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--- pow:problem6s24 [2024/04/22 16:24] – created mazur
+++ pow:problem6s24 [2024/04/29 04:32] (current) – mazur
@@ Line 1: / Line 1: @@
+<box 85% round orange|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 $BCP$:
+\[ \text{area}(\triangle MNP)=\frac{1}{4}\left|\text{area}(\triangle DAP)-\text{area}(\triangle BCP)\right |.\]
+</box>
+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 {{:pow:2024sproblem6.pdf|Solution}}.