Let $ABCD$ be a parallelogram, let $X$ and $Y$ in the segments $AB$ and $CD$, respectively. The segments $AY$ and $DX$ intersects in $P$ and the segments $BY$ and $DX$ intersects in $Q$, show that the line $PQ$ passes in a fixed point(independent of the positions of the points $X$ and $Y$). I guess that the fixed point is the midpoint of $BD$.
Problem
Source: Peru TST CSMO
Tags: geometry
QWERTYphysics
08.02.2018 14:49
Edited....
enhanced
08.02.2018 14:54
see below
enhanced
08.02.2018 15:01
Note that $P,Q\in DX$ which is enough to say that $PQ$ passes through $D$ which is of course a fixed point
jestrada
08.02.2018 16:25
I suppose you meant $BY$ and $CX$ intersect in $Q$ ? Let $M$ be the intersection of $AC$ and $BD$. By Pappus' Theorem , $P, Q, M$ are collinear. We don't even need $ABCD$ to be a parallelogram.
lakecomo224
08.02.2018 16:27
Yeah, I think so too.
xdiegolazarox
08.02.2018 17:28
actually, the ones who made the test(2015) hadn't realized of that solution
MathLuis
02.11.2021 23:55
Let $AC$ meet $BD$ at $M$. Consider the degenerate conic $\mathcal C$ which passes through $A,B,C,D,X$. Note that this conic is actualy the lines $AB$ and $CD$ meaning that $Y$ lies on $\mathcal C$. thus by pascal on $\mathcal C$ we have that $P,Q,M$ are colinear and since $M$ doesn't depend from $X,Y$ we are done.