I have a 600-cell, whose coordinates are given by


$$\begin{array}{ccc}
\text{8 vertices} & \left(0,0,0,\pm1\right) & \text{all permutations,}\\
\text{16 vertices} & \frac{1}{2}\left(\pm1,\pm1,\pm1,\pm1\right),\\
\text{96 vertices} & \frac{1}{2}\left(\pm\tau,\pm1,\pm\tau^{-1},0\right) & \text{even permutations}.
\end{array}$$


Now I would like to isolate a tethraedral cell of the 600-cell and find the hyperplane that passes through it.Probably I should calculate the distances between the vertices to understand which of those are neighbours. Does anyone have a reference or a shorter path to do it?



How can I find the hyperplane passing through a 600-cell
You already could have considered your provided vertices within layers according to their last coord values. Within decreasing order you get:


$(0, 0, 0; 1)$: the single point `o3o5o` at the north pole


$\frac12(0,\pm\tau^{-1},\pm1; \tau)$: a full icosahedron `v3o5o` (with $\tau^{-1}$-sized edges)


$\frac12(\pm1, \pm1, \pm1; 1)$ and $\frac12(\pm\tau^{-1}, 0, \pm\tau; 1)$: a full dodecahedron `o3o5v` (with $\tau^{-1}$-sized edges)


$\frac12(\pm1, \pm\tau, 0; \tau^{-1})$: a full icosahedron `x3o5o` (with unit edges)


$(\pm1, 0, 0; 0)$ and $\frac12(\pm\tau, \pm1, \tau^{-1}; 0)$: a full icosidodecahedron `o3v5o` (with $\tau^{-1}$-sized edges) at the equator


The remainder then is the mirror image on the southern hemisphere (or rather -globe).


Next consider the $\tau^{-1}$-scaled icosahedral section in turn within layers according to its 3rd coordinate:


$\frac12(0,\pm\tau^{-1}; 1)$: top-most edge `v o` (of size $\tau^{-1}$)


$\frac12(\pm1, 0; \tau^{-1})$: pair of vertices at the polar circle `o x`


$\frac12(\pm\tau^{-1}, \pm1; 0)$: equatorial rectangle `x v`


Again the remainder would just be the mirror image on the southern hemisphere.


Therefrom one now derives e.g. the following set of 4 vertices belonging to a single tetrahedral facet of the hexacosichoron (= 600-cell), chosen here by means of a single triangle of that icosahedron at height $\frac12\tau$ plus the north pole vertex at height $1$ as its tip:


$(0, 0, 0, 1)$, $\frac12(0, \pm\tau^{-1}, 1, \tau)$, $\frac12(1, 0, \tau^{-1}, \tau)$.


--- rk


