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A die with normal 4-sided polygon faces has 6 faces. A die with normal 5-sided polygon faces has 12 faces.

If a die has normal $n$-sided polygon faces, how many faces does it have? Is there a general formula for this, and is there a rigorous proof?

Intuitively, if $n$ is odd, then the number of faces is $2n+2$, but I can't think of a rigorous proof for this. The hard cases seem to be when $n$ is even.

dant
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    If you are restricting the die to (convex) regular polyhedra, there are exactly 5: a tetrahedron ($n = 3$, 4 faces), a cube ($n = 4$, 6 faces), an octahedron ($n = 3$, 8 faces), a dodecahedron ($n = 5$, 12 faces), and an icosahedron ($n = 3$, 20 faces). As you can see, there are 3 different cases with triangular faces ($n = 3$) which have a different number of faces, so there isn't a general formula. Edit: for reference, these are called the Platonic solids. – theyaoster Aug 17 '16 at 00:05
  • @BrianYao I didn't know that, thanks. Is there a geometric reason why there can't be, for example, a regular hexagon faced die?

    Conversely, are there any particular properties that $n=3,4,5$ satisfies? Edit: just saw your edit come through. Thanks for the tip.

     – dant Aug 17 '16 at 00:07
  • Yes, there is. Try to construct a vertex of this hexagon-faced die. The problem is, 3 hexagons can share a vertex perfectly on a flat plane, so there's no way to create a 3-dimensional object. On the other hand, 3 pentagons sharing a vertex, for example, occupy only $324^\circ$ out of $360^\circ$, so we have a $36^\circ$ "gap." When we close the gap by bringing the pentagons together, the vertex will be pushed outward, creating a convex shape. This is what allows us to have a convex polyhedron. – theyaoster Aug 17 '16 at 00:16
  • That makes a lot of sense, especially when considering polygons of even more sides. Thanks! – dant Aug 17 '16 at 00:21
  • Yes; when you attempt to do this with three regular polygons with vertex larger than a hexagon (eg three heptagons), their isn't a gap, but an overlap. – Graham Kemp Aug 17 '16 at 00:21
  • Ah, I should have touched on that. It is because in a cube, each vertex is shared by 3 faces, not 4 (which would be planar). Something similar occurs with the triangle-faced platonic solids: if each vertex is shared by 3 triangles, you get a tetrahedron. If each vertex is shared by 4 triangles, you get an octahedron. Shared by 5 triangles? Icosahedron. Shared by 6 triangles? Oops, no more "gap," and we're back to a plane. – theyaoster Aug 17 '16 at 00:40

1 Answers1

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Most relevant aspects have already been discussed in comments.

The most typical generalization of a “die” to other faces would be a platonic solid. There are exactly five platonic solids, including the dodecahedron with its pentagonal faces which you already mentioned in the question. Three of these have the same face shape, namely the tetrahedron, the octahedron and the icosahedron. They all have triangular faces, but four, eight or twenty faces respectively.

So at this point you can already see that there can be no formula to deduce the number of faces from the number of corners per face alone.

In my opinion the most useful formula you can get for computing the number of faces is Euler's polyhedron formula which states that for each convex polyhedron you get $$V-E+F=2$$ where $V$ is the number of vertices, $E$ the number of edges and $F$ the number of faces.

In a comment you asked why there are only five platonic solids, but that is probably better discussed in the question Platonic Solids. Two of the current answers there make use of Euler's polyhedron formula, too.

One could generalize the question further, e.g. by dropping the convexity constraint (and therefore considering Kepler–Poinsot polyhedra as well) or by dropping one of the transitivity requirements of regular polyhedra. For example, the ten-sided die is convex and face-transitive but not edge- or vertex-transitive, and its faces are not regular polygons. So it all depends on what you call a “die”.

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