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Problem 6 (due Monday, April 22)

Let $ABCD$ be a convex quadrilateral whose diagonals $AC$ and $BD$ intersect at a point P. Let $M,N$ be the midpoints of the sides $AB$ and $CD$ respectively. Prove that the area of the triangle $PMN$ is equal to the quarter of the absolute value of the difference between the area of the triangle $DAP$ and the area of the triangle $BCP$: \[ \text{area}(\triangle MNP)=\frac{1}{4}\left|\text{area}(\triangle DAP)-\text{area}(\triangle BCP)\right |.\]

We received only one solution, from Sasha Aksenchuk. Sasha's solution uses analytic geometry and is similar to one of our in-house solutions. For a complete solution see the following link Solution.

pow/problem6s24.txt · Last modified: 2024/04/29 00:32 by mazur