1999 JBMO Problems/Problem 3

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Let $S$ be a square with the side length 20 and let $M$ be the set of points formed with the vertices of $S$ and another 1999 points lying inside $S$. Prove that there exists a triangle with vertices in $M$ and with area at most equal with $\frac 1{10}$.


Triangulate $S$ into triangles with vertices being the vertices of $S$ and the members of $M$. There are $2*(1999 + 1) = 4000$ triangles thusly formed, so by the pigeonhole principle, at least one of the holes has to have area at most $\frac{20^2}{4000} = \frac{1}{10}$, and we are done.

See also

1999 JBMO (ProblemsResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5
All JBMO Problems and Solutions