On Sun, Oct 24, 2004 at 03:54:57PM -0500, David Gale wrote:
I would like to know whether the following stuff is known.
Well known: There does not exist a set of rectangles which will tile both a 1 by 2 rectangle and a square of side sqrt(2) (although they both have the same area). We say these sets are not EQUI-TILABLE.
GENERAL PROBLEM: Let S be a union of a finite set of rectangles and let S' be another such set. Under what conditions are S and S' equi-tilable.
ANSWER: Think of reals as a vector space over Q. For a rectangle of width a and height b we can form their TENSOR PRODUCT a(x)b. The TENSOR AREA of the set S is summation(a_i(x)b_i) summed over the rectangles of S.
THEOREM. The sets S and S' are equi-tilable if and only if they have the same tensor area.
This is surely known. There's a harder problem, one of Hilbert's problem, solved by Dehn: Question. When are two polyhedra in R^3 equidecomposable into polyhedra? [Allow arbitrary polyhedra, not just rectangles] Answer. For each edge, consider l(x)theta, where l is the length of the edge and theta is the angle of the edge, as an element of R(x)(R/2pi), tensor product over Q. The sum of this quantity over all edges of a polyhedron P is called the Dehn invariant of P. Then P and P' are equidecomposable iff they have the same area and same Dehn invariant. I don't know the references, sorry. It was solved early last century. Peace, Dylan