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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.

  2. Top | #12
    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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