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Consider the setup described here.

Let us focus on a special case: $n=6$. The goal is to place $6$ points $x_1,\dots,x_6$ on the unit sphere $S^2\subset\mathbb R^3$ such that $$\min_{i\neq j\in\{1,\dots,6\}}|x_i-x_j|$$ is maximal. Call this maximum $M_6$.

It seems intuitive to place the $6$ points at the centres of the faces of a cube, giving $M_6=\sqrt 2$. I have difficulties arguing why this is the case, i.e. showing that no matter how the $6$ points are place, their minimal distance will never exceed $\sqrt 2$.

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    $\begingroup$ Since the cube is dual to the octahedron, it might help to consider these points as the vertices of an octahedron. $\endgroup$
    – robjohn
    Commented Sep 29, 2022 at 16:14
  • $\begingroup$ One can always find a half sphere containing $4$ points. Maybe this can be useful? $\endgroup$
    – Zuy
    Commented Sep 29, 2022 at 16:42

1 Answer 1

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Position any $6$ points on the sphere. Seeking for a contradiction, assume that the minimal distance between any two points is greater than $\sqrt 2$. Take any two points $p_1$ and $p_2$ and consider the great circle of the sphere passing through them. One of the two half spheres defined by this great circle contains $4$ points.

For concreteness, assume that the considered half sphere is $S^2_+=\{(x,y,z)\in S^2\mid z\geq 0\}$, and that $p_1=(1,0,0)$. The other point $p_2=(x_2,y_2,0)$ on the great circle must satisfy $x_2<0$, else $p_1$ and $p_2$ would be too close. By symmetry, we may even assume $y_2\geq 0$.

The remaining two points $p_3,p_4\in S^2_+$ cannot belong to the quarter sphere $\{(x,y,z)\in S^2\mid x,z\geq 0\}$ because they would otherwise be too close to $p_1$. Further, they cannot belong to the quarter sphere $\{(x,y,z)\in S^2\mid y,z\geq 0\}$ because they would otherwise be too close to $p_2$.

Hence both $p_3$ and $p_4$ belong to the "eighth sphere" $\{(x,y,z)\in S^2\mid x,y<0 \text{ and } z\geq 0\}$. But now $p_3$ and $p_4$ are too close, contradiction.

This also shows that $M_5=\sqrt 2$?

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