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Thread: Question about Archimedean solids

  1. Top | #11
    Administrator lpetrich's Avatar
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    For hexagons, the only non-hyperbolic solutions have r = 3 for all vertices, making a planar shape: -

    For different kinds of vertices, one has 121212, alternation between the two kinds.

    For higher n-gons, all solutions are hyperbolic.

    I have thus derived all 5 of the Platonic solids, all 13 of the Archimedean and Catalan solids, the 4 infinite families of axially-symmetric relatives, and also their plane-tiling counterparts.

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    Contributor repoman's Avatar
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    Wenzel Jamnitzer not only had a great name, he also made and designed some formidable geometric shapes for the 1560s.

    https://www.georgehart.com/virtual-p...jamnitzer.html

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    Contributor repoman's Avatar
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    Are there 4-D analogues to Archimedean solids?

    Can't find any, but I may not be looking with the correct terminology.

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    Quote Originally Posted by repoman View Post
    Are there 4-D analogues to Archimedean solids?

    Can't find any, but I may not be looking with the correct terminology.
    "Archimedean Polychora" are a subcategory of "Uniform Polychora" aka "Uniform 4-Polytopes". I believe there are 47 of them in 4-D, depending on how you count.
    Last edited by beero1000; 11-10-2019 at 05:47 PM.

  5. Top | #15
    Administrator lpetrich's Avatar
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    Where to look:
    Uniform 4-polytope
    The Uniform Polychora

    The classification of them is rather complicated, I must note.

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