Archivename: scimathfaq/sphere
Lastmodified: February 20, 1998 Version: 7.5 See reader questions & answers on this topic!  Help others by sharing your knowledge Cutting a sphere into pieces of larger volume Is it possible to cut a sphere into a finite number of pieces and reassemble into a solid of twice the volume? This question has many variants and it is best answered explicitly. Given two polygons of the same area, is it always possible to dissect one into a finite number of pieces which can be reassembled into a replica of the other? Dissection theory is extensive. In such questions one needs to specify * What is a ``piece"? (polygon? Topological disk? Borelset? Lebesguemeasurable set? Arbitrary?) * How many pieces are permitted (finitely many? countably? uncountably?) * What motions are allowed in ``reassembling" (translations? rotations? orientationreversing maps? isometries? affine maps? homotheties? arbitrary continuous images? etc.) * How the pieces are permitted to be glued together. The simplest notion is that they must be disjoint. If the pieces are polygons [or any piece with a nice boundary] you can permit them to be glued along their boundaries, ie the interiors of the pieces disjoint, and their union is the desired figure. Some dissection results * We are permitted to cut into finitely many polygons, to translate and rotate the pieces, and to glue along boundaries; then yes, any two equalarea polygons are equidecomposable. This theorem was proven by Bolyai and Gerwien independently, and has undoubtedly been independently rediscovered many times. I would not be surprised if the Greeks knew this. The HadwigerGlur theorem implies that any two equalarea polygons are equidecomposable using only translations and rotations by 180 degrees. * Theorem 5 [HadwigerGlur, 1951] Two equalarea polygons P, Q are equidecomposable by translations only, iff we have eqaulity of these two functions: phi_P() = phi_Q(). Here, for each direction v (ie, each vector on the unit circle in the plane), let phi_P(v) be the sum of the lengths of the edges of P which are perpendicular to v, where for such an edge, its length is positive if v is an outward normal to the edge and is negative if v is an inward normal to the edge. * In dimension 3, the famous ``Hilbert's third problem" is: If P and Q are two polyhedra of equal volume, are they equidecomposable by means of translations and rotations, by cutting into finitely many subpolyhedra, and gluing along boundaries? The answer is no and was proven by Dehn in 1900, just a few months after the problem was posed. (Ueber raumgleiche polyeder, Goettinger Nachrichten 1900, 345354). It was the first of Hilbert's problems to be solved. The proof is nontrivial but does not use the axiom of choice. References Hilbert's Third Problem. V.G. Boltianskii. Wiley 1978. * Using the axiom of choice on noncountable sets, you can prove that a solid sphere can be dissected into a finite number of pieces that can be reassembled to two solid spheres, each of same volume of the original. No more than nine pieces are needed. The minimum possible number of pieces is five. (It's quite easy to show that four will not suffice). There is a particular dissection in which one of the five pieces is the single center point of the original sphere, and the other four pieces A, A', B, B' are such that A is congruent to A' and B is congruent to B'. [See Wagon's book]. This construction is known as the BanachTarski paradox or the BanachTarskiHausdorff paradox (Hausdorff did an early version of it). The ``pieces" here are nonmeasurable sets, and they are assembled disjointly (they are not glued together along a boundary, unlike the situation in Bolyai's thm.) An excellent book on BanachTarski is: The BanachTarski Paradox. Stan Wagon. Cambridge University Press, 985 Robert M. French. The BanachTarski theorem. The Mathematical Intelligencer, 10 (1988) 2128. The pieces are not (Lebesgue) measurable, since measure is preserved by rigid motion. Since the pieces are nonmeasurable, they do not have reasonable boundaries. For example, it is likely that each piece's topologicalboundary is the entire ball. The full BanachTarski paradox is stronger than just doubling the ball. It states: * Any two bounded subsets (of 3space) with nonempty interior, are equidecomposable by translations and rotations. This is usually illustrated by observing that a pea can be cut up into finitely pieces and reassembled into the Earth. The easiest decomposition ``paradox" was observed first by Hausdorff: * The unit interval can be cut up into countably many pieces which, by translation only, can be reassembled into the interval of length 2. This result is, nowadays, trivial, and is the standard example of a nonmeasurable set, taught in a beginning graduate class on measure theory. * Theorem 6. There is a finite collection of disjoint open sets in the unit cube in R^3 which can be moved by isometries to a finite collection of disjoint open sets whose union is dense in the cube of size 2 in R^3. This result is by Foreman and Dougherty. * A square cannot be rearranged into a disk, if one is allowed finitely many pieces with analytic boundaries, glued at edges. * A square can be rearranged into a disk, with translations only, if one is allowed to use finitely many pieces with unconstrained shape (not necessarily connected), and disjoint assembly. References Boltyanskii. Equivalent and equidecomposable figures. in Topics in Mathematics published by D.C. HEATH AND CO., Boston. Dubins, Hirsch and ? Scissor Congruence American Mathematical Monthly. ``Banach and Tarski had hoped that the physical absurdity of this theorem would encourage mathematicians to discard AC. They were dismayed when the response of the math community was `Isn't AC great? How else could we get such counterintuitive results?' ''  Alex LopezOrtiz alopezo@unb.ca http://www.cs.unb.ca/~alopezo Assistant Professor Faculty of Computer Science University of New Brunswick User Contributions:Comment about this article, ask questions, or add new information about this topic:
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