October 6–October 11, 2009
Problem
Show that if 16 points are placed in an equilateral triangle with side length 1, then two of the points must have a distance strictly less than 1/4.
Solution
Perhaps this problem was trickier than I suspected. Nearly all the solutions involved a variation of the following: 15 points may be placed in a triangular lattice so that closest points are at distance 1/4, and that it is impossible to add another point without reducing the smallest distance. If the first 15 points points aren't precisely on this lattice, then there must have been two close points among the first 15.
The shortcoming of this approach is formalizing the last sentence to account for any arrangement of 16 points, especially when some points are not on the lattice. This requires careful casework to be thorough.
In contrast to this, here is an elegant approach using the pigeonhole principle.
List of solvers
Kazuo Thow, Mimi Fung, Matt Inouye, Tyler Cooper, Huy Hoang-Nguyen (undergrad); Lloyd Sakazaki, Alejandro Perez, Lincoln Atkinson, Haowei, Jake Bryant, James Becker, Congpa You (outside).
Alejandro Perez wins the prize!