Formula for the curve of regular polygons and surface of regular polyhedra

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

\lim_{h\to\infty}    x^{2h} + y^{2h} = R_{i}^{2h}

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

\lim_{h\to\infty}    x^{2h} + y^{2h} + z^{2h} = R_{i}^{2h}

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

The reason the exponent is 2n 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.

\lim_{h\to\infty}    \sum_{n=0}^{N-1} (\hat{q_{n}} \cdot \vec{s})^{2h}= R_{i}^{2h}

where \vec{s}=(x,y)=(s \cdot cos \theta,s \cdot sin \theta) and \hat{q_{n}}=(cos (\frac{2 \pi n}{N}),sin (\frac{2 \pi n}{N})) are the unit normals to the edges of the polygon. Also note that \hat{s}=(cos \theta,sin \theta).

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

\lim_{h\to\infty}    s(\theta)=\frac{R_{i}}{(\sum_{n=0}^{N-1} (\hat{q_{n}} \cdot \hat{s})^{2h})^{\frac{1}{2h}}}


\lim_{h\to\infty}    s(\theta)=\frac{R_{i}}{(\sum_{n=0}^{N-1} (cos (\frac{2 \pi n}{N}) cos \theta + sin (\frac{2 \pi n}{N}) sin \theta)^{2h})^{\frac{1}{2h}}}

Using the trig identity: cos (a) cos (b) + sin (a) sin (b) = cos (a - b), one can write the equation for the curve describing a 2N-gon as

\lim_{h\to\infty}    s(\theta)=\frac{R_{i}}{(\sum_{n=0}^{N-1} cos^{2h} (\frac{2 \pi n}{N}-\theta))^{\frac{1}{2h}}}

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.

\lim_{h\to\infty}    s(\theta)=\frac{R_{i}}{(\sum_{n=0}^{N-1} cos^{2h} (\frac{\pi n}{N}-\theta))^{\frac{1}{2h}}}

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 \theta 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.

\lim_{h\to\infty}    s(\theta)=\frac{R_{i}}{(\sum_{n=0}^{N-1} cos^{4 \cdot 2h} (\frac{\pi n}{N}-\frac{\theta}{2}))^{\frac{1}{2h}}}

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

\lim_{h\to\infty}    s(\theta)=\frac{R_{i}}{(\sum_{n=0}^{N-1} cos^{4h} (\frac{\pi n}{N}-\frac{\theta}{2}))^{\frac{1}{h}}}

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

\lim_{h\to\infty} \sum_{n=0}^{N-1} (\hat{q_{n}} \cdot \vec{r})^{2h}= R_{i}^{2h}

where \vec{r}=(x,y,z)=(r \cdot cos \theta sin \phi,r \cdot sin \theta sin \phi,r \cdot cos \phi) and \hat{q_{n}} are the unit normals to the faces of the polyhedron. Also note that \hat{r}=(cos \theta sin \phi,sin \theta sin \phi, cos \phi).

Putting it into spherical, we have

\lim_{h\to\infty} r(\theta,\phi)=\frac{R_{i}}{(\sum_{n=0}^{N-1} (\hat{q_{n}} \cdot \hat{r})^{2h})^{\frac{1}{2h}}}

Describing q_{n} 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 q_{n} in one half-chunk is opposite one corresponding q_{n} in the other half-chunk (the q_{n} 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?

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