This is a follow-up question to my previous question, which I felt was too unrelated to what I wanted: Expansion from 3D tetrahedron to 4D 5-cell via combining smaller parts

Anyway, if you don't know already, I'll say it again: I'm making n-dimensional maze puzzles. These are basically mazes where you move around in n-dimensional space (n is specified in the puzzle) and you have to get from point A to point B. The n-dimensional solid is split into its Platonic components (e.g. a 5D hypercube that's 3 cubes wide would become nine 3D cubes) and then those are cut into smaller cubes of the same width (e.g. width 3 would be 27 smaller cubes). After this, I make each mini-cube a wall or a space, and it is a maze then.

I need to have some more solids instead of just n-dimensional cubes. For example, the 5-cell. Problem is, I don't know how to represent a 5-cell.

The 4D hypercube can be represented as a point that extends by translating itself left/right in the first dimension, that extends by translating itself up/down to a square, that extends by translating itself forward/backward to a cube, that extends by translating itself in 4D directions to a hypercube. However, I don't know how this works for a 5-cell, or a 120-cell, or a whatever-cell. I need a way to take a 5-cell and turn it into tetrahedra, then make it into smaller tetrahedra, so I can have a base for my maze. If the cube can be turned into a hypercube by translating it in a 4D direction, how does the tetrahedron translate into a 5-cell?

If the tetrahedron wouldn't make a great maze using this method, let me know of a better method for a tetrahedron in any dimension. Thanks so much in advance. :)

(Sorry for reposting, but I think that previous question would not help, and I don't think editing my first question into a whole new question would be a good idea!)


Start with a regular tetrahedron represented in three dimensions with the vertices





in which each edge has length $\sqrt{8}$.

Add a zero for the fourth coordinate and then specify the fifth vertex as indicated below:






where $a$ is chosen so that the distance from the fifth vertex to the other four matches the $\sqrt{8}$ value in the base tetrahedron. Thereby, $a=\pm\sqrt{5}$.

If you want rational coordinates only, you need five dimensions (compare with an equilateral triangle needing three dimensions, see here). Here is one possible five-dimensional set of vertices (edge length =$\sqrt{2}$):






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  • $\begingroup$ What I need is how to represent a 5-cell as a bunch of tetrahedra. Perhaps I said my question wrong. I will edit it. $\endgroup$ – Jonah Ryman Feb 17 '19 at 19:33
  • $\begingroup$ Are you looking for a projection into three dimensions? $\endgroup$ – Oscar Lanzi Feb 17 '19 at 19:52
  • $\begingroup$ No. A 4D cube can be split into the important "parts:" the position at every unit of length. A 4D cube that is 5 inches long in 4D space can be turned into 5 5x5x5 3D cubes, omitting the inbetweens. This is how I make my mazes. Otherwise there would be infinite blocks to work with, and that would be a daunting task. I'm asking, how would you do this same thing with a 5-cell? $\endgroup$ – Jonah Ryman Feb 17 '19 at 20:02

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