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Topic: World's Largest Icosahedron
Replies: 1   Last Post: Aug 8, 1994 2:45 PM

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John Sullivan

Posts: 14
Registered: 12/6/04
Re: World's Largest Icosahedron! (Probably)
Posted: Aug 8, 1994 2:45 PM
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In article <Cu85HM.JL@news.cis.umn.edu>,
Bob Hesse <hesse@diophantus.geom.umn.edu> wrote:
>the length from vertex to opposing vertex of an icosahedron is equal
>to 5^(1/4)*tau^(1/2)*s where tau is the golden ratio and s is the side

length.
>... "I have a very elegant proof of this result, but am unable to fit
>it in the margin." :-)
>
>If anyone knows of the proof that Rick discovered, maybe they should
>post a reply to this article.


Well, I'll give a proof, but this is surely not the one Rick discovered,
since this one _would_ have fit in the margin.

The icosahedron has 15 pairs of opposite and parallel edges.
Each pair forms the short ends of a golden rectangle, whose diagonal
is the length we are looking for. By the pythagorean theorem, this
diagonal equals $\sqrt{1+tau^2}$ times the side length, equivalent to
Rick's formula. So, you might ask, how do we know that the rectangle
mentioned is a golden rectangle? Well, we can view the icosahedron
as a pentagonal antiprism with two pentagonal pyramids attached to the
pentagons. Remove the two pyramids, and concentrate on the antiprism.
Look at two of opposite edges among its zig-zag edges (the "stitching"
on the "drum"). They form, of course, the short sides of one of our
rectangles. But the long sides of this rectangle are diagonals of the
pentagons, and of course a pentagon diagonal is in golden ratio to its
side.

There's a nice exposition of facts like this about the icosahedron,
dodecahedron, and rhombic 30hedron in the first issue of the journal
Symmetry 1:1 (1990), p29--36,
"On the Icosahedron, the Pentagonal Dodecahedron, and the Rhombic
Triacontahedron" by Arthur Loeb, Jack Gray and Philip Mallinson.
This includes lots of nice stereo pictures of the objects described.

-John Sullivan






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