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A tetrahedral expanded to a icosahedron

  1. Sep 9, 2012 #1
    This is really starting to bug me.

    I built an object out of tooth picks and glue.
    It's simple. An equilateral triangle pyramid. A four sided dice.
    And then when I add a tetrahedral from each of the four faces ... it becomes an object that I just can't seem to find a description of in any of the dozens of math books I've read or on the internet.

    I've been investigating this thing since I independently discovered my first hexaflexagon.
    It took ten years of library search and the spread of the internet to let me know about Arthur Stone and his studies. I now have many 3,4,5,6 faced examples. And a neat example in 3d of the... Tuckerman traverse
    The six sided ones led me to the models that show they are parts of icosahedron.

    continuing searches : http://www.princeton.edu/~mudd/findi...oral/pmc41.htm [Broken]
    Tukey: Arthur, yes. When did Arthur come? He must have been here by '39. Arthur, Dick Feynman, Bryant Tuckerman—who went to IBM—and I were the people who invented hexaflexagons. This came about because Arthur had an English-size notebook. Woolworth sold only American-size paper. He had to cut strips off the edges. He had to do something with the strips, so he started folding polygons. When he folded the hexagon he had the first hexaflexagon. Later came the Feynman diagram, the Tuckerman traverse, and so on.

    It is an interesting object to me. I use it to create 20 sided dice with extensions that are 90 degree offset and they just lead me to see the next iteration using each face as the contact to the next icosahedron.

    My problem is that I don't know any of the skills needed to make my 'simple math object' into a series of frames that show the growth of the iterations.

    any help or direction that would help me to further my desire to understand and describe this object would be appreciated.

    Thanks : Alfi

    Attached Files:

    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Sep 28, 2012 #2
    I think this is what I was looking for


    . If you spread out the hypertetrahedron, you get 5 tetrahedrons as its net. Together the five tetrahedrons have 5*4=20 triangles. 2*4=8 triangles are bound. If you build a hypertetrahedron, you must stick the remaining 12 triangles in pairs.
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