# Equidistance of Points in a Sphere?

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• shintashi

#### shintashi

I've been trying to wrap my head around equidistant points, like platonic solid vertices inside a sphere where the points touch the sphere surface. This led me to the strange and unusual world of mathematical degeneracy, henagons, dihedrons, and so on, along with the lingering question of superposition.

Its easy enough to see poles, but if you have 2 poles, those points are equidistant along one dimension, but the other two are missing (im trying not to get off topic by imagining the two points as projecting a north and south hemisphere), while 3 points creates a plane triangle, which like the 2 pole object occupies infinite superpositions along one dimension of rotation.

But, we get to tetrahedron, octohedron, cube, dodecahedron, etc., and we have stable equidistant points relative to the sphere surface.

But what about an object with five, six, or seven vertices? Are such solids even possible for equidistant values on the surface of a sphere? Would such objects have to necessarily have superpositioned vectors approximating the space where they might be?

would the set of points you imagine give rise to the vertices of a regular polyhedron? If so, you may know that all possible such solids are known.

would the set of points you imagine give rise to the vertices of a regular polyhedron?
That's the difficult part of the question. Points being equidistant from all their nearest neighbours (nearest neighbour equidistance or NNE) is not sufficient to make a polyhedron regular. We can glue square pyramids to faces of a cube to make a non-regular NNE polyhedron. We can even satisfy a convexity requirement if we only glue the pyramids on the top and bottom faces.

I suspect that requiring all vertices to lie on the surface of a sphere will restrict us to only the Platonic Solids, but I don't have a proof of that. Maybe Euclid did.

It does get me wondering what configuration of five points on a sphere would maximise the minimum distance between any two points (MMD). A quick guess is that we put points at the North and South Poles and at three equidistant points around the equator. Such a shape is not regular, nor even NNE, and I have no proof that it is MMD.

To find a MMD configuration of five points on a sphere, I thought of writing a formula for the minimum distance between any two of the five points, as a function of seven real variables, which is sufficient to specify the location of five points on a sphere, modulo rotation. We could then set all partial derivatives to zero and solve, to find the minimum distance configuration. But unfortunately, since the minimum distance is the minimum of 5 x 4 / 2 = 10 distances, and the function ##(x,y)\mapsto \min(x,y)## is not differentiable at crossover points, the formula would not be everywhere differentiable, so usual calculus-based minimum-finding approach would not work.

A simple but ugly and time-consuming alternative is a brute force search through the seven-dimensional space of possible locations of the five points, to find MMD configuration(s).

Perhaps there's a more elegant approach that I'm failing to see.

the OP's question is not sufficiently precise for me to know the answer to my question to him, which is why I posed it in that way. If the only condition is that the distance from each point to all its nearest neighbors should be the same, for all points, then one can inscribe an octahedron in a sphere and then remove one vertex. there will be 5 remaining vertices which do satisfy the requirement as stated. The trick is that not every vertex will have the same number of closest points, but that was not required. So we need to know exactly what requirements the OP intends to ask for. One can obviously alo obtain figures with 6 and 7 and other numbetrs of vertices in the same way by removing vertices from various platonic solids.

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every vertex must have the same number of closest points. That's the problem. A Pyramid can seem equidistant with a length of one, but doesn't conform to sphere geometry like a tetrahedron would.

The icosidodecahedron, the snub cube, and the rhombic triacontahedron satisfy what I think your working definition is, but are not regular polyhedra. You may find the snub cube interesting in particular, as it has five edges per vertex.

So, you may want to look at Catalan solids, Archimedean solids, and uniform polyhedra.

• shintashi