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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 lenght of a side. But, on the other hand, it seems that no regular polihedron (3-polytope) fits this condition.

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  • $\begingroup$ I don't see why you put that "Noob" in the title. This is a good question! $\endgroup$ – Janik Dec 24 '16 at 11:36
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    $\begingroup$ Since there are a finite number of "exceptional" polytopes in dimensions 3 and 4 and then only the standard 3 (simplex, hypercube, and cross-polytope) this should be easily answerable with a little conputation. $\endgroup$ – Steven Stadnicki Dec 24 '16 at 11:46
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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 $(1,0,1,0)$.)

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    $\begingroup$ What about the 24-cell? $\endgroup$ – pregunton Dec 24 '16 at 12:16
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    $\begingroup$ @pregunton I wasn't sure about that case (and didn"t have a good ref handy for coordinates), but fortunately OP asked about dimensions and not specific polytopes, so the 4-cube provides a positive answer there regardless. $\endgroup$ – Steven Stadnicki Dec 24 '16 at 14:56
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You might be interested to know, that, when loosening the condition of being a regular polytope, then not only sporadically some oddballs do fit your property of having circumradius = edge length, but rather there is a series of Wythoffian polytopes, which throughout any dimension would have this property!

In fact that would be the expanded simplex with Dynkin diagram $x3o3o...o3x$ (where I've used typewriter friendly replacements, $x$ meaning a ringed node, $o$ represents an unringed node).

  • $x3x$ is the hexagon
  • $x3o3x$ is the cuboctahedron
  • $x3o3o3x$ is the small prismated decachoron (aka: runcinated pentachoron)
  • $x3o3o3o3x$ is the small cellated dodecateron (aka: stericated hexateron)
  • etc.

Indeed all these have the common property that its equatorial section happens to be nothing but the former item of this list. In fact, in this very orientation they all can be described layerwise as a simplex atop the expanded simplex atop the dual simplex:

$$x3o3o3o...o3x = \text{hull}( x3o3o...o3o\ ||\ x3o3o...o3x\ ||\ o3o3o...o3x )$$

--- rk

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    $\begingroup$ Check me on this: these can also be described abstractly as the convex hull of the root vectors of the $A_n$ lattice, yes? $\endgroup$ – Steven Stadnicki Dec 31 '19 at 20:24
  • $\begingroup$ yes, indeed, they can. $\endgroup$ – Dr. Richard Klitzing Jan 1 at 10:41
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Asking for regular polytopes is quite restrictive, as there aren't too many in higher dimensions (only three in any dimension $d\ge 5$).

As mentioned by Dr. Klitzing, the uniform polytopes aren't too far from the regular ones, but provide a much richer class. In particular, a uniform polytope has all edges of the same length, and so the question makes sense to be asked for these. I want to complement Dr. Klitzing's answer with some actual numbers.

His $d$-dimensional example is a certain non-regular modification of the $d$-simplex (which is regular). Its vertices are the coordinate permutations of the following vector:

$$(-1,\underbrace{0,...,0}_{d-1},1).$$

Clearly, each of these has distance $\sqrt 2$ from the origin. The shortest possible distance between any two of these vertices is also $\sqrt 2$, e.g.

$$\|(-1,\underbrace{0,...,0}_{d-2},0,1) - (-1,\underbrace{0,...,0}_{d-2},1,0)\|=\|(0,\underbrace{0,...,0}_{d-2},-1,1)\|=\sqrt 2.$$

It is not hard to see that any two such vertices of shortest distance form an edge. Hence edge length and circumradius are the same.


The key property exploited here is that the vertex set $V$ of the polytope has the following special property: if $v,w\in V$ are vertices of shortest distance, then $v-w\in V$. In particular, this is true if $V$ is chosen to be the vectors of the $A_d$ root system as stated by Steven in the comments. If $V=A_d$, we obtain the polytopes described above.

The same holds for other root systems, which give further exceptional polytopes with this property, as e.g. the 24-cell (regular) in 4-dimensional space, and the $4_{21}$-polytope (not regular, also called $E_8$-polytope) in 8-dimensional space.

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