Math Problem: Evenly spaced points on a sphere - Part 1

If place points around the circumference of a circle, it is not difficult to position them so that they're evenly spaced.

n	angle	angle	name
3	2π / 3	120°	triangle
4	π / 2	90°	square
5	2π / 5	72°	pentagon
n	2π / n	360 / n	n-gon

But suppose you want to position points around the surface of a sphere so that they're evenly spaced.

What is the angle between adjacent points? Is it possible to position the points so that the angle is the same between all adjacent points on the sphere?

Let's look at some examples. For n=4, you have a tetrahedron.


According to wikipedia, the angle between adjacent points is arccos(-1/3) or about 109.4712°.

For n=6, you have an octahedron. The angle between adjacent points is π / 2 or 90°.



To summarize and expand:

n	faces	tri. faces	edges	tri. edges	angle	angle	name
4	4	4	6	6	arccos(-1/3)	109.4712°	tetrahedron
6	8	8	12	12	π / 2	90°	octahedron
8	6	12	12	18	multiple angles		cube
12	20	20	30	30	2 arctan(2/(1+√5))	63.435°	icosahedron
n	?	?	?	?	?	?

So what's the answer? I don't know, but one thing we have learned is that a cube's points are not all evenly spaced. The angles between them follow a regular pattern. There are twelve edges with an angle between the points of 2 arcsin(1/√3) ≈ 70.528°, and if you include all triangle edges, there are six edges with an angle between the points of 2 arcsin(√2/√3) ≈ 109.4712°.

Continue reading Part 2