1. Dehn’s Theorem on Tetrahedron and the cube.
Rosy Gupta
Mount Holyoke College
Hudson River Undergraduate Conference.
April 28, 2001

2. In the list of 23 problems proposed by David Hilbert during the Second International Congresss of Mathematicians held in Paris 1900, the third problem plays a special role, and this is why:
In contrast to other problems, the third one deals with elementary geometrical questions about the foundations of geometry. In 1899, Hibert finished writing a book, and in this process, he mentioned that - “contrary to the planar case - volume computations in three-dimensional Euclidean geometry are always based on some limiting process and on methods of exhausation.” He also said that one cannot construct a theory of polyhedral volume without the continuity axiom.
Hilbert’s Third Problem: Is it possible to cut a regular tetrahedron into infinitely many pieces by planes and paste the pieces into a cube?

3. In other words, this problem wonders if tetrahedra (type of pyrmaid), which have the same base and the same height, can be made from the same “building blocks”. If yes, then this implies that tetrahedra with equal base and height have the same volume. However, Max Dehn proved that there does not exist a tetrahedra with equal base and height which could have the same “building blocks.”
Max Dehn studied under Hilbert’s supervision and obtained his doctrate in He wrote one of the first systematic expositions on Topology in 1907 and later formulated important problems on group presentation. From 1921 until 1935 he held the chair of Pure and Applied Mathematics at the University of Frankfurt, however he was forced to leave his country by the Nazi regime in After travelling for two years, he came to the United States and taught at several universities and colleges.

4. Dehn’s solution comes from analyzing the dihedral angles of a polyhedron, the angles between its faces.
The dihedral angle of tetrahedron is:
The dihedral angle of a cube is: 90 degrees.

5. D(P) = l11 1 + l22 + l33 + … + lkk
If a tetrahedron is cut up and the pieces are pasted back into a cube, then it implies that the resultant angles have to be right angles, (the dihedral angles of a cube) from the dihedral angles of a tetrahedron.
Dehn drew two conclusions:
volume is not the only thing that is conserved by cutting and pasting a polyhedron;
the dihedral content, “an object that encodes the dihedral angles and ties them to the lengths of the corresponding edges” is also conserved.
Let l be the length and  be the dihedral angle and thus the dihedral content can be written as: l. Thus the total dihedral content can be written as:
D(P) = l11 1 + l22 + l33 + … + lkk
Where l1 , l2, …lk are the lenghts of the edges and 1 , 2 ,…,k are the respective dihedral angles.

6.  D(cube)  D(Tetrahedron)
To solve Hilbert’s problem, Dehn came up with three rules and these three rules or axioms are called Dehn’s invariant:
Rule 1: l( + ) = l + l
Rule 2: (l + m)  = l + m
Rule 3: l = 0
Dehn’s invariant of the unit cube is 0.
Dehn’s invariant of the unit tetrahedron is 6cosh (1/3).
 D(cube)  D(Tetrahedron)

7. Why the dihedral content of a cube is zero?
The cube has 12 edges, each of length 1 and of dihedral angle /2. This gives us:
D(Cube) = 12 /2.
Using Rule 1,
1 /2 + 1 /2 = 1  (/2 + /2) = 1  = 0 (using Rule 3).
Why the dihedral content of a polyhedron is 6cosh (1/3)?
A tetrahedral has 12 edges, each of length 1 and of dihedral angle cosh (1/3). This gives us:
D(Tetrahedral) = 12 cosh (1/3).
1 cosh (1/3) + 1 cosh (1/3) = 1  (cosh (1/3) + cosh (1/3) ) = 1 2cosh (1/3)  0.