Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.1 AIME, harder Prove it Czech-Polish-Slovak Mathematical Match

Let $ABCD$ be a given convex quadrilateral. Determine the locus of the points $P$ lying inside the quadrilateral $ABCD$ and satisfying
$[PAB] \cdot [PCD] = [PBC] \cdot [PDA],$
where $[XYZ]$ denotes the area of triangle $XYZ$ .

Solution

If $P$ lies on one of the diagonals $AC$ or $BD$ , let's say on $AC$ , then
$\frac{[PAB]}{[PBC]} = \frac{AP}{PC} = \frac{[PDA]}{[PCD]},$
which is the desired equality. We prove that no other point lying inside $ABCD$ satisfies the conditions of the problem.

Denote by $O$ the point of the intersection of the diagonals $AC$ and $BD$ and suppose that $P$ lies inside the triangle $ABO$ . Let moreover $BP$ and $AC$ meet at $Q$ and $DP$ and $AC$ meet at $R$ . Then
$\frac{[PAB]}{[PBC]} = \frac{AQ}{QC} \quad \text{and} \quad \frac{[PDA]}{[PCD]} = \frac{AR}{RC},$
which, since $Q \neq R$ , implies that the given equality cannot hold.

Therefore the desired locus of the points $P$ consists of the diagonals $AC$ and $BD$ .

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