I'm not too sure about the exact terminology since Wikipedia is throwing me all over the place. I'm looking for a formula to find out if for n faces a 'regular polyhedron' can exist. In case that's not the right term, I'll specify as to say a three dimensional shape comprised of vertices that are equidistant from each other and equiangular. I'm sort of looking for the extension of platonic solids.

I'm trying to answer a question that was posted on StackOverFlow here https://stackoverflow.com/questions/14805583/dispersing-n-points-uniformly-on-a-sphere/14807728#14807728

about arranging point on a sphere so that each one will be equidistant form each other. Essentially so that its a perfectly uniform distribution. Well I have an idea for an algorithm to do this, and given the research I've done I just need to find a formula so I can round n faces to the nearest platonic solid extension in order to beable to distribute the points perfectly evenly.

Is there such a formula? Or a formula to find for n faces a shape that would come close? (like a soccer ball).

On, a side note, any better ideas to tackle this?

  • 1
    $\begingroup$ It sounds like what you're after might be the Archimedean Solids ( en.wikipedia.org/wiki/Archimedean_solid ) - though if you want vertices that are equidistant and equiangular then you're at precisely the Platonic solids. In either case, there's a finite list, so it's very easy to enumerate them and just check if your $n$ (and note that '$n$ faces' is not the same as '$n$ vertices'!) is on the list. More generally, there's been a lot of research into the equidistribution problem - the most common approaches are generally through subdivision. $\endgroup$ – Steven Stadnicki Feb 11 '13 at 19:41
  • $\begingroup$ The Archimedean solids does seem interesting as well, and a probable solution. Would you know a formula to find the configuration of one given n faces? (or n vertices, yes I realize the difference). Secondly, am I to understand that though only those 5 platonic solids exist, there is no formula to determine compounds of them? i.e. to find out if a polyhedron can be constructed using tetrahedrons or dodecahedrons? $\endgroup$ – Klik Feb 11 '13 at 21:04
  • $\begingroup$ At most four points can be equidistant from each other in 3-space. That is probably not the property you are looking for. $\endgroup$ – Nick Matteo Mar 17 '14 at 2:27

You don't need a formula. Remember that there are precisely $5$ regular convex polyhedrons:

  • The tetrahedron has $4$ vertices, $6$ edges, $4$ faces
  • The octahedron has $6$ vertices, $12$ edges, $8$ faces
  • The cube has $8$ vertices, $12$ edges, $6$ faces
  • The icosahedron has $12$ vertices, $30$ edges, $20$ faces
  • And the dodecahedron has $20$ vertices, $30$ edges, $12$ faces

So the most faces that a regular convex polyhedron can have is $20$, which is the icosahedron. If you want to find an estimate of the number of faces you can divide a sphere in to, given the number of points (vertices), you can make your own imaginary polyhedron using Euler's polyhedra formula. Which states that the number of vertices minus the number of edges plus the number of faces equals $2$. $$ V - E + F = 2 $$

Notice that the number of edges equals one half of the number of sides of the face multiplied by the number of faces. This is true because when the faces are connected to form a polyhedron, each side is shared with another face, so each side must only be counted once as an edge. So if we let s equal the number of sides of whatever face you decide to use (remember you can only choose from a triangle, square, or pentagon) then we can write: $$ \frac12 sF = E $$

Plug this in to Euler's formula to get: $$ V - \frac12 sF + F = 2\\ V + F \left(-\frac12 s + 1\right) = 2\\ F = \frac{2 - V}{-\frac12 s+1} $$ Plug in your number of vertices (points) and the number of sides of the face you decide to use to obtain an estimated number of faces you can divide your surface area in to. Note that the larger the number of points on the sphere, the more accurate your answer will be. This is true because this solution was obtained from assuming that whatever face you chose to use is planar. So the more points you have, the more faces you have, so the faces take up a smaller area of the sphere, in turn making each face less hyperbolic and more planar.

  • $\begingroup$ This made my day! For a completely (or not so much, now that I think of it) unrelated problem, this idea was the key to solve it. Amazed at how math concepts pop up everywhere and in so many different contexts. $\endgroup$ – Ioannes Oct 19 '16 at 10:33

Well there is a generalized formula for regular polyhedrons that will certainly help you.

Kindly go through HCR's Formula for regular polyhedrons


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