Why can't you make a regular solid with an arbitrary number of faces?

Question

In scool we were taught that there are only a few regular solids, and they've all been known since Classical Greek times.  Imagine this experiment.  You've got a transparent, insulating hollow sphere, made out of, say, polycarbonate.  It's in two hemispheres, which you can separate, so you can open it.  You've got a whole lot of light pith balls, coated with metal foil, all with the same mass.  You give all of them the same electrical charge, place them inside your sphere and close it.  You do this in microgravity, say, in the ISS.  They all repel each other, so they're all the same distance apart, so they're the vertices of a regular solid.  You could do this with any number of balls.  So why couldn't you join them up with straight lines and make a line drawing of a regular solid with any number of faces ?

? · Accepted Answer

You can do that, but the "faces" you make that way will either not all be identical, or will not be regular polygons.  So it won't be a regular solid.

As an example, imagine 5 vertexes.  If you put 4 in your container, you get a regular tetrahedron.  But if you put in 5, you'll get 2 points at opposite ends and then 3 points in a ring between the first 2.  The points at the "poles" each have exactly 3 neighbors, but the points on the ring have 4 neighbors.  So the faces formed in this way are not regular.  Even so, it is the minimal energy distribution for the charges.

There are several proofs of the existence of only 5 regular solids.  For example, see: http://en.wikipedia.org/wiki/Platonic_solid#Geometric_proof

? · Answer

You have a fallacy in your reasoning.

They will not all be the same distance from each other.  They might not even ever stop moving (assuming you somehow keep them all identically and non-neutrally charged).

4 balls is the most you will have that will all be equidistant from each other.  To do it with more than 4, you need more than 3 dimensions.

Even the Platonic solids don't satisfy the equidistance condition.

Take, for example, a cube.  8 balls.  There are 3 different distances that a given ball can be from another ball.  The shortest would be the length of one edge.  The medium distance would be the square root of 2 times the shortest distance, for the distance from one ball diagonally across a single face to the opposite corner of that same face.  The longest would be the square root of 3 times the shortest, going through the center of the cube.



Try to imagine a solid with 5 vertices, and all faces have the same number of edges.  What does that solid look like?  Can the faces be quadrilaterals?  No, because once 4 of those vertices are coplanar, you only have 1 left, and the most you can construct a quadrilateral plus one more point is a collection of triangles.  If the faces aren't identical, then it can't be regular.  So, OK, let's try triangles.  We start with 1 vertex, and we put 3 more non-collinear points out there, so that we have 3 triangles.  Now we have used up 4 vertices, and we have to figure out what to do with the fifth one.  We could put it on the other side, giving us a solid that has 5 vertices, 9 edges, and 6 faces, but you can't make that solid regular.  You might think you have (put two tetrahedra together, leaving the 6 faces on the outside), but you've missed part of the definition of a regular solid.  All the angles have to be equal, but the angles in that glued-together middle and the angles at the extreme points are different, and you can't make them the same.  Some vertices have 3 edges (and 3 faces) that meet there, and others have 4.  Squeeze and stretch the shape all you like, but it can't work out to be regular.

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Why can't you make a regular solid with an arbitrary number of faces?

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