1999 JBMO Problems/Problem 3
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 |