How many tetrahedra, octahedra and icosahedra meet at a given vertex of icosa tetrahedral tessellation?

I am not a professional christallographer so that my view of icosa tetrahedral tessellation (see this) is based on guess work guided by physical and biological intuition based on what I call icosatetrahedral model for the genetic code. One basic question is how many tetrahedra, icosahedra and octahedra emerge from a given vertex.

I realized that one can answer this question just from the knowledge of the solid angles associated with vertices of these Platonic solids. The solid angles are naturally defined as ratios of spherical areas to the radial distance squared and at the limit of very small hyperbolic radial distance approaching Euclidian distance, the total solid angle is 4π as in the Euclidian case. Hyperbolic 3-space H³ has 3-D rotation group as isometries so that the notion of Platonic solid applies also in H³.

The lines emanating from the vertex are shared by neighboring T, O, and I emanating from the vertex. Two neighboring lines are associated with a triangular face shared by two Platonic solids involved.

The basic condition for the numbers n(I) of the Platonic solids involved is ∑ _{i ∈{T,O,I}} n_iΩ_i =4π. One can find the general formulas for the solid angles from Wikipedia (see this).

1. Platonic solids are classified by 2 integers {q,p} stating that q p-polygons meet at a given vertex. In the recent case one has only 3-polygons, that is triangles, for all Platonic solids involved. One has

[q(T), q(O), q(I)] =[3,4,5] , [p(T),p(O), p(I)]=[3,3,3] .

Dihedral angle angle is the interior angle between the faces of the Platonic solid and satisfies the general formula

θ(q,p)= 2asin(cos(π/q)/sin(π/p)) .

The solid angle at the vertex is given as

Ω(q,p)= qθ(q,p)- (q-2)π .

Suppose that all vertices are identical as the fact that there is only single vertex figure, rhombicosadodecahedron (see this). The icosatetrahedral vertex involves n(T)==n(3) tetrahedrons, n(O)==n(4) octahedrons and n(I)==n(5) icosahedrons. The sum of the solid angles equals to 4π, which gives

∑_q∈{3,4,5} n(q)[qθ(q,3) - (q-2)π]= 4π .

This gives

∑_q∈{3,4,5} n(q) q asin(cos(π/q)/sin(π/p)- (n(3) +2n(4)+3n(5))π= 4π .

One can try to guess the solution to the condition by starting from the icosa tetrahedral model (see this and this) for the genetic code, which is a fusion of 3 icosahedral codes associated with Hamiltonian cycles with symmetry groups Z₆,Z₄,Z₂ and of a single tetrahedral code defined by the unique tetrahedral Hamiltonian cycle. In the proposed model based on icosatetrahedral tessellations (see this and this), the octahedron is passive and does not contribute to the code. A reasonable guess based on this model is n(I)=3 and n(T)=1.

The normalized vertex solid angles are

[Ω(3),Ω(4),Ω(5)]/4π =[0.043871, 0.1082, 0.2097].

The consistency condition is

[n(T)Ω(T) +n(O)Ω(O) +n(I)Ω(I)]/4π = 1 .

This leaves only the guess [n(T),n(O),n(I)]= [1,3,3] under consideration giving for the sum the value 0.9974 in the accuracy used partially determined by the approximation pisimeq 3.14159. It would seem that the heuristic guess makes sense. It is not yet clear whether this result is consistent with the rather heuristic model of the icosa tetrahedral tessellation discussed in (see this).

Source: https://matpitka.blogspot.com/2024/09/how-many-tetrahedra-octahedra-and.html