Pigeonhole Principle in Equilateral Triangle

By selecting 10 points on an equilateral triangle with side length 1, show that there are at least two points with distance less than or equal to $\frac{1}{3}$ apart.

This problem is a classic application of the pigeonhole principle, which is a fundamental concept in combinatorics. The pigeonhole principle asserts that if you distribute more items than containers, at least one container must hold more than one item. In a geometric context, as in this problem, it can be used to show that among a certain number of points within a bounded shape, at least two points must be closer together than a specified distance.

In solving this problem, you are asked to demonstrate that among ten points placed on an equilateral triangle with sides of length one, at least two of them must be within a distance of one third or less from each other. To understand the distribution of these points, consider subdividing the triangle into smaller, equally sized regions. The choice of these divisions can be guided by the distance constraint, suggesting a partitioning that highlights clusters of points.

The problem encourages understanding how the spatial distribution of points relates to distance constraints, a topic often encountered in discrete geometry and optimization problems. By strategically dividing the triangle into regions that make the positions of points more predictable relative to each other, you can efficiently apply the pigeonhole principle to solve this problem. This approach is especially significant in computational geometry and various optimization problems, where similar principles can provide insights into the optimal arrangements of points or objects.