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Miles
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12 pentagons
« on: Aug 24th, 2006, 2:38pm » |
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The faces of a polyhedron are regular pentagons and regular hexagons (think of a soccer ball as real-life example). Prove that there must be 12 pentagons.
Supplementary question: where is the regularity of the faces required in the proof?
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Icarus
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Re: 12 pentagons
« Reply #1 on: Aug 24th, 2006, 3:30pm » |
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on Aug 24th, 2006, 2:38pm, Miles wrote:| Supplementary question: where is the regularity of the faces required in the proof? |
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It isn't. Regular or irregular, if all faces have 5 or 6 sides, then there must be exactly 12 faces with 5 sides.
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Eigenray
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Re: 12 pentagons
« Reply #2 on: Aug 24th, 2006, 7:24pm » |
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If you're allowed to subdivide faces, you could have any even number of (non-regular) pentagons.
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towr
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Re: 12 pentagons
« Reply #3 on: Aug 25th, 2006, 12:28am » |
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on Aug 24th, 2006, 3:30pm, Icarus wrote:| It isn't. Regular or irregular, if all faces have 5 or 6 sides, then there must be exactly 12 faces with 5 sides. |
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Only if the pentagons aren't triangles or squares.. (i.e. with one or two pairs of adjacent sides having 180 degrees angle betwene them)
A tetrahedron has only 4 irregular pentagons 
(or three pentagons and a triangle, if you don't want one side adjoined to two [e](of course, a triangle isn't a hexagon, unless three of it's side get length 0; but there's always the hexa or octahedron)[/e])
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| « Last Edit: Aug 25th, 2006, 2:40am by towr » |
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Miles
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Re: 12 pentagons
« Reply #4 on: Aug 25th, 2006, 1:32am » |
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Not sure what you mean by sub-dividing, but I didn't have that in mind.
The fairly simple proof that I have uses regularity to make one step straightforward. Of course, if you show us your proof, Icarus (or towr), we can see for ourselves how you can do it without assuming regularity - excluding the "sides meeting at 180 degrees" case you have already described. (Not saying it can't be proved for irregularity, just haven't come up with a proof myself).
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| « Last Edit: Aug 25th, 2006, 1:33am by Miles » |
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Barukh
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Re: 12 pentagons
« Reply #5 on: Aug 25th, 2006, 2:03am » |
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Miles, try to use Euler's polyhedral formula.
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towr
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Re: 12 pentagons
« Reply #6 on: Aug 25th, 2006, 2:43am » |
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on Aug 25th, 2006, 2:03am, Barukh wrote:| Miles, try to use Euler's polyhedral formula. |
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That doesn't necessarily work for non-convex polyhedra
(Of course, not knowing Miles' proof I don't know whethe rhis does. Nor in fact, if the proposition holds at all in that case)
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Miles
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Re: 12 pentagons
« Reply #7 on: Aug 25th, 2006, 3:37am » |
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I avoided being completely precise about the polyhedron, but I was thinking of a "normal" convex polyhedron sufficient for the theorem hinted by Barukh to hold (which I indeed used in my proof). I expected others would consider more exotic cases if they wished (and you haven't let me down in that expectation).
In my proof, I need to know that only 3 faces meet at each vertex. Making the pentagons and hexagons regular ensures that. I am sure there are other routes that can relax regularity, but I'm still waiting to hear how you do it...
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| « Last Edit: Aug 25th, 2006, 3:38am by Miles » |
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towr
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Re: 12 pentagons
« Reply #8 on: Aug 25th, 2006, 3:51am » |
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If the pentagons don't need to be regular, or convex, you can replace one regular pentagon by 5 non-regular non-convex ones. Fairly trivial even..
So in that case you can have more than 12 pentagons.
And if all pentagons need to be convex, you can still replace a hexagon with 10 convex pentagons and a hexagon.
So I'd say that either they have to be regular, or perhaps at least all the same shape and size. Or some other condition.
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| « Last Edit: Aug 25th, 2006, 4:02am by towr » |
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Barukh
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Re: 12 pentagons
« Reply #9 on: Aug 25th, 2006, 4:38am » |
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Good point, Miles. I think you are right.
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Grimbal
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Re: 12 pentagons
« Reply #10 on: Aug 25th, 2006, 6:00am » |
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If the polygons are really regular there are only 2 polyhedra that can be constructed with it. The one is a football ball the other is a dodecahedron. Indeed, both have 12 pentagons.
