Geometry question - points on a sphere

[Ladewig]

If one were to place two points as far as possible from each other on a sphere, they would be 180° apart. Similarly, three points could be placed as far as possible from each other by positioning them 120° from each other

The optimal solition for four points would be to imagine a tetrahedron inside the sphere with the of the points placed at the corners of the tetrahedron. Using Platonic solids also provide the solutions for 6, 8, 12 and 20 points by using octahedrons, cubes, icosahedrons, and dodecahedrons.

points angle
 2       180
 3       120
 4       109.5
 5       ?
 6       90

My question is what is the furthest apart one can place five points? Obviously, one could point three points on a equator and one point at each of the poles which would mean some of the points are 120° from each other and others are 90° from each other. Can one use a different configuration in which the smallest angle is greater than 90°? If so, what would that angle be?


I know some people in this subforum sometimes shoot from the hip. I am hoping for an answer more useful than "I don't think so."

[LucyR]

Sorry to shoot from the hip but the answer's no. 120 degrees for equilateral triangle on a diameter, other two at corresponding poles 180 degress, remainder at 90.

[to.by]

Depends on what you mean by " as far from each other as possible". Do you mean the shortest distance between two of the points, the sum of the distances between any two points, the sum of the shortest distances from the points to their nearest neighbour? Do you mean the distances measured along a great circle or along a straight line between two points?

[LucyR]

Originally posted by to.by
Depends on what you mean by " as far from each other as possible". Do you mean the shortest distance between two of the points, the sum of the distances between any two points, the sum of the shortest distances from the points to their nearest neighbour? Do you mean the distances measured along a great circle or along a straight line between two points?

I think this is what is being asked:

Imagine five identical point charges (electrons if you like) on the surface of a conducting sphere. Under the action of the repulsive Coulombic potential what will be their equilibrium positions?

[Matabiri]

Also shooting from the hip; I don't think there is a solution. The solution would have to form the points of a solid form in which all sides were the same shape. For four points, the form is a tetrahedron, for six an octahedron (and for eight, a cube). I don't think there is an equivalent solid form for five (or seven) points.

ETA: Thinking, it is possible to construct such a form for five points - two tetrahedra attached by one face would do it, but the corners of such a figure do not lie on the surface of a sphere.

[69dodge]

Originally posted by Ladewig
Can one use a different configuration in which the smallest angle is greater than 90°?I don't think so.

:p

But I'll tell you why I don't think so.

Imagine we have a sphere directly in front of us. We'll put points on it, one by one, trying to keep them all slightly more than 90 degrees away from each other.

The first point can go anywhere. Let's put it at the bottom, at the south pole. That eliminates the entire southern hemisphere; all the rest of the points need to go in the northern hemisphere.

Ok, now we need to put the second point somewhere. Due to the symmetry of the sphere-with-a-single-point-at-the-bottom, the longitude of the second point doesn't matter; the important question is what its latitude will be. Since we can pick any longitude we want, let's decide that the point will go somewhere on the meridian of longitude that passes through the rightmost position of the sphere (the east pole?).

Just as the first point did, the second point also will eliminate a hemisphere centered on it. If we place it exactly at the east pole, it will eliminate the right half of the sphere. Sliding it upwards will eliminate more of the northern hemisphere, while freeing up an equal amount of the southern hemisphere. But the entire southern hemisphere is already out anyway, due to the first point. So, if we want to leave as much room as we can for the remaining three points, the second point should be placed as low as possible. So, we place it just a bit above the equator.

Now we're left with the upper left quarter of the sphere free, and three more points to place. It should be clear that the best we can do is place two of them directly in front of us, one on the near side of the sphere and the other on the far side, with the last point halfway between them. (The reasoning is as follows: if the point nearest to us is not directly in front of us, it can be moved farther from the other two by moving it yet nearer to us; and, similarly, if the point farthest from us can be moved yet farther from us, doing so will move it farther from the other two points.)

But even in this best possible scenario, the last point is no further than 90 degrees from the third and fourth points. QED

[JamesM]

Geometrically shooting from the hip: uniformly spaced points on a sphere would have to form a regular polyhedron, wouldn't they? There are only five regular polyhedra in 3-dimensions: with 4, 6, 8, 12 and 20 faces, familiar to all D&D players, I'm sure. Therefore, these, in addition to 1 and 2, are the only number of points that can be placed on the surface of a sphere with equal spacing between all points. (ETA: although it's clear you already know this, Ladewig).

Chemically shooting from the hip: I strongly suspect that a geometry "better" than the trigonal bipyramidal does not exist, or it would almost certainly have been observed in the geometry of a 5-coordinate transition metal complex, ML₅ (or in phosphorus pentachloride, PCl₅), where the M is at the centre of the sphere and the ligands seek to minimise the repulsion between their electrons. Assuming the ligands are identical and the M-L bond lengths are identical, this is similar to the scenario envisaged by LucyR.

In fact, the complex will take up a trigonal bipyramidal structure, which is the geometry you described in your post.

[Matabiri]

Originally posted by JamesM
Geometrically shooting from the hip: uniformly spaced points on a sphere would have to form a regular polyhedron, wouldn't they?

Yeah, that's what I was intimating, but forgot the words, for reasons which now escape me.

[Ladewig]

69 dodge, thank you.

