Math Problem: Evenly spaced points on a sphere - Part 3

Continued from Part 2

Let’s try to find a way to systematically divide the triangles on a sphere into smaller triangles in such a way that the resulting triangles are as evenly spaced as possible, starting with either a tetrahedron, an octahedron, or an icosahedron. One thing we can do is place a new point in the center of each triangle. This is called a Kleetope. It is not possible to create a distribution of points that is as evenly spaced as possible by producing successive Kleeptopes because the resulting triangles will become less and less like equilateral triangles. Some points will be very close to each other and other points will be further apart.

   

Another thing we can do is split each triangle into four triangles by placing new points at the midpoint of each triangle edge.

   

When you subdivide a triangle into four triangles on the surface of a sphere, the four triangles are not all the same size. The center one is larger. But as you keep subdividing and the triangles get smaller, the curvature of the sphere makes less and less of a difference and the size of the subdivided triangles become more uniform. So if you can show that a set of points after some number of subdivisions are as evenly spaced as possible, then further subdivisions will also be as evenly spaced as possible. Let’s find the size of the center triangle after subdividing.



Use the spherical law of cosines twice, once for the bigger triangle, and once for the smaller triangle. We’ll assume the radius of the sphere is 1.





Solve for cos(t).



Try to simplify using the pythagorean trigonometric identity and the double angle formula.





Substitute for cos(t) in the smaller triangle.









When the big triangle (2 a) is very small, x is barely bigger than a. When 2 a is 2 pi / 9 (or 40°), x is about 5% bigger than a. When 2 a is 4 pi / 9 (or 80°), x is about 24% bigger than a. Finally, when 2 a is 2 pi / 3 (or 120°), x is twice the size of a, which is the same size as the original triangle. This is the case when the three original points are all evenly spaced around the equator, and the three new points are placed on the midpoints, which are also on the equator.

My guess is that you could take a tetrahedron, an octahedron, or an icosahedron, and iteratively split each triangle into four triangles, and you’d end up with an arrangement of points at each iteration that is as evenly spaced as it could be. Additionally, you could form the Kleeptope one time for each of these iterative arrangements and the arrangement of points would be as evenly spaced as they could be.

Iteratively splitting the tetrahedron

points	edges	faces	distance between adjacent points
4	6	4	109.4712°
10	24	16	54.7356° and 90°
34	96	64	various

Forming the Kleetope for each iteration of the tetrahedron

points	edges	faces	distance between adjacent points
8	18	12	109.4712° and something
26	72	48	various
98	288	192	various

Iteratively splitting the octahedron

points	edges	faces	distance between adjacent points
6	12	8	90°
18	48	32	45° and 60°
66	192	128	various

Forming the Kleetope for each iteration of the octahedron

points	edges	faces	distance between adjacent points
14	36	24	90° and something
50	144	96	various
194	576	384	various

Iteratively splitting the icosahedron

points	edges	faces	distance between adjacent points
12	30	20	63.435°
42	120	80	31.717° and 36°
162	480	320	various

Forming the Kleetope for each iteration of the icosahedron

points	edges	faces	distance between adjacent points
32	90	60	63.435° and something
122	360	240	various
482	1440	960	various

So it looks like we can make a lot of arrangements of points that are probably as evenly spaced as possible, and we can make them larger and larger without limit.

However, it may not possible to mathematically specify an arrangement of points that is as evenly spaced as possible for any number of points n, because when n is a large prime, the optimal arrangement of points is probably unrelated to any patterns of points for any smaller n.