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Eigenray
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Re: 12 pentagons
« Reply #11 on: Aug 25th, 2006, 6:30am » |
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If the faces are all regular, each interior angle is either 108 or 120 degrees, meaning [edit]-- if the polyhedron is convex --[/edit] no more than three can meet at a point. And you can't have exactly two faces meeting at a point, unless you want to get silly about what you call a polyhedron. So 3 faces meet at each point, hence 3V = 2E. If there are p pentagons and h hexagons, we have V-E = -1/3 E = -1/6 (5p+6h), and F = p+h, so
V - E + F = chi
gives p = 6*chi. Since p>0 (you can't do much with just hexagons except tile the plane), chi>0. Since the polyhedron is "obviously" a CW complex for an orientable compact surface, chi = 2(1-g) > 0 means g=0 [edit]Since the polyhedron is convex, chi=2[/edit], and p=12.
If the faces don't have to be regular, and your edges/faces don't have to be connected components of intersections with lines/planes, you can start with a cube, and add 6 points, on 3 pairs of opposite edges, to get 6 hexagons. Then a hexagon can be subdivided into 2 pentagons, and a pentagon can be subdivided into a pentagon and a hexagon. By repeating these, you can get any nonnegative even number of pentagons.
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| « Last Edit: Aug 25th, 2006, 3:23pm by Eigenray » |
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towr
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Re: 12 pentagons
« Reply #12 on: Aug 25th, 2006, 12:54pm » |
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on Aug 25th, 2006, 6:00am, Grimbal wrote:| If the polygons are really regular there are only 2 polyhedra that can be constructed with it. |
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There's only two convex ones. There's infinitely many non-convex ones. Just stick any two together at a similar face, and the surface forms a new polyhedron.
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towr
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Re: 12 pentagons
« Reply #13 on: Aug 25th, 2006, 12:57pm » |
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on Aug 25th, 2006, 6:30am, Eigenray wrote:| By repeating these, you can get any nonnegative even number of pentagons. |
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You can also replace a pentagon by six pentagons, so you can get most odd numbers of pentagons as well.
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Eigenray
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Re: 12 pentagons
« Reply #14 on: Aug 25th, 2006, 3:14pm » |
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on Aug 25th, 2006, 12:57pm, towr wrote:
You can also replace a pentagon by six pentagons, so you can get most odd numbers of pentagons as well. |
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And have each edge incident to an even number of faces?
on Aug 25th, 2006, 12:54pm, towr wrote:
There's only two convex ones. There's infinitely many non-convex ones. Just stick any two together at a similar face, and the surface forms a new polyhedron. |
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Excellent point! Now excuse me while I go edit my previous post to assume the obviously necessary assumption of convexity...
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| « Last Edit: Aug 25th, 2006, 3:23pm by Eigenray » |
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Icarus
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Re: 12 pentagons
« Reply #15 on: Aug 25th, 2006, 6:52pm » |
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I am assuming that the polyhedron is spherical (i.e., it is homeomorphic to a sphere). Convexity is not required. I am also assuming that all vertices are non-degenerate (have at least 3 edges). Then you can get at least 12 pentagons for polyhedra with 5 or more sides to every face.
Let E be the number of edges, V the number of vertices, F the number of faces, and for all n, let F(n) be the number of faces with n sides.
Since each vertex has 3 or more edges, 2E >= 3V. Also, 2E = (5F(5) + 6F(6) + ...). Euler's formula states that F+V = E + 2.
F + (2/3)E >= E + 2
6F >= 2E + 12
6F(5) + 6F(6) + ... >= 5F(5) + 6F(6) + ... + 12
F(5) >= F(7) + 2F(8) + ... + 12.
Hence F(5) >= 12.
Of course, when each vertex has exactly 3 edges, you can replace the >= with equality in the calculation, and get F(5) = 12 when only polygons and hexagons are present.
It is interesting that there is no limitation on the number of hexagons in this formula, or its extension to cases with smaller numbers of sides per face. Spherical trivalent polyhedra cannot have more than 12 faces unless they have a face of size 6 or larger. They cannot have more than 12 faces of size 5 or less unless they have faces of size 7 or greater.
Note also that the limitations still hold if you allow non-planar faces and non-linear edges - as long as the resulting "polyhedra" is still homeomorphic to a sphere, so that Euler's formula applies.
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towr
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Re: 12 pentagons
« Reply #16 on: Aug 26th, 2006, 2:19pm » |
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on Aug 25th, 2006, 6:52pm, Icarus wrote:| I am assuming that the polyhedron is spherical (i.e., it is homeomorphic to a sphere). |
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Could you elaborate on what exactly 'homeomorphic to a sphere' means?
(Well, maybe not ''exactly', but rather in terms I can understand which might be slightly less precise ;))
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Icarus
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Re: 12 pentagons
« Reply #17 on: Aug 27th, 2006, 2:45pm » |
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Literally, it means that there is a one-to-one correspondence between points of the polyhedron and the sphere which is continuous in both directions.
Less literally (much less), it means that the polyhedron can be stretched and pulled into a spherical shape.
A polyhedron that is orientable (required if it exists in 3D space) and has no "holes" like a torus has, is going to be homeomorphic to a sphere.
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