Nieuwland numbers for classical solids (Octahedron, Dodecahedron, Icosahedron)

Prove that (i) the Octahedron has Nieuwland number 3√2/4, and (ii) the Dodecahedron and Icosahedron both have Nieuwland number ν ≈ 1.0108, where ν is a root of P(x) = 2025x^8 − 11970x^6 + 17009x^4 − 9000x^2 + 2000.

Background

The Nieuwland number quantifies the maximal scaling factor achievable in Rupert-type passages. While numerical or heuristic values are known or conjectured for several Platonic solids, rigorous proofs for these specific values remain elusive.

The authors highlight these as representative open cases in the broader program of determining exact Nieuwland numbers for classical polyhedra.

References

It is, for instance, still open to prove that the Octahedron has Nieuwland number $3\sqrt{2}/4$ or that the Dodecahedron and Icosahedron both have Nieuwland number $\nu \approx 1.0108$, a root of ( P(x) = 2025x8 - 11970x6 + 17009x4 - 9000x2 + 2000 ).

— A convex polyhedron without Rupert's property  (2508.18475 - Steininger et al., 25 Aug 2025) in Section 9: Discussion, Open Problems

As it currently seems, neither our approach, nor the one of can be used to prove that \nu(\text{Icosahedron}) = \nu(\text{Dodecahedron}) = \xi as given in (\ref{eq:conj:dod}). Thus, this dual pair of Platonic solids remains open.

— The Nieuwland number of the Octahedron  (2609.10788 - Steininger et al., 9 Sep 2026) in Section 4, Remarks and discussion, first item

We conjecture that any projection of the Octahedron satisfies

\sup_\pi \frak{o}(\pi(\OOO)) \leq \frac{3\sqrt{2}{4} \inf_\pi \frak{o}(\pi(\OOO)),

which would yield another proof of the Nieuwland number of the Octahedron.

— The Nieuwland number of the Octahedron  (2609.10788 - Steininger et al., 9 Sep 2026) in Section 4, Remarks and discussion, fifth item