The Dehn Test Picks the Lone Tilers: Sweeps and Truncated Tetrahedra That Fill Space

A tetrahedron with its four corners cut off by fractions c1, …, c4 of the edge has Dehn invariant 6a(1 − Σc_i) ⊗ arccos(1/3) (Proposition 2.1); so it can fill space alone only when the cuts add up to one edge, and that is the same as cutting the tetrahedron T by the inverted tetrahedron of twice its size, 2(−T), placed anywhere (Proposition 2.2). For cuts by whole layers of the octet it is also the same as keeping the octet's ratio of two tetrahedra to every octahedron (Proposition 2.3). Theorem 2.4 (computer-checked, exact cover): all twelve Dehn-zero integer truncations with n ≤ 5, and (5, 3, 0, 0) at n = 8, tile space, each with two pieces per period, the piece and its point inverse; we conjecture this for every rational cut (Conjecture 2.5) and list the seven golden cut vectors that are the next to try (Proposition 2.6). On the sweeps the same test picks the lone tilers: on the cube–octahedron sweep the Dehn invariant vanishes exactly at Kelvin's truncated octahedron and the cube (Theorem 1.1); on the CO–RD sweep it vanishes on the stretch 1 ≤ μ ≤ 4/3 and at the B3 permutohedron, and the angle count leaves only the rhombic dodecahedron (Theorem 1.2); on the tetrahedron sweep it vanishes only at μ = 2, the known β-tin tiler (Proposition 1.3). DICTO proposed filling the recorded gaps of the Concordance with a combination of new and current methods; these results fill the Dehn column of its grid of tests and families and the row of the tetrahedron sweep. Computation, proofs, figures and drafting with Claude (Anthropic).