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I have two convex polyhedra such that their sums of side areas are equal. It is true that I can cut one of them and flatten it on the plane, then fold the flattened polygon to reach the other polyhedron?

For special case in which these two polyhedra are cube and regular tetrahedron, The answer in Yes I think. But I think the answer is "No, not necessarily" in general case, and I think the problem is difficult!

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What do you mean by "equal side areas"? Do you mean the SUMs of the areas are equal? –  Igor Rivin Jul 28 at 22:40
    
Yes, I mean the sums of areas are equal, this is a necessary condition, my question is "Is it sufficient?". –  Morteza Jul 28 at 22:58
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I would be interested to learn of details of your {cube, regular tetrahedron} unfold/refold construction, for that would constitute a partial solution to one of the open problems in my book with Erik Demaine, Geometric Folding Algorithms. –  Joseph O'Rourke Jul 29 at 23:24
    
@JosephO'Rourke: I'm not sure about the regularity, because I have no useful software to check, I will try to find the way. –  Morteza Jul 30 at 7:59
    
OK, Morteza. Regularity is key, for as you see in my answer, cube to tetrahedron is not rare. Cube to regular tetrahedron would be quite interesting. –  Joseph O'Rourke Jul 31 at 0:16

2 Answers 2

The general answer is sometimes Yes, various unfold/refold pairs of convex polyhedra exist, but not always, depending on what is meant by "cut." Here is an example that Erik Demaine, Marty Demaine, Anna Lubiw, and I worked out carefully:

The Latin-cross unfolding of the cube can refold into precisely 23 distinct convex polyhedra, as displayed below (all of the same surface area): the cube, two doubly covered flat quadrilaterals, seven tetrahedra, three pentahedra, each with one or more quadrilateral faces, four hexahedra, and six octahedra:


  LatinCrossRefoldings
  Fig. 25.30, p.408 in Geometric Folding Algorithms: Linkages, Origami, Polyhedra, 2007.
Here is a "movie" of one of the refoldings to a tetrahedron:
                    Latin2Tetra
But more generally, in the 2012 paper Refold Rigidity of Convex Polyhedra, we showed that every convex polyhedron can be cut open and refolded to an incongruent convex polyhedron.

But if the cutting is restricted to follow edges of the convex polyhedron ("edge unfoldings"), then there are "refold-rigid" polyhedra. For example, each of the 43,380 edge unfoldings of a dodecahedron may only fold back to the dodecahedron.


Added. I apologize for not initially understanding Morteza's precise question, which asks whether or not, for any two given pair of convex polyhedra $P$ and $Q$ with the same surface area, is it always possible to cut open $P$ and refold to $Q$. This remains an open question as far as I know. It is a version of Open Problem 25.31 (Fold/Refold Dissections) in Geometric Folding Algorithms.

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Yes, I mean the free cut, not only on edges. Thanks. –  Morteza Jul 29 at 7:26

If this says what I think it says, the answer is "NO" for even the example cited, since you are prescribing the intrinsic metric on the polyhedron, and it is a celebrated theorem of A. D. Alexandrov (which is a slightly tweaked theorem of Cauchy) that the intrinsic metric determines the convex polyhedron uniquely. For more, see Alexandrov's "Convex Polyhedra"

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Do you mean that no two polyhedrons have this property? I have some examples, for example I can reach a tetrahedron from a cube, maybe the tetrahedron is not regular! –  Morteza Jul 28 at 22:48
    
Actually, the answer is sometimes Yes. I will try to find time to post... –  Joseph O'Rourke Jul 28 at 22:51
    
Then I obviously don't understand the question. –  Igor Rivin Jul 28 at 22:55
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I think the confusing aspect here is what does it mean to "refold" the flattened surface. If it must be reglued exactly the same, then Alexandrov's theorem settles it. But if one reglues differently, (in general) many different convex polyhedra can be obtained. –  Joseph O'Rourke Jul 28 at 23:21
    
@JosephO'Rourke Ah, OK! –  Igor Rivin Jul 29 at 13:36

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