Appendix: Towards Goldberg-Coxeter operation on mixed polyhedra

The methods discussed up to now only handle polyhedra with a single type of face, either 3- or 4-sided. This section discusses the extension of the Goldberg-Coxeter operation to polyhedra with mixed triangle and quad faces. It’s possible to divide a polyhedron by applying a certain \(\Delta\) operator to the triangle faces, a certain \(\Box\) operator to the quadrilateral faces, and dealing with what happens over the edges of the original polyhedron.

Two operators are ‘’compatible’’ if:

  1. The same number of vertices lie along the base edge.
  2. The same number of edges cross the base edge, excluding ones that meet a vertex at the edge.
  3. The same number of edges lie on the base edge.
  4. Edges that cross the base edge do not create digons (faces with 2 edges).

For the operators \(\Delta(a,b)\) and \(\Box(c,d)\), the first 3 requirements can be distilled to these equations:

  • \(\gcd(a, b) = \gcd(c, d) = g\)
  • \(c + d + g = 2(a + b)\). This equation has no solution if c and d are both odd numbers.

The 4th requirement is more complicated to put in equational form.

Consequences of these rules include:

  1. \(\Box(1,0)\) is compatible with \(\Delta(1,0)\)
  2. Iff \(\Box(a,b)\) is compatible with \(\Delta(c,d)\), then \(\Box(na,nb)\) is compatible with \(\Delta(nc,nd)\) for all positive integers n.
  3. \(\Box(n,0)\) is compatible with \(\Delta(n,0)\) (Class I)
  4. Iff \(\Box(a,b)\) is compatible with \(\Delta(c,d)\), then \(\Box(b,a)\) is compatible with \(\Delta(d,c)\).
  5. \(\Delta(n,n)\) is compatible with \(\Box(n,2n)\) and \(\Box(2n,n)\) (Class II triangles)
  6. \(\Box(n,n)\) is not compatible with any \(\Delta\). (Class II squares)
  7. If \(\Box(a_1,b_1)\) is compatible with \(\Delta(c_1,d_1)\), and \(\Box(a_2,b_2)\) is compatible with \(\Delta(c_2,d_2)\), it does not follow that the composed subdivision \(\Box(a_1,b_1)\Box(a_2,b_2)\) is compatible with \(\Delta(c_1,d_1)\Delta(c_2,d_2)\). \(\Delta(1,1)\Delta(1,1) = \Delta(3,0)\); \(\Box(1,2)\Box(1,2) = \Box(4,3)\); \(\Box(1,2)\Box(2,1) = \Box(5,0)\).