The diameter of the circum-circle than has d² = 4/(4-a²) or 4=v²+4/d² = 4/d² = 4-v²

The vertex-figure of a regular polyhedron {P,Q} has v² = 4p²/(4-q²), which is a pQo (ie a Q-gon, of side p.

We find its corresponding diameter2 at 4/d² = 4 - 4p²/(4-q²), or multoplying through by 4-q²/4, 4/d² = ((4-q²) - p² )/(4-q²) -> d² = 4(4-q²)/(4-p²-q²).

If v² represents the vertex-figure, and w² represents the edge-figure, that is vertex-of-vertex, then we get the vertex-figure becomes p²v², and the new vertex-figure denominator is 2v²-p²w², where the numerator is 2q²v²

------------------------------------------------------------------------------------------------------

we use v² = 1, 2, Ø² amd 3 for {3.4,5,6} respectively.

- Code: Select all

(1) 1 always

(2) 2 always

(3) 3-b 4-a² = 2(2) - a²(1)

(4) 4-2b 2(3-b)-1(2) 1 = sc(3)

(5) 5-3b 2(4-2b)-1(3-b) 1 = sc(3)

Here we find the Schlafli function for {p,3,3,3...}, where a²=1+b, and b is the third chord (first parallel to the edge).

When P=2, 3, 4, 5, this gives (for 2d+), 2n-2, n, 4, and 2-(n-2)/Ø. In the hexagon, b=2, so the 3d polytope at (4-2b) gives 0.

The diameter² of any regular polytope is then 2sch(v)/sch(f), where v is the vertex-figure symmetry and f is the figure symmetry. It does not even touch the polytope.

The value of sch(ab) = sch(a)sch(b), so this allows us to find the diameter of any marked node figure, and we can deal with branching groups.