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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.

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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