For which dimensions does there exist a regular $n$-polytope such that the distance of its vertex to its center equals the length of its side?

Question

For example, it is easy to see that an hexagon, a regular $2$-polytope, is an example of what I am looking for: the distance between a vertex and its centre equals the length of a side. But, on the other hand, it seems that no regular polyhedron ($3$-polytope) fits this condition.

Answer 1

We can work with standard embeddings of the $n$-cube (as $[-1,1]^n$) and cross-polytope (vertices $(\pm1,0,0,\ldots) $, $(0,\pm1,0,\ldots) $, etc.) and the embedding of the $n$-simplex in $\mathbb{R}^{n+1}$ given by $0\leq x_i\leq 1$ and $\sum_{i=0}^{n+1}x_i=1$.

In these coordinates, the distance from the cube's vertices to its center is $\sqrt{n} $ while its edge length is just $2$, so it's a solution for $n=4$. The distance from the cross-polytope's vertices to its center is always $1$ and the edge length is always $\sqrt2$ so it's never a solution. The $(n-1)$-simplex is a face of the cross-polytope so it also has edge length $\sqrt2$ while the distance to the center is $\sqrt {(1-\frac1n)^2+\frac1{n^2}+\ldots+\frac1{n^2}}$ $=\sqrt{1-\frac2n+1}$ $=\sqrt{2-\frac2n}$, so these are never equal either. This just leaves the exceptional cases in dimension 3, which as OP notes are inequal.

(ETA: per the comments below, it's worth noting that the 24-cell provides another example in four dimensions. This is easiest to see in the embedding with vertices all permutations of $(\pm1,\pm1,0,0)$; then the distance from any vertex to the center is $\sqrt2$, as is the distance between two adjacent vertices like $(1,1,0,0)$ and $(1,0,1,0)$.)