I am familiar with a proof that the cube and tetrahedron are not scissors congruent along the following lines:
- Given a polyhedron or collection of polyhedra $\mathcal{P}$ whose edges form a set $E$, let the length and dihedral angle of an edge $e\in E$ be denoted $l_e\in \mathbb{R}$ and $\alpha_e\in [0,2\pi]$ respectively. Let $\alpha_e'$ be the image of the dihedral angle in $\mathbb{R}/\pi\mathbb{Q}$. The Dehn invariant $\sum_{e\in E}l_e\otimes\alpha_e'$ is a tensor in $\mathbb{R}\otimes\mathbb{R}/\pi\mathbb{Q}$ and is invariant under dissections and rearrangements.
- The Dehn invariant of the cube is $0$, as all angles are a rational multiple of $\pi$.
- The Dehn invariant of the unit tetrahderon is $6\otimes\arccos{\frac{1}{3} }$ which is non-zero because $\arccos{\frac{1}{3} }$ is not a rational multiple of $\pi$. Hence the cube is not scissors congruent to the tetrahedron.
While Dehn invariant is useful in many contexts, I do not think it is needed to prove that these particular polyhedra are not scissors congruent. Consider the following argument:
- Take $A(\mathcal{P})=\sum_{e\in E}\alpha_e$. Then if $\mathcal{Q}$ is a dissection of $\mathcal{P}$, the sum $A(\mathcal{Q})$ differs from $A(\mathcal{P})$ by a natural multiple of $\pi$. This follows as the dihedral angles introduced in $\mathcal{Q}$ can lie on an edge $e$ of $\mathcal{P}$ and so sum to $\alpha_e$ which they replace; they can lie on a face and hence sum to $\pi$ or they can lie internally to $\mathcal{P}$ and sum to $2\pi$.
- Thus the quantity $A'(\mathcal{P})=\sum_{e\in E}\alpha_e'\in\mathbb{R}/\pi\mathbb{Q}$ satisfies $A'(\mathcal{P})=A'(\mathcal{Q})$ when $\mathcal{Q}$ is scissors congruent to $\mathcal{P}$.
- For a cube this quantity is 0, while for a regular tetrahedron this quantity is $6\arccos{\frac{1}{3}}$.
- Hence the cube is not scissors congruent to the tetrahedron.
Is there a flaw in this line of reasoning from the sum of the dihedral angles?
