On existence of polyhedra with a fixed number of edges per face

Question

Let $P$ be a convex polyhedra such that each face has exactly $A$ edges. Denote with $V$, $E$ and $F$ the number of vertices, edges and faces of $P$ respectively. Since each face has $A$ edges we get that $E=\frac{AF}{2}$. This means that $A$ and $F$ cannot be both odd. The graph $G$ induced by $P$ is a planar graph such that each vertex has at least degree $3$. This means we can use Euler's formula to get the amount of vertices: $$V=2+E-F=\frac{4 + AF -2F}{2}.$$ Question

• For which values of $A$ and $F$ does such a convex polyhedron exist?
• If such a polyhedron exists, how many such non-isomorphic polyhedra exist? By isomorphism I mean that their respective skeleton graphs are isomorphic as graphs.

Partial results

The average degree $d$ of $G$ is equal to $d=\frac{2E}{V}=\frac{2AF}{4+(A-2)F}$. Since each vertex has at least degree $3$ we get that $3 \leq d$. Since $G$ is planar we also have that $d < 6$. This two inequalities give restrictions to the possibilities of $A$ and $F$: $3 \leq d$ gives that $$A \leq \frac{6F-12}{F}.$$ The inequality $d < 6$ gives that $$\frac{3F-6}{F} < A.$$ This gives us that $A < 6$. We also have that $3 \leq A$ since a face of a polyhedron has at least $3$ edges. The above inequalities and the fact that $F \geq 0$ give

• If $A=3$, then $F \in [4, \infty)$
• If $A=4$, then $F \in [6, \infty)$
• If $A=5$, then $F \in [12, \infty)$

Denote with $\mathcal{E}$ the set of all even numbers. Since $A$ and $F$ cannot be both odd we have that

• If $A=3$, then $F \in \mathcal{E} \cap [4, \infty) $
• If $A=4$, then $F \in [6, \infty)$
• If $A=5$, then $F \in \mathcal{E} \cap [12, \infty)$

Answer 1

These are the duals of the convex polyhedra where each vertex is $A$-valent, which you might have an easier time finding information on. The graphs of these dual polyhedra are all the $A$-regular 3-connected planar graphs.

Case $A = 3.$

You want polyhedra with all faces being triangles, which are the simplicial polyhedra. Wikipedia has a list of several classes of these. They exist for any even number of faces greater than two. One example with $2n$ faces for any $n > 2$ is the $n$-gonal bipyramid.

The graphs of the duals are 3-regular (aka cubic) 3-connected planar graphs. The number of non-isomorphic such graphs with 10 to 20 vertices can be found at Gordon Royle's page, or as OEIS A000109.

\begin{array}{| c | c |} \hline F & \text{#} \\ \hline 4 & 1 \\ 6 & 1 \\ 8 & 2 \\ 10 & 5 \\ 12 & 14 \\ 14 & 50 \\ 16 & 233 \\ 18 & 1249 \\ 20 & 7595 \\ 22 & 49566 \\ 24 & 339722 \\ 26 & 2406841 \\ 28 & 17490241 \\ 30 & 129664753 \\ \hline \end{array}

As you can see, there are a lot.

Case $A = 4.$

You want polyhedra with all faces being quadrilaterals. (These are the 3-dimensional cubical polytopes.) Such polyhedra exist with $2n$ faces for any $n \geq 3$: the trapezohedra, aka deltohedra, which are duals to antiprisms.

Quadrilateral-faced polyhedra also exist with $3n$ faces for any $n \geq 3$, by taking an $n$-gonal bipyramid and truncating each vertex on the equatorial $n$-gon to the midpoints of the equatorial edges (making a ring of rhombi around the equator), as described in this answer of achille hui for $n = 5$. When $n=3$ you get the Herschel enneahedron:

There are no quadrilateral-faced heptahedra.

Given a quadrilateral-faced polyhedron, we can "glue" an irregular cube to one of its faces to get a quadrilateral-faced polyhedron with four more faces. Starting from the Herschel enneahedron, you can thus get $F$ faces for any $F \geq 9$ equivalent to 1 mod 4.

The program plantri will generate all non-isomorphic 3-connected planar quadrangulations, which are equivalent to quadrilateral-faced convex polyhedra, and finds:

\begin{array}{| c | c |} \hline F & \text{#} \\ \hline 6 & 1 \\ 7 & 0 \\ 8 & 1 \\ 9 & 1 \\ 10 & 3 \\ 11 & 3 \\ 12 & 11 \\ 13 & 18 \\ 14 & 58 \\ 15 & 139 \\ 16 & 451 \\ 17 & 1326 \\ 18 & 4461 \\ 19 & 14554 \\ 20 & 49957 \\ 21 & 171159 \\ 22 & 598102 \\ 23 & 2098675 \\ 24 & 7437910 \\ 25 & 26490072 \\ 26 & 94944685 \\ 27 & 341867921 \\ 28 & 1236864842 \\ 29 & 4493270976 \\ 30 & 16387852863 \\ 31 & 59985464681 \\ \hline \end{array}

(OEIS A007022).

Since there are quadrilateral-faced 11-hedra, we can do the same cube-gluing operation to get any $F \geq 11$ equivalent to 3 mod 4.

Thus, for $A=4$ we can have 6 faces, or any number greater than 7.

Case $A = 5.$

There are again lots of examples. Suppose that the polyhedron has $v_k$ vertices of valence $k$. You can specify the number of vertices of each valence, $(v_3, v_4, v_5, \dotsc)$, and such a polyhedron exists with all pentagonal faces so long as $v_4 \geq 6$ and $v_3 = 20 + \sum_{k \geq 4} (3k - 10) v_k$, by a result of J.C. Fisher in Five-valent convex polyhedra with prescribed faces. So the total number of vertices is $$ V = 20 + \sum_{k \geq 4} (3k - 9) v_k, $$ and since $V = \frac{3F}{2} + 2$, the total number of faces is $$ F = 12 + 2\sum_{k \geq 4}(k-3)v_k. $$ By setting $v_4$ to any integer at least 6, and all other $v_k = 0$, we can guarantee the existence of pentagonal-faced polyhedra with any even number of faces greater than or equal to 24.

There are also pentagonal-faced polyhedra with 12 faces (the dodecahedron), 16 faces (the dual of the snub square antiprism), 18 or 20 faces (the polyhedra with planar graphs shown below), and 22 faces (the result of gluing two regular dodecahedra together along a face, as described in this answer of Oscar Lanzi.) (The 20-faced pentagonal planar graph is the dual of the so-called 20-quintic graph 1.)

The same answer of Oscar Lanzi, in the question Possible all-Pentagon Polyhedra, asserts that there are no pentagon-faced polyhedra with 14 faces. The Fisher article cited above provides @Oscar's reference for the nonexistence of a pentagonal 14-hedron.

Using plantri and countg (from the nauty gtools), we can count the number of 5-regular 3-connected planar graphs, to enumerate the number of isomorphism classes:

\begin{array}{| c | c |} \hline F & \text{#} \\ \hline 12 & 1 \\ 14 & 0 \\ 16 & 1 \\ 18 & 1 \\ 20 & 6 \\ 22 & 14 \\ 24 & 96 \\ 26 & 518 \\ 28 & 3917 \\ 30 & 29821 \\ \hline \end{array}