# Given seven points inside a hexagon with side length $1$ prove that there exists two points with distance at most $1$

Seven points are given inside a regular hexagon whose sides have length $1$. Prove that there are two among these seven points such that the distance between them is at most $1$.

Now if I divide the hexagon into $6$ regions, since we have $7$ points, by the pigeon hole principle there is a region with at least two points in it. The distance between two points is at most $1$ because each region is an equilateral triangle with sides of length $1$.

Would this be sufficient or any other approaches that would work better.

• You can't do better! Your approach is as good as it gets for this problem. Nice going! Commented Sep 14, 2017 at 20:24
• I will add that answer to this question also shows a picture: Pigeonhole principle, choosing point in a region. Commented Jul 1, 2020 at 14:29

The distance between two points in an equilateral triangle with side length $1$ is at most $1$ because this triangle lies in a circle of radius $1/2$.
• Over the top! Enclosing the equilateral triangle in circle of radius $1$ is easier and is all you need. Commented Sep 14, 2017 at 20:29
• @ThePortakal: Good point! (+1). While it might seem obvious that two points in an equilateral triangle can't be more than a distance of $1$ from each other, it does require proof. Commented Sep 14, 2017 at 20:30
• @Rob Arthan: Why is radius $1$ sufficient? Commented Sep 14, 2017 at 20:33
• @quasi: Because the equilateral triangle is contained in the circle of radius $1$ about any of its points. Commented Sep 14, 2017 at 20:40