If a polyhedron is made only of pentagons and hexagons, how many pentagons can it contain? With the assumption of three polygons per vertex, one can prove there are 12 pentagons.
Let's not make that assumption, and only use pentagons.
12 pentagons: dodecahedron and tetartoid.
24 pentagons: pentagonal icositetrahedron.
60 pentagons: pentagonal hexecontahedron.
72 pentagons: dual snub truncated octahedron.
132 pentagons: 132-pentagon polyhedron.
180 pentagons: dual snub truncated icosahedron.
Here's what the 132 looks like.
In that range of 12 to 180, what values are missing? For values missing here where an all-pentagon polyhedron exists, what is the most symmetrical polyhedron for that value?
Edit: According to Hasheminezhad, McKay, and Reeves, there are planar graphs that lead to 16, 18, 20, and 22 pentagonal faces, but I've never seen these polyhedra.
16 would be the dual of the snub square antiprism.
20 would be the dual of this graph:
2 Answers
$\begingroup$You can get $4k$ pentagonal faces for any $k≥3$. Take $k$ pentagons meeting at a point (in the plane, or on a sphere). Between each adjacent pair of pentagons, draw another pentagon, making a second ring of $k$ pentagons. Each of the original $k$ pentagons has a single exposed edge left; to this edge and the adjacent two edges of second-ring pentagons, add two more to form a new pentagon. This makes a third ring of $k$ pentagons. Joining the $k$ new vertices of this ring to a single point completes a fourth ring of $k$ pentagons.
Having drawn the planar graph, Steinitz's theorem says such a polyhedron exists.
The initial vertex, and the final vertex, have degree $k$; all other vertices have degree three. When $k=3$ this is a standard construction of the regular dodecahedron.
The polyhedron can be made with an axis of $k$-fold rotation through the two vertices of degree $k$; we also have reflections and a half-turn taking one such vertex to the other, for at least $4k$ symmetries.
On the other hand, clearly the number of faces must be even (since 5 times the number of faces is twice the number of edges). So, the remaining possibilities for the number of faces are equivalent to 2 mod 4.
$\endgroup$ 1 $\begingroup$There is an easy way to get a polyhedron with $10n+2$ pentagons only.
Start with a regular dodecahedron. Take a congruent dodecahedron and merge it face to face with the first one, removing the merged faces to get an increment of $10$ faces. Repeat as desired with additional regular dodecahedra.
Yes, it's ugly, in that we lack convexity and don't have elegant symmetry (for $n \ge 2). But it systematically makes infinitely many numbers of faces work, and the faces themselves are regular.
Addendum:
Following up on @Kundor's answer, we can interpret this in terms of graph theory. When we merge an additional dodecahedron into the figure, we are dividing one of the pentagonal faces of the graph into $11$ faces, such that the adjacent faces are undisturbed (they remain pentagonal). Such a division can be applied to any "base" ployhedron, so for example a base polyhedron with $16$ faces guarantees polyhedra with $10n+6$ faces, $n\ge 2$, as well.
Combining this result with @Kundor's implies that almost any even number of faces $\ge 12$ can be accessed. Only $14$ and $18$, which are not multiples of $4$ and too small to be reached via the $10$-face incrementation, require further analysis. For $18$ we have a planar graph corresponding to a polyhedron (basically a trigonal bipyramid where each face is divided in thirds) but for $14$ there are no planar graphs and thus no solutions!
So the possible numbers of faces turn out to be $12$ and all even numbers greater than or equal to $16$.
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