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Math Help - Polyhedron made from 12 pentagons and 60 hexagons

  1. #1
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    Polyhedron made from 12 pentagons and 60 hexagons

    It's been about 30 years since I took high school geometry, so my math skills are a little rusty

    What is the name of a polyhedron made from 12 pentagons and 60 hexagons (see http://www.georgehart.com/Puzzle2.html) , and how do I calculate the edge length of the faces, from the in-diameter of the polyhedron?

    For example, if I want to make such a polygon, with an in-diameter of 3 meters, what would be the edge length of the pentagons/hexagons?
    Last edited by LoremIpsum; April 1st 2010 at 08:38 PM.
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    Quote Originally Posted by LoremIpsum View Post
    It's been about 30 years since I took high school geometry, so my math skills are a little rusty

    What is the name of a polyhedron made from 12 pentagons and 60 hexagons (see Puzzle) , and how do I calculate the edge length of the faces, from the in-diameter of the polyhedron?

    For example, if I want to make such a polygon, with an in-diameter of 3 meters, what would be the edge length of the pentagons/hexagons?
    In fact, the figure illustrated at http://www.georgehart.com/Puzzle2.html is not a genuine polyhedron at all. It contains several vertices where three hexagons meet. But the angle of a regular hexagon is 120, and three of those add up to 360. That means that three hexagons can only meet in that way as part of a flat surface. Yet the picture shows them as forming part of a convex solid. That in turn means that the hexagons must be distorted in some way so as to give the appearance of convexity.

    In a genuine convex polyhedron, the sum of the angles at each vertex must be strictly less than 360. The greatest number of faces possible in a polyhedron consisting of regular pentagons and hexagons is 32 (12 pentagons and 20 hexagons), in a truncated icosahedron.
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    Quote Originally Posted by Opalg View Post
    In fact, the figure illustrated at Puzzle is not a genuine polyhedron at all. It contains several vertices where three hexagons meet. But the angle of a regular hexagon is 120, and three of those add up to 360. That means that three hexagons can only meet in that way as part of a flat surface. Yet the picture shows them as forming part of a convex solid. That in turn means that the hexagons must be distorted in some way so as to give the appearance of convexity.

    In a genuine convex polyhedron, the sum of the angles at each vertex must be strictly less than 360. The greatest number of faces possible in a polyhedron consisting of regular pentagons and hexagons is 32 (12 pentagons and 20 hexagons), in a truncated icosahedron.
    I do not need the pentagons and hexagons be regular, if they are predictably irregular, that would be ok too, as long as they were as close to regular as possible. I just need to know, the right measurements, to build it.
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