Worked solution: Dehn's invariant: the tetrahedron and cube are not scissors-congruent (1900)

Step 4 of 7: Compute: every dihedral angle of a cube is rational, so its invariant vanishes

In plain words

A cube is the friendliest possible test case: wherever two faces meet, they meet at a perfectly square corner. Since a right angle is one quarter of a full turn — as rational a fraction of $\pi$ as it's possible to be — Dehn's construction erases it completely, no matter how long the cube's edges are.

\operatorname{Dehn}(\text{cube}) = 0, \qquad \theta_{\text{cube}} = \frac{\pi}{2}

Detailed analysis

Every dihedral angle of a cube is $\pi/2$ , a rational multiple of $\pi$ , so $[\theta(e)] = 0$ in $\mathbb{R}/\pi\mathbb{Q}$ for every edge and hence $\operatorname{Dehn}(\text{cube}) = 0$ ; this holds regardless of the cube's side length, since the length only multiplies a term that is already zero. Because $\operatorname{Dehn}$ is additive under dissection (Step 3), every solid that actually is scissors-congruent to a cube — for example, a rectangular box, or a right prism over a Bolyai–Gerwien-decomposed polygon base — must likewise have Dehn invariant $0$ .

Knowledge used in this step

Polyhedra and their volumes