By messing around, I found the formula for a square of inner radius ` (the radius of the largest circle fitting inside, lying tangent with all the edges) to be of the form:`

` `

and the cube of inner radius ` (the radius of the largest sphere fitting inside, lying tangent with all the faces) to be of the form:`

` `

If one takes ` to be less than infinity, the corners of the object are rounded.`

The reason the exponent is ` is because one of these shapes centered at the origin is only a closed curve/surface if our equation is an even function of each variable.`

Finding these forms made me wonder what the forms for the regular N-gons (there are infinity of these) and N-hedra (there are five of these called the Platonic Solids) are.

I noticed that my form for the square fit a more generalized form. This is how I first found the form for the even N-gons, the 2N-gons.

` `

where ` and `

` are the unit normals to the edges of the polygon. Also note that `

`.`

The equation is implicit, and difficult to deal with. It can easily be put into polar explicit form.

` `

or

` `

Using the trig identity: `, one can write the equation for the curve describing a 2N-gon as`

` `

This form only holds when N is even. If I input odd N, the form is exactly equal to if I input twice that N, returning a polygon with 2N number of sides instead of N. This can be fixed with a few steps.

First, I will replace the N in the previous equation with 2N, so that, no matter what N you input, you will get the formula for the corresponding 2N-gon, whether N is odd or even.

` `

Now, we have the form for the 2N-gons when we input N. We can easily cut the figure in half and connect the edges by dividing ` by 2. The resulting figure will have, correctly, N number of sides, but will be warped. This can be corrected by increasing the exponentiation on each term in the sum by a factor of 4, or by decreasing the exponentiation of the final sum by a factor of 4.`

` `

In the limit that h approaches infinity, this is equivalent to the expression

` `

So, this is the form for any regular N-gon in polar. I do not know how to get a cartesian form for the odd (2N+1)-gons.

Here are some example plots (of course, taking h to be less than infinity):

Now, I’ll follow the same logic, minus the last few steps, to arrive at the form for four of the five platonic solids.

Beginning at the form for the surface of a cube (written above, near top), we see that it also fits a more generalized, 3D version of the previous relationship

` `

where ` and `

` are the unit normals to the faces of the polyhedron. Also note that `

`.`

Putting it into spherical, we have

` `

Describing ` is not as easy in 3D. Find an axis of symmetry that points through the center of your solid and either a vertex or the center of a face. Make that axis the z axis, and work in spherical coordinates.`

The form above only works for solids which have a particular property (mentioned below). However, 4 of the 5 platonic solids have this property, so it is close to the solution that I want. Here is an example plot of a dodecahedron:

I’m not the best at MATLAB, so I’m sure there are better ways to plot surfaces. Anyway, here is an edge-traced version:

Here are two properties that are interesting about the platonic solids:

1. They can all be divided into two identical interlocking chunks that contain half the total number of faces each.

2. In all but the tetrahedron, a ` in one half-chunk is opposite one corresponding `

` in the other half-chunk (the `

` belonging to the face directly across from it’s face)`

This second property makes the octahedron, cube, dodecahedron, and icosahedron similar to the even 2N-gons from before when we were in 2D. However, it makes the tetrahedron like the (2N+1)-gons, and I don’t understand how to go through a similar process to arrive at the form for a tetrahedron.

So, my question is: What is the formula for the surface of a tetrahedron?