1999 JBMO Problems/Problem 3

Revision as of 10:50, 12 August 2020 by Duck master (talk | contribs) (created page w/ solution & categorization)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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.

See also

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