An Application of Logic to Combinatorial Geometry: How Many Tetrahedra are Equidecomposable with a Cube?

It is known (Rapp, 1985) that elementary geometry with an additional quantifier “there exist uncountably many” is decidable. We show that this decidability helps in solving the following problem from combinatorial geometry: does there exist an uncountable family of pair-wise non-congruent tetrahedra that are n−equidecomposable with a cube? DEFINITION 1. By an elementary geometry, following [Tarski 1951], [Seidenberg 1954], we mean the following theory: the variables x, y, z, ... run over real numbers; terms are constructed from variables by using addition and multiplication; elementary formulas are of the type t = t or t > t or t < t, where t, t are terms, and arbitrary formulas are constructed form the elementary ones by adding logical connectives & (“and”), ∨ (“or”), → (“implies”), ¬ (“not”), and quantifiers ∀x and ∃x. The well-known result of Tarski is that elementary geometry is decidable [Tarski 1951]. DEFINITION 2. By an elementary geometry with an additional quantifier “there exist uncountably many” we mean a theory with the same elementary formulas, in which a formula is obtained from elementary ones by adding the logical connectives, quantifiers, or the expression (Q1a1...an)P , where P is an arbitrary formula and a1, ..., an is an arbitrary finite set of variables. The semantics of this expression is “there exist uncountably many vectors (a1, ..., an) for which the formula P is true”. It is known that this theory is decidable [Rapp 1985]. APPLICATION. One of the remaining open problems connected with Hilbert’s Third Problem (on polyhedra) [Hilbert 1902] is as follows [Boltianskii 1978]. In a plane, every triangle (and moreover, every polygon) is equidecomposable with some square in the sense that they can be both decomposed into the same finite number of pair-wise congruent polygonal pieces. In 3-dimensional space, only for some tetrahedra it was known that they are equidecomposable with a cube [Hill 1896]. Solving a problem posed by Hilbert, Dehn proved [Dehn 1900] that there exist tetrahedra that are not equidecomposable with any cube. So, a natural problem arises: what tetrahedra are equidecomposable with a cube?