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Is it possible to form a closed loop by joining regular (platonic) tetrahedrons together side-to-side, with each tetrahedron having two neighbours? It should be a loop with a hole in, as can be done with 8 cubes, or 8 dodecahedrons as shown below. What is the minimum number of tetrahedrons needed?

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Edit: Could it be that it is possible to create such a ring by allowing the tetrahedrons to extend in one more spatial dimension (R^4)?

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up vote 9 down vote accepted

No, it isn't possible. Here is the reference, which I haven't found online (maybe someone can find a link):

Mason, J. (1972). "Can Regular Tetrahedra Be Glued Together Face to Face to Form a Ring?" Mathematical Gazette 56 (397) p.194-197.

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If you have a JSTOR subscription (or access to a library which does), the link is – Willie Wong Jan 28 '11 at 20:38
Interesting to note that the author also allows immersions: the tetrahedra are allowed to cut into one another. It may seem strange that this "unphysical case" is allowed at first, but in fact by doing so the question becomes simpler (one can re-word the problem purely group theoretically). – Willie Wong Jan 28 '11 at 20:44

One can get interesting geometrical "rings" of tetrahedra when they are attached edge to edge: On this page some of the examples use regular tetrahedra and others less regular ones.

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To the follow up question about allowing embeddings in $\mathbb{R}^4$: the answer is yes. One simple example is to start with the hexadecachoron or 600-cell and remove some of the constituent tetrahedra.

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Do you have any idea what the minimum would be? I can imagine 3 could work, but I dont know if that would elave a hole. Also could the number 8 for the other solids be reduced to 3 or even 2 by embedding in yet higher R^n ? – TROLLKILLER Jan 29 '11 at 10:18
@solomoan I don't think 3 would work. If you start with the 4-simplex (4D version of the tetrahedron), you need 4 to surround a vertex (without leaving a hole, sort of like 3 panes of the tetrahedron around a vertex). I suspect (without proof) that this is a lower bound. What the actually minimum is I don't know. – Willie Wong Jan 30 '11 at 1:23

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