For me, a *polytope* is the convex hull of finitely many points. It is said to be *vertex-transitive* / *edge-transitive* if its symmetry group acts transitively on its vertices / edges. Let's call a polytope *symmetric* if it is simultaneously vertex- and edge-transitive.


I wonder what are the symmetric polytopes with maximal edge lengths (relative to their circumradius). I suspect that the longest edges are found for simplices, followed by [cross-polytopes](https://en.wikipedia.org/wiki/Cross-polytope), but I have no proof for this statement.


Especially, I wonder whether there are any such polytopes of edge length $\ge\sqrt{2}\times$circumradius besides simplex and cross-polytope.



Maximal edge length of symmetric polytopes
The vertices of a symmetric (i.e. vertex- and edge-transitive) polytope form a spherical code, where the edge-length defines the minimal distance between two points. In fact, the same can be said about any polytope for which all vertices are on a sphere and all edges have a common length.


In the book


* Conway, John Horton, and Neil James Alexander Sloane. *Sphere packings, lattices and groups*.


they introduce $A(n,\theta)$ to denotes the maximal number of points on an $(n-1)$-dimensional sphere (in $\Bbb R^n$) so that any two points have an angle of at least $\theta$ between them.
There is then stated that $A(n, \pi/2)=2n$ (the vertices of the crosspolytope), and $A(n,\theta)\le n+1$ for $\theta>\pi/2$ (the vertices of a simplex).


This seems to imply that indeed there are no other such polytopes with long edges.


