Problem 13 – February 22, 2019
Some Are Not That Far Away
Consider any five points P1, P2, P3, P4, and P5 in the interior of a square of side length 1. Denote by dij the distance between points Pi and Pj when i is not equal to j. Prove that at least one of the distances dij is less than 
.
Your answer must be submitted as a proof!
Submit your answers to mathpotw@acu.edu. Details for submissions can be found here.
Correct answers were submitted by Dr. Mark Riggs!
Solution to Some Are Not That Far Away
The unit square may be divided into four square quadrants, each with side length 1/2. Each point must be in one of these quadrants, and since there are five points, two must be in the same quadrant.
The farthest apart two points in a square can be is the length of a diagonal, which is achieved when the points are at opposite corners. But two opposite corners of a quadrant must have at least one on the border of the unit square. So for both to be in the interior of the unit square, they must be even closer. Therefore the distance between two of the points is less than the length of a diagonal of a square of side 1/2, which is .
