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I had some troubles with this problem :

Let $ABCD$ be a convex quadrilateral. $M$ and $N$ are the midpoints of the diagonals $AC$ and $BD$. The sides $AB$ and $CD$ are extended until they intersect. The intersection point is $E$. The sides $AD$ and $BC$ are extended until they intersect. The intersection point is $F$. Let $P$ be the midpoint of the segment $[EF]$. Prove that $M$, $N$, $P$ are collinear.

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First, I found that a quadrilateral in which the opposite sides interesect is also known as a complete quadrilateral. Then, the line $M-N-P$ is known as Newton-Gauss line and the problem above as Newton's Problem.

I've taught about solving it using areas. I've used the property that median divides the triangle in two echivalent triangles (with the same area). Many properties can be derived from it.

I haven't figure out, but I'm interested in a proof using areas. I would appreciate some suggestions.

Thanks!

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  • $\begingroup$ I remember this is a IMO shortlist problem .. I am forgetting the year. $\endgroup$ – Sawarnik Aug 18 '15 at 9:43
  • $\begingroup$ @Sawarnik Probably before 2000. $\endgroup$ – scummy Aug 18 '15 at 10:02
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Let $S$ denote the area of $ABCD$.

Lemma 1: The locus of points $X$ in the interior of $ABCD$ such that $|XAB|+|XCD|=\frac{S}{2}$ is a line segment.

In triangle $ADE$, consider a translation of $AB$ to $ES$ and $CD$ to $ET$. Thus, $|XES|=|XAB|$ and $|XET|=|XCD|$. Thus, $|XAB|+|XCD|=|XES|+|XET|=|XSET|=|EST|+|XST|$. Since $S$ and $T$ are fixed points, it follows that $|XST|$ is constant, so the locus of $X$ is the segment made by the intersection of the line parallel to $ST$ with $ABCD$. Further note that if $X$ is outside $ABCD$, then we have $|XAB|-|XCD|=\frac{S}{2}$ or $|XCD|-|XAB|=\frac{S}{2}$.$_\square$

Note that since $M$ is the midpoint of $AC$, $|AMB|=|BMC|$ and $|AMD|=|CMD|$, so $|AMB|+|CMD|=\frac{S}{2}$, so $M$ lies on the locus of $X$. Similarly, $N$ lies on the locus of $X$.

Now note that since $EP=PF$, $|PAB|=\frac{1}{2}|FAB|$ since the distance between $P$ and $AB$ is half of the distance between $F$ and $AB$. Similarly, $|PCD|=\frac{1}{2}|FCD|$. Thus, we have $|PAB|-|PCD|=\frac{1}{2} (|FAB|-|FCD|) = \frac{S}{2}$, so $P$ also lies on this line.

Therefore, $M$, $N$, $P$ are collinear.

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