Dehn’s theorem
We all know the elementary formula^{} to compute the area of a triangle^{}: basis times height divided by two. This formula can be justified with a scissor type argument: one divides the triangle into smaller polygons^{} and rearranges these polygons to obtain a rectangle^{} which should have the same area.
Can we use the same argument to compute the volume of a pyramid^{}? This is the third Hilbert’s problem. Quite surprisingly the answer is negative, as states the theorem below. This means that the formulae to compute the volume of polyhedra cannot be proved without a limiting process (for example using integrals).
Definition 1.
We say that two polyhedra $P$ and $Q$ are scissorequivalent if there exists a finite number ${P}_{\mathrm{1}}\mathrm{,}\mathrm{\dots}\mathrm{,}{P}_{N}$ of polyhedra and ${\theta}_{\mathrm{1}}\mathrm{,}\mathrm{\dots}\mathrm{,}{\theta}_{N}$ isometries such that

1.
$P={\bigcup}_{k=1}^{N}{P}_{k}$ and $Q={\bigcup}_{k=1}^{N}{\theta}_{k}({P}_{k})$;

2.
${P}_{j}\cap {P}_{k}$ and ${\theta}_{j}({P}_{j})\cap {\theta}_{k}({P}_{k})$ have empty interior for every $k\ne j$
The properties given above assure that two scissorequivalent polyhedra must have the same volume. It is also simple to prove that the scissorequivalence is indeed an equivalence relation^{}.
Theorem 1.
The regular tetrahedron^{} is not scissorequivalent to any parallelepiped^{}.
Title  Dehn’s theorem 

Canonical name  DehnsTheorem 
Date of creation  20130322 16:18:04 
Last modified on  20130322 16:18:04 
Owner  paolini (1187) 
Last modified by  paolini (1187) 
Numerical id  11 
Author  paolini (1187) 
Entry type  Theorem 
Classification  msc 51M04 
Classification  msc 52B45 
Related topic  BanachTarskiParadox 
Related topic  HilbertsProblems 
Related topic  RegularTetrahedron3 
Defines  scissorequivalent 