A reader of this blog, in a comment on the last post here, asked what would happen if each face of an icosahedron were augmented by another icosahedron. I was also asked what the convex hull of such an icosahedron-cluster would be. Here are pictures which answer both questions, in order.
While the icosahedron augmented by twenty icosahedron forms an unusual non-convex shape, its convex hull is simply a slightly “stretched” version of the truncated dodecahedron, one of the Archimedean solids.
The reader who asked these questions did not ask what would happen if the icosahedron-cluster above were to be augmented, on every face, by yet more icosahedra. However, I got curious about this, myself, and created the answer: the following cluster of even-more numerous icosahedra. This could be called, I suppose, the “reaugmented” icosahedron.
Finally, here is the convex hull of this even-larger cluster. No one asked for it; I simply got curious.
To accomplish the polyhedron-manipulation and image-creation for this post, I used a program called Stella 4d: Polyhedron Navigator, which is available at http://www.software3d.com/Stella.php. A free trial download is available there, so you can try the software before deciding whether or not to purchase it.
RobertLovesPi.net 
Nice, Spectacular…
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Thanks!
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Nice, spectacular…
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Thank you!
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First of all, thank you for bringing my suggestion to life! I must get the Stella application myself, so I do this with other polyhedra!
Second, about the results of this article. The truncated dodecahedron as the convex hull is very surprising to me. Sorry to give you more suggestions, (I should probably buy the software already) but what would the convex core of the figure be? A pentakis icosahedron?
A truncated icosahedron? A truncated dodecahedron again? I don’t know enough about convex hulls/cores to have an intuition for this sort of thing.
Finally, the bicluster (clustered twice) icosahedron’s convex hull is a very familiar to me. I can clearly see the 12 pentagons from the dodecahedron, and the 20 triangles from the icosahedron, so there is probably a way to access the figure sans augmentation. Here’s my guess on an alternative way to make the figure.
1.Start with a dodecahedron.
2.Edge-chamfer said dodecahedron.
3.Somehow, split each hexagonal face
of the chamfered dodecahedron into two trapezoids. (I have no idea how this step would be done.)
4.Expand the figure, but only expand edges adjacent to the long side of the trapezoid. (I also have no idea how this step would be done.)
Hopefully you could gleam something from my ramblings.
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Trust me, you should get Stella yourself! It’s a free trial download at the url in this blog-post. =)
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