# Placing points on sphere as far apart as possible

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 $$\sqrt 2$$.

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

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 $$M_5=\sqrt 2$$?