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I suspect it is impossible to split a (any) 3d solid into two, such that each of the pieces is identical in shape (but not volume) to the original. How can I prove this?

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On a related note: Kimmo Eriksson proves in [The American Mathematical Monthly Vol. 103, No. 5 (May, 1996), pp. 393-400] that a convex polygon is splittable in two properly congruent pieces iff it has rotational symmetry. –  Mariano Suárez-Alvarez Sep 7 '10 at 17:16
    
A couple of questions. 1) What exactly does split mean? 2) Why is this tagged topology? –  Aryabhata Sep 7 '10 at 17:25
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Is this connected to the Banach-Tarski paradox? –  Isaac Sep 7 '10 at 18:36
    
Was thinking of a 3d analogue to Pythagoras' Theorem. –  Guillermo Phillips Sep 7 '10 at 19:02
    
Isaac, I don't think so, because that paradox involves dividing a sphere into non-spherical pieces. But I am also curious if there are any solutions involving pathological shapes or division methods. –  Dan Brumleve Sep 8 '10 at 1:13
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2 Answers

You can certainly take a rectangular box, $2^{1/3} \times 2^{2/3} \times 2$ and slice it into two boxes of size $1 \times 2^{1/3} \times 2^{2/3}$.

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I'm confused, doesn't this contradict the theorem in the answer Mariano Suárez-Alvarez wrote? –  anon Sep 7 '10 at 18:47
    
Not impossible then. –  Guillermo Phillips Sep 7 '10 at 18:59
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@muad: Boxes aren't strictly convex. –  whuber Sep 7 '10 at 19:51
    
@whuber, thank you - I understand now –  anon Sep 7 '10 at 20:24
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Also can we make the pieces different sizes and connect them sideways, like a Golden Rectangle? How many different ways are there to do this in n-space? –  Dan Brumleve Sep 8 '10 at 1:37
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It seems that Puppe and others proved that this is impossible for any strictly convex solid. See [B. L van den Waerden, Aufgabe Nr 51, Elem. Math. 4 (1949) 18, 140]

The reference comes from Unsolved problems in geometry by Hallard T. Croft, K. J. Falconer and Richard K. Guy.

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Is it also true that no strictly convex solid can be divided into any two strictly convex pieces, whether or not they are identical with each other or either with the original? –  Dan Brumleve Sep 8 '10 at 1:06
    
@Dab, I would imagine that that is true (for at a point where the two pieces touch which is not at the boundary of the original body at most one of them will be strictly convex) –  Mariano Suárez-Alvarez Sep 8 '10 at 9:51
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