The Delian Bond: Samson, Delilah and the Octet Between Kepler's Rhombic Dodecahedron and the Cube
- 1Independent artist, Osaka, Japan
Abstract
In short. Regular tetrahedra and octahedra fill space together: the octet. Join each tetrahedron's centre to each octahedron's centre through the face they share, cut every such double cone by one flat plane parallel to that face, and give each side to its own centre. Slide the cut, and every position still fills space with two convex solids, from Kepler's rhombic dodecahedra through the octet to cubes. At one position the two solids have exactly the same volume: the tetrahedron's piece has doubled and the octahedron's halved, the doubling of the ancient Delian problem in both directions at once. DICTO named this pair Samson and Delilah. Main theorem. For every cut position t from 0 to 1 the two pieces are convex and fill space face to face: for t ≤ 1/3 a regular tetrahedron of edge 3t and a rhombic dodecahedron with its eight three-fold corners cut, for t ≥ 1/3 a regular octahedron of edge 3(1 − t)/2 and a cube with four alternate corners cut, with exact volumes. Theorem 2: at t = (2√6 − 3)/15 and t = 5/3 − (2/3)√2 every edge is equal; at t = 1 − ∛6/3 the two kinds each fill half of space. Theorem 3 (the Delian bond): at t = 1 − (2/3)·2−1/3, and only there, the two pieces have equal volume: Samson, the cut cube, twice the octet's tetrahedron, and Delilah, a regular octahedron, half the octet's octahedron. Exact in ℚ(√6), ℚ(√2), ℚ(∛2) and ℚ(∛6). With paper #48 this gives DICTO's six DICTONIC SOLIDS in three DICTONIC BONDS (golden, keyed, Delian). Not all six are new as shapes; what is new is each bond. A remark shows that the same weighted cut for the cube meets only known solids (truncated cubic, rectified cubic and Kelvin's honeycombs), so the count stays at three bonds; another remark gives the polar pair of #48. The family through the octet, its equal-edged and half-and-half members and the Delian bond were not found in the sources checked. The ideas, and the names Samson, Delilah, DICTONIC SOLIDS and DICTONIC BONDS, are DICTO's; computation, exact proofs and drafting with Claude (Anthropic). More at dictospheres.com. View in the app: https://kaleidohedra.dictospheres.com/?open=dictonic:delian
Related
- A Pythagoras Tree Grown from a Dodecahedron
- The rhombic dodecahedral conjecture
- Linear and Nonlinear Phase Mixing for the Gravitational Vlasov–Poisson System Under an External Kepler Potential
- Dynamics of planar integrable Kepler billiards with a focused hyperbolic branch
- Hidden Symmetries of the Kepler Problem and Jacobi's Theorem
BibTeX
@misc{dicto2026delian,
title = {{The Delian Bond: Samson, Delilah and the Octet Between Kepler's Rhombic Dodecahedron and the Cube}},
author = {DICTO},
howpublished = {Zenodo (CERN European Organization for Nuclear Research)},
year = {2026},
month = oct,
doi = {10.5281/zenodo.23606828},
url = {https://doi.org/10.5281/zenodo.23606828},
note = {CC BY-SA},
}
RIS
TY - UNPB
TI - The Delian Bond: Samson, Delilah and the Octet Between Kepler's Rhombic Dodecahedron and the Cube
AU - DICTO
PY - 2026
DA - 2026/10/24
PB - Zenodo (CERN European Organization for Nuclear Research)
DO - 10.5281/zenodo.23606828
UR - https://doi.org/10.5281/zenodo.23606828
AB - In short. Regular tetrahedra and octahedra fill space together: the octet. Join each tetrahedron's centre to each octahedron's centre through the face they share, cut every such double cone by one flat plane parallel to that face, and give each side to its own centre. Slide the cut, and every position still fills space with two convex solids, from Kepler's rhombic dodecahedra through the octet to cubes. At one position the two solids have exactly the same volume: the tetrahedron's piece has doubled and the octahedron's halved, the doubling of the ancient Delian problem in both directions at once. DICTO named this pair Samson and Delilah. Main theorem. For every cut position t from 0 to 1 the two pieces are convex and fill space face to face: for t ≤ 1/3 a regular tetrahedron of edge 3t and a rhombic dodecahedron with its eight three-fold corners cut, for t ≥ 1/3 a regular octahedron of edge 3(1 − t)/2 and a cube with four alternate corners cut, with exact volumes. Theorem 2: at t = (2√6 − 3)/15 and t = 5/3 − (2/3)√2 every edge is equal; at t = 1 − ∛6/3 the two kinds each fill half of space. Theorem 3 (the Delian bond): at t = 1 − (2/3)·2−1/3, and only there, the two pieces have equal volume: Samson, the cut cube, twice the octet's tetrahedron, and Delilah, a regular octahedron, half the octet's octahedron. Exact in ℚ(√6), ℚ(√2), ℚ(∛2) and ℚ(∛6). With paper #48 this gives DICTO's six DICTONIC SOLIDS in three DICTONIC BONDS (golden, keyed, Delian). Not all six are new as shapes; what is new is each bond. A remark shows that the same weighted cut for the cube meets only known solids (truncated cubic, rectified cubic and Kelvin's honeycombs), so the count stays at three bonds; another remark gives the polar pair of #48. The family through the octet, its equal-edged and half-and-half members and the Delian bond were not found in the sources checked. The ideas, and the names Samson, Delilah, DICTONIC SOLIDS and DICTONIC BONDS, are DICTO's; computation, exact proofs and drafting with Claude (Anthropic). More at dictospheres.com. View in the app: https://kaleidohedra.dictospheres.com/?open=dictonic:delian
ER -