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I am currently writing a geometry paper "Rectifications of Convex Polyhedra" and I am confused to have discovered what appears to be a remarkable discrete geometric fact:

Conjecture: Let $P$ be a convex polyhedron. Then $P \cong P^{\circ} \sqcup R_{1}[P]$.

The relation $\cong$ is to denote scissors congruence, the dual polyhedron is $P^{\circ}$, and $R_{1}[P]$ is the first rectification of $P$, defined as $$R_{1}[P] = \text{conv} \left\{\frac{x_{i} + x_{j}}{2} \; | \; (i,j) \in E(P) \right\}.$$

I have been drawing many examples of when this happens to work out (cube decomposing into a rectified cube (cuboctahedron) and eight pieces (orthoschemes) which rearrange to be an octahedron; octahedron decomposing into a rectified octahedron (cuboctahedron) and six pieces which rearrange to be a cube; tetrahedron decomposing into a rectified tetrahedron (octahedron) and four pieces which rearrange to be a tetrahedron. I've also tried various examples of prisms and bipyramids which all appear to work, I believe if a counterexample exists it will be non-centrally symmetric and I am working on a proof in the case of centrally symmetric convex polyhedra.

I am interested in classifying which convex polyhedra this holds for if there is a counterexample to the conjecture.

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    $\begingroup$ How does the volume work out? Are you scaling the RHS pieces? $\endgroup$ – Douglas Zare May 8 '15 at 18:45
  • $\begingroup$ As @DouglasZare says, you need to specify how the polyhedra are scaled. If you only had one polyhedron on the RHS, the scaling would be unique to get the volumes to be the same; but as it stands, you haven't specified the relative scaling between $P^\circ$ and $R_1[P]$. $\endgroup$ – Dylan Thurston May 8 '15 at 19:59
  • $\begingroup$ I think that scaling is important but for any unit tetrahedron there is not a scissors decomposition into a smaller regular tetrahedron and a regular octahedron see my revised answer. $\endgroup$ – Kristal Cantwell May 8 '15 at 20:20
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If the conjecture is true then the for the unit tetrahedron it will have a unit tetrahedron decomposible into a smaller regular tetrahedron and a regular octahedron. The regular octahedron of side $a$ will have Dehn invariant equal to $-2$ times the Dehn invariant of the tetrahedron side $a$. Hence its Dehn invariant will be -$2a$ times the Dehn invariant of the unit tetrahedron. the tetrahedron will have side $b$ less than one and its Dehn invariant will be $b$ times the Dehn invariant of the tetrahedron of side $1$. So one one side of the equation we will have the Dehn invariant of the unit tetrahedron and on the other side $b-2a$ times the Dehn invariant of the unit tetrahedron where $b$ is less than one and $a$ is positive and we cannot have equality.

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  • $\begingroup$ I think you need to pay attention to the relative size of the pieces, as indicated in Doug Zare's comment. $\endgroup$ – Dylan Thurston May 8 '15 at 19:15
  • $\begingroup$ The only thing that matters in three dimensions is the dehn invariant. $\endgroup$ – Kristal Cantwell May 8 '15 at 19:20
  • $\begingroup$ Thank you for your counterexample. Do you think there is any class of polyhedra for which the conjecture holds? Maybe simple/simplicial polyhedra, or self-dual polyhedra. $\endgroup$ – Samuel Reid May 8 '15 at 19:20
  • $\begingroup$ I think the simple polyhedra include the dodecahedron and in that case the same argument gives a contradiction. $\endgroup$ – Kristal Cantwell May 8 '15 at 19:25
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    $\begingroup$ @Dylan Thurston: There are times when it seems intuitive to project from the natural domain to $\mathbb{R}$, such as in the oft-proved result (of Debrunner) that if a polyhedron tiles $\mathbb{R}^3$, its Dehn invariant is $0$. If you think of the Dehn invariant living in $\mathbb R$, then it is natural to use the square-cube ratio to say that the projected Dehn invariant must be arbitrarily small, hence $0$. $\endgroup$ – Douglas Zare May 8 '15 at 21:44

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