Can a bipyramidal figure be achieved with 5 points on a sphere?

[Alkatran]

Say you have a sphere, and n points to place on it. You want each of the points to be equivalent to the others if you allow relabeling and the average distance between points should be as great as possible. Essentially the most spread out 'fair' distribution of points when no point is special compared to the others.

For example: when n = 4, the points would be the vertices of a tetrahedron. When n = 6, the points would be the centers of the faces of a cube. When n = 3 the points would be the vertices of a triangle.

My question is: with 5 points, the only possible layout I have found is the vertices a pentagon. Is there a better one?

 

[fopc]

Have you considered the five vertices of a double tetrahedron?

 

[Alkatran]

I don't think a double tetrahedron will work. The three points at the joining point have all the other points equidistant from them, but the two points away from it do not.

  1
 /|\
2-3-4
 \|/
  5

1 is further away from 5 than any point is from 4, therefore no relabeling can make 1 equivalent to 4. However, this may not be true if we take arc length into account (which is what really matters in this case) or move 1 and 5 closer together... but I have no experience with this type of geometry.

 

[D H]

Based on your rejection of the double tetrahedron, there is no arrangement of five points (or almost any value of N) other than planar that satisfies your criteria. The exceptions are the vertices of the convex regular polyhedra. There are only five convex regular polyhedra, aka the Platonic solids: the regular tetrahedron, hexahedron (cube), octahedron, dodecahedron, and icosahedron.

 

[Alkatran]

Based on your rejection of the double tetrahedron, there is no arrangement of five points (or almost any value of N) other than planar that satisfies your criteria. The exceptions are the vertices of the convex regular polyhedra. There are only five convex regular polyhedra, aka the Platonic solids: the regular tetrahedron, hexahedron (cube), octahedron, dodecahedron, and icosahedron.

I assumed this might happen. However, I do know there are at least two arrangements for every even number of points. Just make two parallel (n/2)-gons out of the points (the n=6 and n=8 solutions examples of this). This will have a higher average distance than the simple n-gon. There's also the trivial "all points at the same spot" solution, haha.

 

[Werg22]

From what I know about molecular geometry, what you are looking for is a bipyramidal figure, where you basically have two superposed tetrahedron.