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, 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.


2 Answers 2


Just a partial answer.

Within what you define to be "symmetrical polytopes" clearly is the subset of "quasiregular polytopes", i.e. those which can be described by a Coxeter-Dynkin diagram (i.e. follow some kaleidoscopical Wythoff construction), where just a single node is being ringed. In fact those are uniform by means of their Wythoff construction, and by virtue of having just a single node being ringed all edges would belong to the same equivalence class.

This moreover excludes reducible Coxeter groups, as those would require for more than one edge type. Obviously the ratio of circumradius to edge size gets increased, whenever any link mark of the diagram gets larger than 3. Therefore you are restricted here to links marked 3 throughout. Similarily that ratio increases, whenever the ringed node gets away from any of its ends (of a possibly bifurcated) diagram.

Running through the possible cases here, and considering the dimensional behaviour, as well as those exceptional cases of the Gosset figures, we have to state that your claim is completely right in the realm of quasiregular polytopes.

Whether you would be able to find symmetrical polytopes outside the quasiregular ones (still asking for convexity for sure) and whether there would be any counterexamples, I don't know.

--- rk

  • $\begingroup$ How can I quickly see that all edges are equivalent when only a single node is ringed in the Coxeter-Dynkin diagram? $\endgroup$
    – M. Winter
    Jul 5, 2019 at 15:47
  • $\begingroup$ Each ringed node effectively is the symmetry equivalence class of the connection between a pair points across a single, specified mirror type. Thus each ringed node is exactly one edge class. In fact it is even that you can size those differently. Thus a x3y3o is a truncated tetrahedron with y-sized triangles and alternatingly x and y sized semiregular hexagons. $\endgroup$ Jul 7, 2019 at 14:59

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.


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