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Let $P$ be a polyhedron which satisfies the following three conditions:

  1. $P$ is built out of regular hexagons and regular pentagons.
  2. Three faces meet at each vertex.
  3. $P$ is topologically a sphere.

An easy Euler characteristic argument tells you that $P$ has exactly twelve pentagonal faces.

An example of a polyhedron like this is a truncated icosahedron (soccer ball for those of us in the States, football for everyone else). In this case, the pentagonal faces are arranged with some nice symmetry, and the polyhedron has icosahedral symmetry.

Another (trivial) example is the regular dodecahedron, which again has icosahedral symmetry.

Here's my question: Is this symmetry forced? What, if anything, can be said in general about the symmetry of a polyhedron which satisfies the above three conditions?

Edit: Since the discussion below points out that there are precisely two polyhedra which satisfy the above conditions, a suitable evolution of the question, which has already begun to be discussed below, is this: What symmetry groups can a polyhedron have if one or more of the above conditions are relaxed?

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In case you're curious, the mathematical name for a soccer ball is a truncated icosahedron: en.wikipedia.org/wiki/Truncated_icosahedron –  Qiaochu Yuan Mar 29 '10 at 18:18
If you drop the condition of regularity, you can have other combinatorial types. For example, you can join 2 copies of 6 pentagons around a hexagon. See fullerenes for more examples. However, I would guess that these can't be made from regular polygons. I recommend using Polydron models to check. –  Douglas Zare Mar 29 '10 at 18:23
Indeed the soccer ball and the dodecahedron both have rotational symmetry group $A_5$: see en.wikipedia.org/wiki/Icosahedral_symmetry. Are you asking whether any other polyhedron meeting your requirements has symmetry group $A_5$, or nontrivial symmetry group, or what? "Very nice symmetry" is not very precise. –  Pete L. Clark Mar 29 '10 at 18:49
Since the question as stated turned out to have a not very interesting answer, would any of the geometers out there like to address what happens when the regularity conditions in 1) are dropped? –  Pete L. Clark Mar 29 '10 at 18:54
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8 Answers

up vote 12 down vote accepted

Only soccer ball or dodecahedron.

Clearly 3 hexagons can not meet at one vertex. Thus we have only 3 choices for one vertex:

  • 3 pentagons
  • 2 pentagons + 1 hexagon
  • 1 pentagons + 2 hexagon

Note that if $[pq]$ is an edge then $p$ has the same type as $q$ (the type is determined by angle at $[pq]$). Thus the polyhedron is completely determined by one vertex. Further:

  • Once you have a vertex of the first type you have a regular dodecahedron.
  • If you have a vertex of the second type then you will get one hexagon surrounded by pentagons. Then it is easy to see that you can not continue.
  • For the third type you will get a soccer ball or "truncated icosahedron" as some people call it :)
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That's assuming all vertices look the same, but it's also possible to have polyhedra in which all faces are hexagons or pentagons and all vertices have degree three but e.g. some vertices have three hexagons while others have one pentagon and two hexagons. –  David Eppstein Mar 29 '10 at 18:53
Anton, you are too brief. I'm afraid that those who can understand this proof do not need it. –  Sergei Ivanov Mar 29 '10 at 19:01
What is missing in Anton's answer is the fact that all vertices must have the same type. This can be proved as follows: let $[pq]$ be a common edge of faces $A$ and $B$. Since $A$ and $B$ are regular polygons, the angle between their remaining edges at $p$ is the same as at $q$. Hence the face meeting $A$ and $B$ at $p$ has the same angles as the one meeting $A$ and $B$ at $q$. Therefore $p$ and $q$ are of the same type. This is so for every pair of adjacent vertices and hence for all vertices. –  Sergei Ivanov Mar 29 '10 at 19:10
I added something, hope it is better now. –  Anton Petrunin Mar 29 '10 at 20:17
I do not see where the conditions (2) and (3) are used in the proof. If you relax the conditions of the problem, an infinite hexagonal lattice would be the third solution. Are there any other solutionos possible, if (2a) more than three polygons may meet at a surface, (3a) P is topologically a closed surface (4) P may self-intersect? –  Tomaž Pisanski Mar 29 '10 at 20:56
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If one gives up on regularity of the hexagonal and pentagonal faces then these graphs (usually called "Fullerene graphs") don't have to have much symmetry. See e.g. this 26-vertex graph with 12 pentagonal faces, 2 hexagonal faces, and only order-four symmetry.

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An easy not-so-symmetric Fullerene to imagine: fit six pentagons around a regular hexagon, and then fit two of these caps together to get what looks like a squashed dodecahedron. Though now that I think about it, I don't think the pentagons end up flat when you do this. –  Anton Geraschenko Mar 29 '10 at 19:14
That's the example I described in the comments on the question. To see that you can make the pentagons flat, start with a nonplanar 12-gon which projects to a regular 12-gon and has alternating vertices in parallel planes. Attach quadrilaterals to each 3 adjacent vertices so that the final vertex is on the line perpendicular to the projected 12-gon through its center. Six of these quadrilaterals meet above at P, and six meet below at Q. Then truncate P and Q. To make this easier to visualize, remember the 10-sided dice from Dungeons and Dragons, –  Douglas Zare Mar 29 '10 at 19:32
which are squashed versions of regular dodecahedra with pentagonal pyramids attached to two opposite faces. –  Douglas Zare Mar 29 '10 at 19:32
Anything that looks combinatorially like a polyhedron can be made with faces that are flat polygons: see Steinitz' theorem, en.wikipedia.org/wiki/Steinitz%27s_theorem –  David Eppstein Mar 29 '10 at 20:43
Thanks, I had forgotten that. –  Douglas Zare Mar 29 '10 at 20:51
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Perhaps you will also find the following interesting:

