# Difference between revisions of "Talk:Polyhedron"

SeanTheSheep (Talk | contribs) |
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Yeah go for it. I was thinking about extending these into the semiregular polyhedra and demiregular to match the stuff I did on tessellations. There are also the truncations and stellations. There is a huge article here, and there is a load of stuff which can be imported. I have to go now, but I'll look in later --[[User:SeanTheSheep|SeanTheSheep]] 11:39, 15 May 2007 (EDT) | Yeah go for it. I was thinking about extending these into the semiregular polyhedra and demiregular to match the stuff I did on tessellations. There are also the truncations and stellations. There is a huge article here, and there is a load of stuff which can be imported. I have to go now, but I'll look in later --[[User:SeanTheSheep|SeanTheSheep]] 11:39, 15 May 2007 (EDT) | ||

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+ | You don't need complex math: | ||

+ | You need at least 3 polygons to meet at a vertex. | ||

+ | |||

+ | So, | ||

+ | if we use triangles, we could have 3, 4 or 5 (6 or more is more than 180 degrees) | ||

+ | if we use squares, we could have 3 (4 or more is more than 180 degrees) | ||

+ | if we use pentagons, we could have 3 (4 or more is more than 180 degrees) | ||

+ | |||

+ | if we have higher order, they won't fit. | ||

+ | |||

+ | So, all you have to show is that under these constraints there are only these 5. | ||

+ | |||

+ | --[[User:SeanTheSheep|SeanTheSheep]] 11:46, 15 May 2007 (EDT) |

## Latest revision as of 10:46, 15 May 2007

Maybe we'd like something on the duality of the regular polyhedra, and something on why there are only five? I'm not an algebraist by training, but I could probably take a shot at it. Brainslug 11:36, 15 May 2007 (EDT)

Yeah go for it. I was thinking about extending these into the semiregular polyhedra and demiregular to match the stuff I did on tessellations. There are also the truncations and stellations. There is a huge article here, and there is a load of stuff which can be imported. I have to go now, but I'll look in later --SeanTheSheep 11:39, 15 May 2007 (EDT)

You don't need complex math:
You need at least 3 polygons to meet at a vertex.

So, if we use triangles, we could have 3, 4 or 5 (6 or more is more than 180 degrees) if we use squares, we could have 3 (4 or more is more than 180 degrees) if we use pentagons, we could have 3 (4 or more is more than 180 degrees)

if we have higher order, they won't fit.

So, all you have to show is that under these constraints there are only these 5.

--SeanTheSheep 11:46, 15 May 2007 (EDT)