No four points with pairwise distance 1 can be contained inside a halfdisk of radius 1.

Question

An open disk $D$ of radius $1$ in the Euclidean plane is the set of points with distance less than $1$ to the center of the disk. An open half disk $H$ of radius $1$ is obtained by "cutting" $D$ into two equal sized parts and taking one of them.

I would like to show that $H$ can not contain 4 points of pairwise minimum distance at least 1. While the statement is intuitively clear, I do not have a good idea as how to tackle the problem. Does anybody have an idea as how to tackle this?

Any pointers would be appreciated.

score 2 · Accepted Answer · answered Sep 29 '14 at 14:26

2

Suppose you have four points in the half disk. Cut the half disk $H$ to three sectors, each opening at an angle $\pi/3$. By the pigeonhole principle there must be two points in one disk. Show that the distance between any two points in such a sector is less than one.

answered Sep 29 '14 at 14:26

Joonas Ilmavirta

25,809

1

Nice solution. In your final sentence I think you mean " less than one", since the half disk $H$ is open. Then the proof is finished. – SPK.z Sep 29 '14 at 14:30
@SPK, thanks, corrected. – Joonas Ilmavirta Sep 29 '14 at 14:31

No four points with pairwise distance 1 can be contained inside a halfdisk of radius 1.

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