1999 JBMO Problems/Problem 3

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 $\frac 1{10}$ .

Solution

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.

1999 JBMO (Problems • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5
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1999 JBMO Problems/Problem 3

Solution

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