Difference between revisions of "1999 JBMO Problems/Problem 3"
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Latest revision as of 11:50, 12 August 2020
Let be a square with the side length 20 and let
be the set of points formed with the vertices of
and another 1999 points lying inside
. Prove that there exists a triangle with vertices in
and with area at most equal with
.
Solution
Triangulate into triangles with vertices being the vertices of
and the members of
. There are
triangles thusly formed, so by the pigeonhole principle, at least one of the holes has to have area at most
, and we are done.
See also
1999 JBMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 | ||
All JBMO Problems and Solutions |