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.



Convex Polyhedra Scissors Congruence Problem
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.


