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Berkeley Math Circle Monthly Contest 4

Geometry Difficulty 5.4 AIME, harder Prove it United States

Problem:

Let $O$ be the intersection of the diagonals $AC$ and $BD$ of the convex quadrilateral $ABCD$ . Let $S_{1}, S_{2}, S_{3}$ , and $S_{4}$ denote the areas of the triangles $ABO$ , $BCO$ , $CDO$ , and $DAO$ .

a) Prove that $S_{1} \cdot S_{3} = S_{2} \cdot S_{4}$ .

b) Does there exist a quadrilateral $ABCD$ such that $S_{1}, S_{2}, S_{3}$ , and $S_{4}$ are consecutive positive integers in some order?

Solution

Solution:

a. Let $M$ and $N$ be feet of perpendiculars from $B$ and $D$ to $AC$ . Then $S_{1} = AO \cdot BM / 2$ , $S_{2} = CO \cdot BM / 2$ , $S_{3} = CO \cdot DN / 2$ , and $S_{4} = AO \cdot DN / 2$ . Now the desired statement follows immediately from the previous four relations.

b. We will prove that the answer to the question is no. Assume the opposite, i.e. that $n, n+1, n+2, n+3$ are the given areas, where $n$ is a positive integer. Since $S_{1} \cdot S_{3} = S_{2} \cdot S_{4}$ and $n < n+1 < n+2 < n+3$ we must have $n(n+3) = (n+1)(n+2)$ which is impossible.

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