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A set of $8$ points in an unit cube is given, such that the distance between any two points belonging to the set is not less than $1$. Find the distance between the center of the cube and the closest to it point from the set.

Intuitively, I think, each of the points in the set is a vertex for the cube and the distance is $\frac{\sqrt3}{2}$, but I don't know how to prove it.

prometheus21
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  • What is to prove if you put all 8 points at 8 cube vertices? Distance between them is either $\sqrt3$, $\sqrt2$ or $1$. Distnace between any of them and the center of the cube is $\sqrt3/2$. I think it would be much more interesting to ask: What is the minimal possible distance between one of the points and the center of the cube. – Saša Jul 16 '18 at 17:56

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A set of 8 points in an unit cube is given, such that the distance between any two points belonging to the set is not less than 1.

In 1966, J. Schaer proved that there is just one such set, which indeed consists of 8 vertices of the cube.

You can find the proof, short and elegant, in that article. See end of page 267, from "The case $k=8$."

Schaer, J., On the densest packing of spheres in a cube, Can. Math. Bull. 9, 265-270 (1966). ZBL0144.21402.

Timur M.
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