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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.

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

2 Answers 2

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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 $1/2$.

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  • $\begingroup$ Over the top! Enclosing the equilateral triangle in circle of radius $1$ is easier and is all you need. $\endgroup$
    – Rob Arthan
    Commented Sep 14, 2017 at 20:29
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    $\begingroup$ @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. $\endgroup$
    – quasi
    Commented Sep 14, 2017 at 20:30
  • $\begingroup$ @Rob Arthan: Why is radius $1$ sufficient? $\endgroup$
    – quasi
    Commented Sep 14, 2017 at 20:33
  • $\begingroup$ @quasi: Because the equilateral triangle is contained in the circle of radius $1$ about any of its points. $\endgroup$
    – Rob Arthan
    Commented Sep 14, 2017 at 20:40
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    $\begingroup$ The maximum distance between two points in any triangle is its longest side-length. $\endgroup$ Commented Sep 14, 2017 at 20:50
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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.

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