Matabiri and JamesM, I apologize for not being clear enough. I didn't expect each of the five points to be equidistant from its nearest neighbor. Some asymetry is necessary when using a number of points (greater or equal to 5) that does not correspond to the vertices of a regular solid.

Speaking of which, how does one handle seven points?

[Ladewig]

Originally posted by to.by
Do you mean the distances measured along a great circle or along a straight line between two points?

Does the answer to this last question change any of the solutions?

[JamesM]

Originally posted by Ladewig
Speaking of which, how does one handle seven points?
I think the answer is pentagonal bipyramidal (5 points at the equator, instead of 3), but I can't prove it analytically.

For the case of an arbitrary number of points, you might find this (http://penguin.ewu.edu/~trolfe/SCCS-96/rolfe_tj.html) interesting - very similar to LucyR's casting of the problem. But again, not analytical.

[69dodge]

Originally posted by JamesM
I think the answer is pentagonal bipyramidal (5 points at the equator, instead of 3), but I can't prove it analytically.Suppose you nudge one of the equatorial points up a bit, leave the two next to it on the equator, and of the two remaining equatorial points, move one up a bit and the other down a bit. Wouldn't all the angles then be greater than 72 degrees?

[Ladewig]

Originally posted by 69dodge
Suppose you nudge one of the equatorial points up a bit, leave the two next to it on the equator, and of the two remaining equatorial points, move one up a bit and the other down a bit. Wouldn't all the angles then be greater than 72 degrees?

That's what I thought at first, but then I realized with an odd number of points on the equator, there is a problem; you can't raise and lower alternating points, because when you get back to the first point, you end up with two adjacent points being raised or lowered - unless I am missing something.

[69dodge]

Right. I didn't raise and lower alternating points. There are three positions, not just two: up, down, and on the equator, shown as +, -, and 0.
+
0 0
- +
You're looking down on the sphere from the north pole.

[CurtC]

Originally posted by LucyR
I think this is what is being asked:

Imagine five identical point charges (electrons if you like) on the surface of a conducting sphere. Under the action of the repulsive Coulombic potential what will be their equilibrium positions?That's not what was asked. That's one way to make an interesting math problem, but it's not *this* math problem. I think what was asked was to make the minimum distance between any two points as large as possible.

[Ladewig]

Originally posted by 69dodge
Right. I didn't raise and lower alternating points. There are three positions, not just two: up, down, and on the equator, shown as +, -, and 0.
+
0 0
- +
You're looking down on the sphere from the north pole.


I like it. Do you have a number for the size of the smallest angle?

[rppa]

This is a famously hard problem. My memory says that the general solution is not known for arbitrary n. That is, there are known good solutions, but no proof that they are optimal.

Check this page out (http://www.ogre.nu/sphere.htm)

The relevant part is the list of papers under "Packing: maximize the least separation". The others give solutions for other packing criteria, of which there are many.

Edited to add: Dave Rusin's summary article (http://www.math.niu.edu/~rusin/known-math/index/spheres.html) (from the link I gave above) is an even better place to start.

[LucyR]

Originally posted by Ladewig

Speaking of which, how does one handle seven points?

On the basis of the analogy I presented earlier you'd get a regular pentagon on a diameter and a point at each of the two corresponding poles.

Btw. is my interpretation of your question correct? CurtC has a different interpretation.

[LucyR]

Too late to edit my post but have a look at this site:

http://www.mathpages.com/home/kmath005.htm

[LucyR]

Bumpity...

bumpity...

bump!

[bronto]

It's always a treat to find a new link to my site (http://www.ogre.nu/)!

rppa wrote: This is a famously hard problem. My memory says that the general solution is not known for arbitrary n. That is, there are known good solutions, but no proof that they are optimal.
Several different statements are bundled in that. No general solution is known, i.e. no algorithm gives the answer for arbitrary n in a finite number of steps. I would bet that no general solution is possible.
Empirical solutions are known for many n, but only a few are proven optimal.
There are several practical algorithms (http://www.ogre.nu/pack/pack.htm) for a "good enough" pack, but their results are never optimal (unless for very small n).

[bronto]

The best pack of eight points is a square antiprism (http://www.enginemonitoring.com/sphere/pages/pack08.htm), not a cube. Nor is the dodecahedron optimal for 20; the solution there is harder to describe, but suffice to say its convex hull is all triangles.

For seven points, the repulsion figure (pentagonal dipyramid) has no resemblance to the best packing (http://www.enginemonitoring.com/sphere/pages/pack07.htm). I'm a bit surprised to find such a dramatic difference for such a small number.

[Art Vandelay]

Originally posted by JamesM
Geometrically shooting from the hip: uniformly spaced points on a sphere would have to form a regular polyhedron, wouldn't they? Isn't a soccer ball a counter example?

Another question: are the solutions unique (under the proper symmetry, of course)? What about in the electrical charge interpretation?

[bronto]

Indeed, the best packings of five and eleven points are the vertices of a square pyramid (octahedron missing one) and a gyroextended pentagonal pyramid (icosahedron missing one), neither of which you'd normally call regular.

Most optimal packings are unique, but there are exceptions (http://home.earthlink.net/~jbuddenh/pack/sphere/mult/multpics.htm). I'm not aware of any such pairs in the electron problem; for some numbers (of which the lowest is 16 (http://mathpages.com/home/kmath005.htm)) there are two or more locally-stable states, but they are not equal.