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

Question

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.

Answer 1

This seems sufficient, but you can add something like this:

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 $1/2$.

Answer 2

It can be seen that the constant 1 cannot be improved, by the example of 6 vertices of the hexagon and the center. However, the number 7 can be reduced to 6, as following : given 6 points inside a disk of radius 1, there exist two of them at a distance at most 1. If one of them is the center, it's clear. Ohterwise, join the points with the center. There exist two of these radiuses that form an angle at most 60. Hence the points lie in a circular sector of angle 60, so the distance between them is at most the radius, 1.