V. Braungardt und D. Kotschick: “The classification of football patterns”, 2006.

(for a popular Summary see: D. Kotschick: „The Topology and Combinatorics of Soccer Balls”, The American Scientist, Juli-August 2006)

ABSTRACT. We prove that every spherical football is a branched cover, branched only in the vertices, of the standard football made up of 12 pentagons and 20 hexagons. We also give examples showing that the corresponding result is not true for footballs of higher genera. Moreover, we classify the possible pairs (k; l) for which football patterns on the sphere exist satisfying a natural generalisation of the usual incidence relation between pentagons and hexagons to k-gons and l-gons.

Here a football (AE: soccer ball) pattern is a map on the two-sphere satisfying the following conditions:

  1. at least three edges meet at every vertex
  2. all faces are pentagons and hexagons
  3. the edges of each pentagon meet only edges of hexagons
  4. the edges of each hexagon alternately meet edges of pentagons and of hexagons
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Thanks for the link to the paper. This is a very interesting result. –  Bill Kronholm Mar 30 '10 at 19:43
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From a combinatorial point of view one can define a fullerene to be a 3-valent 3-connected graph with exactly 12 faces which are 5-gons (pentagons) and h 6-gons (hexagons). By Steinitz's Theorem fullerenes which exist as graphs can be realized by convex polyhedra. Branko Grunbaum and Theodore Motzkin showed, The number of hexagons and the simplicity of geodesics on certain polyhedra, Canadian J. Math., 15 (1963) 744-751, that the admissible values of h for such graphs are all non-negative integers h except h = 1. Other proofs of this, given by construction, show other features than what Grunbaum and Motzkin did. (For references see article listed below.)

What are the symmetry groups which can arise as the automorphism groups of fullerene graphs? There are only 28 such groups and they are listed on page 36 of the book: Geometry of Chemical Graphs: Polycylces and Two-Faced Maps, by Michel Deza, and Mathieu Dutour Sikiric, Cambridge U. Press, 2008. By a theorem of Peter Mani, these fullerene graphs can be realized by 3-dimensional polyhedra with the full automorphisms group of the graph as the group of isometries of the realizing polyhedron.

For further discussion of fullerenes and some open problems about fullerene graphs see:

Malkevitch, J., Geometrical and Combinatorial Questions about Fullerenes, in Discrete Mathematical Chemistry, (P. Hansen, P. Fowler, M. Zheng, eds.), Volume 51, DIMACS Series in Discrete Mathematics and Computer Science, AMS, Providence, 2000, pp. 261-266.

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If you are willing to relax the trivalency requirement (and not require convexity, which was not one of the stated constraints), you can make all sorts of polyhedra, using only regular pentagons and hexagons, which have all sorts of symmetry.

For example, start with two truncated icosahedra, and remove one hexagon from each of them. Now you can glue them together along the removed hexagons (matching up pentagonal and hexagonal faces), forming a new polyhedron with 3 and 4-valent vertices and considerably less symmetry than what you started with. Continuing, you could make long rod-shaped polyhedra, or you could make some spiky polyhedron by replacing multiple hexagons with other truncated icosahedra.

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These are cool molecules! –  Victor Protsak Jun 23 '10 at 8:23
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Lemma: In a polyhedron of this type with polygons $A$ and $B$ sharing an edge $e$, the two other polygons meeting $e$ must have the same number of sides.

Proof: By local symmetry reflecting through the perpendicular bisector of $e$, the angles are equal.

Sergei Ivanov proved the same lemma in the comments.

Since 5 is odd, all of the polygons around a pentagon must have the same number of sides, since you can't have a nonconstant alternating sequence. So, the only possibilities are that all polygons are pentagons, or that each pentagon is surrounded by hexagons, and each of these hexagons is surrounded by 3 pentagons and 3 hexagons. In the latter case, attaching pentagonal pyramids to each pentagon extends each hexagon into an equilateral triangle, producing a polyhedron whose faces are equilateral triangles with 5 meeting at a vertex, an icosahedron, so the original was a truncated icosahedron.

Note that if you have equilateral triangles and squares meeting 4 to a vertex, then there are two possibilities for 3 squares and 1 triangle at a vertex, one with cubic symmetry and one with only dihedral symmetry with a belt which is an octagonal prism.

By contrast, if you require that there are 3 congruent triangles meeting at a vertex, but drop the regularity assumption, you get a family of disphenoids, which generically have the Klein 4-group as symmetries and no reflective symmetry. These are related to ideal hyperbolic tetrahedra.

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A follow up to Tomaz Pisanski's question: What if we allow skew regular hexagons, i.e., non-planar regular hexagons? An example of this sort of thing is the Petrie polygon of the cube.

Note that we can't have non-planar skew regular pentagons because of parity.

(This should be a comment on the question at top, but the site rules don't let me do that yet.)

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An interesting example involving self-intersection, but the valence is 4 is given on page 158 of Wenninger's Polyhedron Models (Model #102), this is Great Dodecahemicosahedron, with 10 hexagons and 12 pentagons. You can check it out as well on wikipedia. A similar example is the small demidecahemicosahedron. This provides a partial answer to Pisanski's follow up question.

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