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Subject: sci.math FAQ: Cutting a Sphere
This article was archived around: 17 Feb 2000 22:55:52 GMT
Last-modified: February 20, 1998
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? Borel-set?
Lebesgue-measurable set? Arbitrary?)
* How many pieces are permitted (finitely many? countably?
* What motions are allowed in ``reassembling" (translations?
rotations? orientation-reversing 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 equal-area polygons are equi-decomposable.
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 Hadwiger-Glur theorem implies that any two equal-area polygons
are equi-decomposable using only translations and rotations by 180
* Theorem 5 [Hadwiger-Glur, 1951] Two equal-area 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
equi-decomposable by means of translations and rotations, by
cutting into finitely many sub-polyhedra, and gluing along
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, 345-354). It was the first of
Hilbert's problems to be solved. The proof is nontrivial but does
not use the axiom of choice.
Hilbert's Third Problem. V.G. Boltianskii. Wiley 1978.
* Using the axiom of choice on non-countable 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
This construction is known as the Banach-Tarski paradox or the
Banach-Tarski-Hausdorff paradox (Hausdorff did an early version of
it). The ``pieces" here are non-measurable 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 Banach-Tarski is:
The Banach-Tarski Paradox. Stan Wagon. Cambridge University Press,
Robert M. French. The Banach-Tarski theorem. The Mathematical
Intelligencer, 10 (1988) 21-28.
The pieces are not (Lebesgue) measurable, since measure is
preserved by rigid motion. Since the pieces are non-measurable,
they do not have reasonable boundaries. For example, it is likely
that each piece's topological-boundary is the entire ball.
The full Banach-Tarski paradox is stronger than just doubling the
ball. It states:
* Any two bounded subsets (of 3-space) with non-empty interior, are
equi-decomposable 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
* The unit interval can be cut up into countably many pieces which,
by translation only, can be reassembled into the interval of
This result is, nowadays, trivial, and is the standard example of
a non-measurable set, taught in a beginning graduate class on
* 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.
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 Lopez-Ortiz firstname.lastname@example.org
http://www.cs.unb.ca/~alopez-o Assistant Professor
Faculty of Computer Science University of New Brunswick