Neighboring solids in tetrahedral-octahedral honeycomb

Question

In the tetrahedral-octahedral honeycomb, each vertex seems to be incident to 6 octahedra and 8 tetrahedra:

Such simple combinatorial fact is probably well-known, or perhaps even obvious. However, coming from a different field, I struggle to justify it mathematically. I thought perhaps one can read this from the Schläfli symbols notation or the Coxeter diagram, but did not succeed at that.

How would one go at arriving at this result if one does not want to use the rigorous method of counting colorful solids in a Wikipedia picture?

Answer 1

You can check that solid angles of those polyhedra add up to $4\pi$. We have (see here for details):

$$ \Omega_{tetr}=2\pi-6\arcsin\sqrt{2\over3}, \quad \Omega_{oct}=2\pi-8\arcsin\sqrt{1\over3}. $$ Hence, when 6 octahedra and 8 tetrahedra meet at a vertex, they cover a solid angle given by: $$ \Omega_{tot}=8\Omega_{tetr}+6\Omega_{oct}= 28\pi-48\left(\arcsin\sqrt{2\over3}+\arcsin\sqrt{1\over3}\right). $$ But the angles between parentheses are complementary (the squares of their sines add up to $1$), hence: $$ \Omega_{tot}=28\pi-48{\pi\over2}=4\pi. $$

Answer 2

From the Wikipedia page referred to in the question:

For an alternated cubic honeycomb, with edges parallel to the axes and with an edge length of $1$, the Cartesian coordinates of the vertices are: (For all integral values: $i,j,k$ with $i+j+k$ even)

$(i, j, k)$

In this construction of a tetrahedral-octahedral honeycomb, vertex $(i, j, k)$ is incident on $12$ edges, given by the $12$ vectors $(0, \pm1,\pm1)$, $(\pm1, 0,\pm1)$, $(\pm1, \pm1, 0)$, lying in the $3$ rectangular Cartesian coordinate planes meeting at $(i, j, k)$:

The $12$ vertices adjacent to $(i, j, k)$ are marked here in blue. The $6$ rectangular Cartesian coordinate semi-axes passing through $(i, j, k)$ are terminated by red dots, marking the centres of the $6$ octahedra incident on $(i, j, k)$.

Vertex $(i, j, k)$ is incident on $8$ cubes, of side length $1$, belonging to the underlying cubic honeycomb. In each of these cubes, $3$ edges of the tetrahedral-octahedral honeycomb extend diagonally across the $3$ faces of the cube meeting at $(i, j, k)$. The far ends of these $3$ diagonals, together with $(i, j, k)$ itself, constitute the vertices of one of the $8$ tetrahedra incident on $(i, j, k)$.

In the next diagram, I have shaded in the triangular faces of these tetrahedra opposite to vertex $(i, j, k)$:

In the next picture (taken from a different point of view), I have instead shaded in the square cross-sectional slices of the $6$ octahedra incident on $(i, j, k)$ (centred on the $6$ red dots in the first picture):

As can be seen from the last image, the $12$ vertices of the tetrahedral-octahedral honeycomb adjacent to vertex $(i, j, k)$ are the vertices of a cuboctahedron. For more information about this Archimedean solid, and for images of higher quality, see for example The cuboctahedron | Hexnet, or Cuboctahedron - Wikipedia.

From the Wikipedia page just referred to:

The Cartesian coordinates for the vertices of a cuboctahedron (of edge length $\sqrt{2}$) centered at the origin are: $$ \begin{array}{c} (\pm1,\pm1,0) \\ (\pm1,0,\pm1) \\ (0,\pm1,\pm1) \end{array} $$

which at least seems to confirm that I haven't misunderstood the construction.

Answer 3

The Coxeter-Dynkin diagram of the tetrahedron-octahedron-honeycomb is a tridental graph, which can be given in typewriter friendly inline notation as

x3o3o *b4o

Here the "*b" part means a virtual revisiting of the b-th (real) node (i.e. the middle one of the left part, which thus becomes the bifurcation point of the graph, as the right part has to be reattached thereon). The "x" here represents a ringed node, while "o" represents an not ringed node of the diagram.

Using this representation, it becomes very easy to provide the total incidence matrix of this honeycomb

x3o3o *b4o   (N → ∞)  

. . .    . | N | 12 | 24 |  8 6  
-----------+---+----+----+-----  
x . .    . | 2 | 6N |  4 |  2 2  
-----------+---+----+----+-----  
x3o .    . | 3 |  3 | 8N |  1 1  
-----------+---+----+----+-----  
x3o3o    . | 4 |  6 |  4 | 2N *  
x3o . *b4o | 6 | 12 |  8 |  * N  

Thus you see that the vertex figure (the super-diagonal part of the first row) is represented by a cuboctahedron, and that the 2 contained cell types are tetrahedra (sub-diagonal part of one-but-last row) and octahedra (sub-diagonal part of last row).

Cf. https://bendwavy.org/klitzing/incmats/octet.htm (and further explanations provided at this website).

--- rk

PS: you even can derive the vertex figure, you've asked for, from that given Dynkin diagram x3o3o *4o as being . x3o *b4o (= o3x4o) i.e. a cuboctahedron. Thereof the 8 triangles . x3o   . are the bases of the vertex emanating pyramids (tetrahedra) x3o3o   . and the 6 squares . x . *b4o are the equatorial facets of the vertex emanating octahedra x3o . *b4o.

--- rk