Two polytopes (in particular, polygons or polyhedra) A and B are equidecomposable if there exist decompositions

A = i=1n Ai
B = i=1n Bi
of them into nearly disjoint (i.e., disjoint or with an intersection of lower dimension, with an empty interior) polytopes such that for all i=1,...,n, Ai is congruent to Bi.

Equidecomposable polygons might be called scissors-and-glue equivalent: given A, some scissors and glue, we can cut up A, rearrange the pieces, and glue them into B.

Naturally, equidecomposable polytopes must have equal measure (area, volume, or what-have-you in d dimensions), since their congruent pieces have equal measure. What about the converse?

In two dimensions, any two polygons of equal area are equidecomposable. In three dimensions, there exist even tetrahedra of equal volume which are not equidecomposable -- this is the substance of Hilbert's third problem.

See also the related concept of equicomplementable polytopes.

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