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8 cubes can be joined face-to-face to form a closed ring with a hole in it, with each cube sharing a face with only two others. The same can be done with 8 dodecahedrons. Is the same possible with the other platonic solids? What is the minimum number of each solid needed to form such a loop?

Given a convex regular d-polytope, is there a general way to determine if it can form a loop, by gluing their d-1 dimensional faces together? (Assuming the loop has a hole and no two objects intersecting, ie no two objects share a d-volume).
And is there a way to compute the minimum number of equal polytopes needed for this?

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Edit: added image

I am also looking for software that can be used to check for atleast small N

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Cross-posted from MSE (under a different username) under the title "Closed Polytope Loop": math.stackexchange.com/questions/19203/closed-polytope-loop One answer there includes a nice image of 8 dodecahedra forming a torus. –  Joseph O'Rourke Jan 28 '11 at 13:49
    
Do you know Geomag? :) –  François Brunault Jan 28 '11 at 17:17
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Maybe the following would answer your question? textodigital.com/P/GG/bp04.php and textodigital.com/P/GG/bp08.php (well, maybe not for the question of the smallest number...) –  François Brunault Jan 28 '11 at 18:43
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@Fran those things arent regular tetrahedrons i think –  fastforward Jan 28 '11 at 19:12
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For the tetrahedron, the answer is no, as noted in math.stackexchange.com/questions/19386/tetrahedral-torus The method used in the article linked there can presumably be also applied for more general reflections. But the computations will be rather messier. –  Willie Wong Jan 28 '11 at 20:47
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2 Answers

up vote 17 down vote accepted

All but the tetrahedron.

As noted in a comment, ultimately referencing a 1972 paper, for tetrahedra this cannot be done. I haven't looked at the paper, but the proof may go as follows: Let the vertices of your base tetrahedron be the standard basis vectors in $\mathbb R^4$ times 4, so their center point is the all-ones vector. Construct the four matrices which reflect through the faces of your tetrahedron (while fixing the sum-to-4 hyperplane). Observe that some entries of these matrices equal $\frac23$. Apply a reduced word in the reflection group to the all-ones vector, and prove by induction that the first generator in the word is always indicated by which entry has the lowest power of 3 in the denominator, with a predictable pattern mod 3 in the numerators. Conclude that no nontrivial product of reflections takes the center point back to itself.

For the octahedron, attaching a pair along opposite faces allows you to continue in the pattern of carbon atoms in the diamond crystal structure, where each atom has tetrahedral bonds in directions all four of which are the negatives of any neighbor's bonds. It is thus possible to glue together 12 octahedra positioned like the carbons and C-C bonds of the "chair" conformation of cyclohexane. Since eight of the twenty faces of an icosahedron are inclined like the faces of an octahedron, the exact same arrangement is possible with icosahedra.

EDIT: In fact, you can do better. Given any polytope that has two pairs of opposite parallel facets such that reflected polytopes may be attached to all four facets simultaneously without overlapping, a thickened parallelogram may be constructed by attaching eight identical polytopes along facets. This works because a pair of parallel reflections amounts to a pure translation, making the same possible attachment directions available at both ends of the double reflection.

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Actually, there's no need to keep track of the numerators mod 3, and the same proof works for regular simplices of any dimension $d$, by noticing which entry of the multiply-reflected all-ones vector has the lowest power in the denominator of $d$, in odd dimensions, or $\frac d2$, in even dimensions greater than $2$. It follows that the representation of the Coxeter group (whose graph is a complete graph with all edges labelled by infinity) is faithful for all $d>2$, and that no such solid torus loop exists for any higher-dimensional regular simplex. I think that finishes the question. –  Tracy Hall Jan 28 '11 at 23:54
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The same answer is also referenced on page 150 of Martin Gardner's 1987 "Time Travel and Other Mathematical Bewilderments". He credits Kurt Schmucker for the 8-polyhedron solutions for the four larger platonic solids and the same paper of J. H. Mason for the impossibility proof for tetrahedra. –  Tracy Hall Jan 29 '11 at 2:56
    
@Tracy-Hall, my eyes must be missing the reference to the 1972 paper on this page. What/where exactly is the paper you're referring to? Thanks for the Martin Garner reference. –  sleepless in beantown Feb 1 '11 at 2:20
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It's not on this page, but found in the top answer to math.stackexchange.com/questions/19386/tetrahedral-torus that Willie Wong linked to in a comment to the question. –  Tracy Hall Feb 4 '11 at 1:28
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A long time ago a similar question arose in the sci.math newsgroup. I worked it out that in order to make a loop out of dodecahedra, one has to use sets of three dodecahedra in a row, as shown (four times) in the picture above -- otherwise one introduces a set of rotations in R^3 which has no relations which we can use to get the first and last faces to line up. See http://www.math-atlas.org/98/dodec_prf

dave

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