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

Frequently asked questions

Why does the plane case (Bolyai–Gerwien) work but not always in 3D?

The 2D proof works by cutting every triangle into a rectangle using a construction that only needs straight cuts and a rotation, a trick that generalises smoothly; the 3D analogue would need to cut every tetrahedron into a cube, but as Dehn's invariant will show, dihedral angles obstruct this in a way that has no counterpart for the angles of a 2D triangle.

Source

Max Dehn, 1900

Über raumgleiche Polyeder

Step 1 of 7: Hilbert's third problem: does the plane's cut-and-reassemble trick survive in 3D?

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In plain words

In school geometry you learn that any polygon can be cut with straight lines into pieces and reassembled into a square of the same area — a triangle, a pentagon, a star, it doesn't matter, as long as the areas match. In 1900 David Hilbert asked, as the third of his famous 23 problems, whether the same magic trick works one dimension up: can any two polyhedra of equal volume always be cut into finitely many solid pieces and reassembled into each other? He suspected the answer was no, and asked for a proof — without waiting for an answer, he even suggested that finding an explicit invariant obstruction would settle it.

\text{2D (Bolyai–Gerwien, 1833): equal area} \iff \text{scissors-congruent} \qquad \text{3D (Hilbert's 3rd problem, 1900): } \overset{?}{\iff}

Detailed analysis

A finite set of polygons $P, Q$ is scissors-congruent if $P$ can be cut by straight-line segments into finitely many polygonal pieces that can be rearranged (by rigid motions) to exactly tile $Q$ . The Wallace–Bolyai–Gerwien theorem (essentially complete by 1833, through the independent work of William Wallace, Farkas Bolyai, and Paul Gerwien) shows this happens if and only if $P$ and $Q$ have equal area: any polygon can be triangulated, each triangle cut into a rectangle, and rectangles of the same total area combined and recut into any target shape.

Hilbert's third problem, posed at the 1900 International Congress of Mathematicians in Paris, asks the three-dimensional analogue: is having equal volume likewise sufficient for two polyhedra to be scissors-congruent (cuttable into finitely many polyhedral pieces reassembled by rigid motions)? Hilbert conjectured the answer was no and suggested that a rigorous proof would require exhibiting some additional invariant, besides volume, that dissection could not change — precisely the strategy his student Max Dehn carried out within the same year.

The rest of this proof constructs Dehn's invariant (Step 2), proves it is unchanged by cutting and reassembling (Step 3), and computes it on two specific solids of equal volume — a cube and a regular tetrahedron — to show they differ (Steps 4–6), giving Hilbert's problem a definitive negative answer (Step 6), with Sydler's 1965 completion of the picture in Step 7.

Terms in this step

scissors-congruence: Two shapes are scissors-congruent if one can be cut into finitely many pieces (polygonal pieces in the plane, polyhedral pieces in space) that can be rearranged by rigid motions to exactly form the other.
invariant: A quantity or algebraic object computed from a shape that does not change under some allowed operation (here, cutting and reassembling); if two shapes have different invariants, no such operation can turn one into the other.