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

Step 3 of 7: Additivity: the invariant is unchanged by dissection

In plain words

When you slice a solid into pieces, every new cut creates new edges inside the solid, but these new edges always come in pairs (or small groups) whose dihedral angles add up to exactly a straight angle $\pi$ or a full turn $2\pi$ — angles that are rational multiples of $\pi$ and so vanish in Dehn's bookkeeping. That is precisely why Dehn engineered the invariant to work modulo $\pi\mathbb{Q}$ : it guarantees that whatever mess of new internal edges a cut creates, they never add anything to the total.

P = P_1 \sqcup \cdots \sqcup P_k \implies \operatorname{Dehn}(P) = \sum_i \operatorname{Dehn}(P_i)

Detailed analysis

If $P$ is cut into finitely many polyhedral pieces $P_1, \ldots, P_k$ that reassemble into $Q$ , one checks directly (tracking how each new interior edge contributes angle pairs summing to $2\pi$ or $\pi$ , which vanish mod $\pi\mathbb{Q}$ , and how each original edge's angle splits additively across pieces) that $\operatorname{Dehn}(P) = \sum_i \operatorname{Dehn}(P_i) = \operatorname{Dehn}(Q)$ . So $\operatorname{Dehn}$ is invariant under scissors-congruence, exactly like volume — but it carries strictly more information.

Knowledge used in this step

Polyhedra and their volumes