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Please have a look at

http://www.cijoint.fr/cjlink.php?file=cj201006/cijHr27640.jpg

the object in question is a truncated icosahedon whose sides are pearls.

It is an interesting little bauble which can be found in some cheap jewelry stores.

Each pearl is drilled and through the hole thus created there are two threads.

Upon exiting the hole (therefore at a vertex) the threads are directed towards adjacent pearls.

Since at each vertex three sides (pearls) come together, there are three threads at at each vertex (these threads are visible if one looks at the picture carefully).

Any idea how this "threaded" truncated icosahedron was put together?

I suspect this may have to do with graph theory, an assumption very easy to make since I know absolutely nothing about the subject.

My ultimate goal is to be able to put together such a polyhedron using small pieces of pipes and string, and who knows may be other polyhedras.

I have built a truncated icosahedron with Zometool parts but I have not been able to find a solution, only conjectures.

Thank you.

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  • $\begingroup$ I see an uncommented vote to close, which is inexplicable to me. I would guess that a substantive (versus coincidental) answer might involve both representation theory and graph theory. The question is sufficiently clear to understand. I hope any other votes to close have a good and articulated reason. $\endgroup$
    – Steve Huntsman
    Commented Jun 4, 2010 at 19:53
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    $\begingroup$ Besides being cubic, the truncated icosahedral graph is Hamiltonian: mathworld.wolfram.com/TruncatedIcosahedralGraph.html $\endgroup$
    – Steve Huntsman
    Commented Jun 4, 2010 at 19:55
  • $\begingroup$ I'm starting to get it. I've been staring at the picture, if we regard the pearls as vertices the figure is what you get if you take the truncated icosahedron and truncate again at each vertex as far as the midpoint of each edge, making for a large number(60) of additional equilateral triangles, while still having 20 hexagons and 12 pentagons. I don't know a name for the solid I am describing...You prefer to view each pearl as being the midpoint of an edge. $\endgroup$
    – Will Jagy
    Commented Jun 5, 2010 at 4:10

1 Answer 1

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The points of the icosahedron are known. Coordinates can be found for each point. From these the coordinates for the truncated icosahedron can be derived. From these the points on the edges where the pearls can be placed can be derived. The resulting polyhedron depends on the properties of the points placed on the edges. Is there a single point bisecting the edge. Are there two points symmetrically placed about the midpoint of the edge? If so how. Once this choice is made the resulting points coordinates can be determined and from this the lengths of the sides which ought to give enough information for construction.

In regards to the question about the easiest way to build a truncated icosahedron in the comments. I would use the fact that all if its faces are equilateral hexagons and pentagons with two pentagons and one hexagon at each vertex. If I could get rigid hexagons and pentagons I could fit them together and I would be able to get the polyhedron relatively easily. It is much easier to construct this because all the faces are regular polygons so they could be constructed first and the rest of the construction is relatively easy.

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  • $\begingroup$ I am well aware of the metric properties of the truncated icosahedon. I rephrase my question. Given 90 idetical pieces of tubing and some length(s) of string, how can I build from these a truncated icosahedron knowing that each piece of tubing contains at most 2 threads of string, that there should be no more than three threads at each vertex (or node) and altogether the fewer number of knots (possibly only one). Thank you. $\endgroup$
    – cartesys
    Commented Jun 4, 2010 at 23:00
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    $\begingroup$ If you are willing to have two threads in each tube, but six at each vertex, then you can do it with one knot. The reason is that, when you double each edge of the graph, each vertex becomes of even degree, hence there is an Eulerian closed path. $\endgroup$
    – John Stillwell
    Commented Jun 5, 2010 at 2:56
  • $\begingroup$ @John--Seems like the desired Eulerian path should be a sort of "double cover" Hamiltonian path, similar in spirit to a de Bruijn sequence. $\endgroup$
    – Steve Huntsman
    Commented Jun 5, 2010 at 13:17

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