, $\angle TKB=\angle NKF$.
$\therefore \triangle KBT\cong\triangle KFN$.
$\Rightarrow \angle KBT=\angle KFN$
$\Rightarrow BT||FN\Rightarrow BC||FN$.
Let $X$ be the intersection point of lines $BE$ and $NF$.
In $\triangle FBX$, $K$ is the midpoint of $BF$ and $KN||BX$.
$\Rightarrow N$ is the midpoint of segment $FX$.
Now, since $FX||BC$ and $N$ is the midpoint of $FX$, the line $NP$ must meet $BC$ at the midpoint of segment $BC$.
Let's define $Q$ to be the mid point of $BC$. $\therefore N,P,Q$ are collinear.
By similar argument above, we can prove that $M,P,Q$ are collinear.
So, lines $MN$ and $PQ$ are the same line. So, line $MN$ must go through the point $Q$, independent of the position of $P$.

$Q.E.D.$
IGO 2020 Elementary P4.png
$M,N,P,Q$ collinear. $Q$ is the midpoint of segment $BC$
IGO 2020 Elementary P4.png (74.68KiB)Viewed 4923 times
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

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