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A convex polyhedron without Rupert's property

Jakob Steininger, Sergey Yurkevich

TL;DR

The paper settles Rupert's property for convex, origin-symmetric polyhedra by constructing the Noperthedron, a 90-vertex polyhedron that provably lacks Rupert's property via a rigorous, computer-assisted, rational proof. The authors develop global and local exclusion theorems that rule out Rupert configurations by analyzing projections in a five-dimensional parameter space, deriving sharp operator bounds, and expressing these with rational approximations to enable SageMath verification. A substantial algorithm builds a solution-tree over millions of regions, with rational checks ensuring completeness and correctness; this confirms the Noperthedron is not Rupert and yields a Rupert-but-not-locally-Rupert example, the Ruperthedron, with Nieuwland number at least $1.003$. The work demonstrates a robust framework for proving geometric-projection properties of 3D solids using symmetry reductions, derivative-based bounds, and exact rational verification, opening avenues for disproving Rupert's property for other solids and refining associated quantitative measures.

Abstract

A three-dimensional convex body is said to have Rupert's property if its copy can be passed through a straight hole inside that body. In this work we construct a polyhedron which is provably not Rupert, thus we disprove a conjecture from 2017. We also find a polyhedron that is Rupert but not locally Rupert.

A convex polyhedron without Rupert's property

TL;DR

The paper settles Rupert's property for convex, origin-symmetric polyhedra by constructing the Noperthedron, a 90-vertex polyhedron that provably lacks Rupert's property via a rigorous, computer-assisted, rational proof. The authors develop global and local exclusion theorems that rule out Rupert configurations by analyzing projections in a five-dimensional parameter space, deriving sharp operator bounds, and expressing these with rational approximations to enable SageMath verification. A substantial algorithm builds a solution-tree over millions of regions, with rational checks ensuring completeness and correctness; this confirms the Noperthedron is not Rupert and yields a Rupert-but-not-locally-Rupert example, the Ruperthedron, with Nieuwland number at least . The work demonstrates a robust framework for proving geometric-projection properties of 3D solids using symmetry reductions, derivative-based bounds, and exact rational verification, opening avenues for disproving Rupert's property for other solids and refining associated quantitative measures.

Abstract

A three-dimensional convex body is said to have Rupert's property if its copy can be passed through a straight hole inside that body. In this work we construct a polyhedron which is provably not Rupert, thus we disprove a conjecture from 2017. We also find a polyhedron that is Rupert but not locally Rupert.

Paper Structure

This paper contains 30 sections, 39 theorems, 180 equations, 6 figures, 1 table.

Table of Contents

  1. Introduction
  2. Setting and definitions
  3. Structure of the paper
  4. Outline of the proof
  5. The Noperthedron
  6. Definition of the Noperthedron
  7. Refined Rupert's property for the Noperthedron
  8. Bounding $R_a(\alpha), R(\alpha), M(\theta, \varphi)$
  9. The global theorem
  10. Definitions and lemmas
  11. Proof of the global theorem
  12. The local theorem
  13. Definitions and lemmas
  14. Spanning points
  15. Locally maximally distant points
  16. ...and 15 more sections

Key Result

Theorem 1

The Noperthedron does not have Rupert's property.

Figures (6)

  • Figure 1: Two projections of the unit cube.
  • Figure 2: The Noperthedron.
  • Figure 3: Two projections of the Octahedron: one in direction $(\overline{\theta}_1, \overline{\varphi}_1) = (0,0)$ (in red) and one in direction $(\overline{\theta}_2, \overline{\varphi}_2) = (\pi/4, \tan^{-1}(\sqrt{2})$ (in black).
  • Figure 4: A polygon $\mathop{\mathrm{\mathcal{P}}}\nolimits = \{P_1, Q_1, Q_2, P_2, P_3\} \subset \mathop{\mathrm{\mathbb{R}}}\nolimits^2$. Here $O$ is the origin. Note that $Q_1$ is $\delta$-LMD but $Q_2$ is not, since $\|A\| > \|Q_2\|$ and $A \in \mathop{\mathrm{\mathrm{Sect}}}\nolimits_\delta(\overline{Q_2})$. As it is apparent from the proof of \ref{['lem:LMD']} this is because $\angle(O Q_2 P_2)$ is obtuse.
  • Figure 5: The projection of the octahedron $\mathop{\mathrm{\mathbf{O}}}\nolimits$ in direction $(\overline{\theta}, \overline{\varphi}) = (\pi/4,\tan^{-1}(\sqrt{2})$.
  • ...and 1 more figures

Theorems & Definitions (93)

  • Theorem 1
  • Definition 2
  • Definition 3
  • Definition 4
  • Definition 5
  • Definition 6
  • Lemma 7
  • proof
  • Corollary 8
  • proof
  • ...and 83 more