An icosidodecahedron has thirty vertices, upon which the symmetry group of the polyhedron acts transitively. Is there a set of fifteen non-intersecting diagonals of the icosidodecahedron (i.e., line segments that are not edges but both of whose endpoints are vertices) such that the symmetry group of the icosidodecahedron acts transitively on these diagonals? You can see the motivation for the question and some failed attempts here.
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$\begingroup$ I enjoyed meeting you at G4G13 this weekend. I posted a few vZome models on SketchFab in response to our discussion at lunch today. Hope they helps. I'd love to see a few pictures of your tensegrity model when you get it finished. David $\endgroup$– DavidApr 16, 2018 at 2:18
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$\begingroup$ It certainly helps to illuminate the situation. But both the "long" set and the "short" set you posted consist of 30 diagonals, with two ending at each vertex. I am looking for a set of 15, with one ending at each vertex; in other words, they should be strictly non-intersecting in that they should not intersect at vertices, either. I don't actually think it is possible; if you have any insight into proving that I would be glad to hear it. $\endgroup$– Glen WhitneyApr 18, 2018 at 0:34
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$\begingroup$ If i understand the question correctly, the requirement for the diagonals to act transitively under icosahedral symmetry implies that they are all the same length. If that's the case, then I can assure by my own empirical inspection of all of the possible diagonals, that it's impossible because the sets I posted are the only ones with no internal intersections, but as you noted, they do intersect at the vertices. $\endgroup$– DavidApr 19, 2018 at 1:18
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$\begingroup$ If you're willing to accept 5 sets of diagonals, each having 3-fold rotational symmetry around a common axis, then the model at skfb.ly/6ysQF may do the trick for your tensegrity model even though it doesn't meet your requirement for being transitive under icosahedral symmetry. Best of luck. $\endgroup$– DavidApr 19, 2018 at 2:37
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$\begingroup$ That's a lovely construction, thanks! If you follow the link in the OP (and maybe go back or forward a posting or two), you'll see there is also a set of 10 equivalent diagonals and a disjoint set of five, so that the overall collection has fivefold symmetry. But I am more interested in your impossibility assertion. Are the two sets of 30 that you posted each entire orbits under the icosahedral group? Or are there other diagonals of one or both types? If it is easy for you to post the orbit decomposition of the body diagonals (leaving out the very longest ones), we could finish this off.. $\endgroup$– Glen WhitneyApr 19, 2018 at 5:15
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