Equality of Nieuwland numbers for the tetrahedron, cuboctahedron, and truncated octahedron

Prove that \(\nu(\text{Tetrahedron})=\nu(\text{Cuboctahedron})=\nu(\text{Truncated Octahedron})\).

Background

The authors define the difference body of a polyhedron and observe that the difference body of the Tetrahedron is the Cuboctahedron. They then conjecture equality of the Nieuwland numbers of the Tetrahedron, Cuboctahedron, and Truncated Octahedron. The manuscript explicitly states that the authors are still working on proving this conjecture and do not establish it.

References

Coincidentally, D(\text{Tetrahedron}) = \text{Cuboctahedron} and we actually conjecture that \begin{align} \label{eq:conjtetra} \nu(\text{Tetrahedron}) = \nu(\text{Cuboctahedron}) = \nu(\text{Truncated Octahedron}). \end{align}

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