5 Points on a Sphere

  • Thread starter Alkatran
  • Start date
Alkatran
Science Advisor
Homework Helper
942
0

Main Question or Discussion Point

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?
 

Answers and Replies

90
0
Have you considered the five vertices of a double tetrahedron?
 
Alkatran
Science Advisor
Homework Helper
942
0
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.

Code:
  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.
 
Last edited:
D H
Staff Emeritus
Science Advisor
Insights Author
15,329
681
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
Science Advisor
Homework Helper
942
0
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.
 
1,424
1
From what I know about molecular geometry, what you are looking for is a bipyramidal figure, where you basically have two superposed tetrahedron.
 

Related Threads for: 5 Points on a Sphere

  • Last Post
Replies
4
Views
2K
Replies
16
Views
7K
Replies
7
Views
793
Replies
3
Views
722
  • Last Post
Replies
1
Views
1K
Replies
7
Views
2K
Replies
10
Views
791
  • Last Post
Replies
6
Views
9K
Top