# geometric problem solved with Pigeon Hole Principle

The problem is:

Show that among any 5 points in a equilateral triangle of unit side length, there are 2 whose distance is at most 1/2 units apart.

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Break the equilateral triangle into four smaller triangles by joining the midpoints of the sides. Now we have four holes and five pegions hence three must be at least one triangle which contains at least two points. The distance between the two points cannot exceed the side of the triangle.Also the side of the triangles formed is $\frac1{2}$ and the result follows . Hence among any 5 points in a equilateral triangle of unit side length, there are 2 whose distance is at most 1/2 units apart.