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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.

Martin Sleziak
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HighSchool15
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2 Answers2

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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 $1/2$.

ThePortakal
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  • Over the top! Enclosing the equilateral triangle in circle of radius $1$ is easier and is all you need. – Rob Arthan Sep 14 '17 at 20:29
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    @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. – quasi Sep 14 '17 at 20:30
  • @Rob Arthan: Why is radius $1$ sufficient? – quasi Sep 14 '17 at 20:33
  • @quasi: Because the equilateral triangle is contained in the circle of radius $1$ about any of its points. – Rob Arthan Sep 14 '17 at 20:40
  • @Rob Arthan: That seems circular to me. How do you know that? An in any case, that's not just one circle. – quasi Sep 14 '17 at 20:41
  • No: if $X$ is any point in the equilateral triangle, and $U_1$, $U_2$ and $U_3$ are its vertices, then the triangles $XU_iU_j$ ($i\neq j$) all have $U_iU_j$ as their longest side. But let's not worry about it too much. ThePortakal's answer is fine as it stands. – Rob Arthan Sep 14 '17 at 20:44
  • @Rob Arthan: Maybe that works, but frankly, the single enclosing circle -- the circumscribed circle with radius $\sqrt{3}/6$ (hence less than $1/2$) is more instantly clear. – quasi Sep 14 '17 at 20:47
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    The maximum distance between two points in any triangle is its longest side-length. – Angina Seng Sep 14 '17 at 20:50
  • @Lord Shark the Unknown: It's a question of what's known, and what requires proof (at a given level). – quasi Sep 14 '17 at 20:51
  • My point was that you don't need to estimate the radius of the circumcircle. – Rob Arthan Sep 14 '17 at 20:52
  • Ok, I get your points (both yours and Lord Shark's), but I felt ThePortakal's argument defended nicely against any possible doubts. – quasi Sep 14 '17 at 20:55
  • Umm, my calculation of the radius of the circumcircle is $R = \frac{1}{2} \frac{a}{\sin A} = \frac{1}{2} \frac{1}{\sqrt{3}/2} = \frac{1}{\sqrt{3}} > \frac{1}{2}$. In any case, the diameter has to be longer than the side 1 (it's the hypotenuse of a right triangle with one of its legs being 1). – Daniel Schepler Sep 14 '17 at 21:16
  • @Daniel Schepler: Oops, you're right. So the circumcircle doesn't work. Back to Lord Shark's argument then. Of course, unless that's considered known, it still requires proof. – quasi Sep 14 '17 at 21:34
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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.

orangeskid
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