Minimal data required to determine a convex polytope

Question

Let $P\subset \Bbb R^d$ be a convex polytope. Suppose that I know

1. its combinatorial type (aka. the face-lattice),
2. the length $\ell_i$ of each edge, and
3. the distance $r_i$ of each vertex from the origin.

Question: Does this already determine $P$ (up to orthogonal transformation)?

This is the case if all $\ell_i$ are the same, and all $r_i$ are the same (see this question). But what if they are not the same? What if I do not know the combinatorial type but only the edge-graph?

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Update

I am not sure whether the formulation of my question was too vague, so below I added a second equivalent version of what I am asking:

Given two combinatorailly equivalent polytopes $P_1,P_2\subset\Bbb R^d$, and a corresponding face-lattice isomorphism $\phi:\mathcal F(P_1)\to\mathcal F(P_2)$. Now suppose that each edge $e\in\mathcal F_1(P_2)$ has the same length as $\phi(e)\in\mathcal F_1(P_2)$, and that each vertex $v\in\mathcal F_0(P_1)$ has the same distance from the origin as $\phi(v)\in\mathcal F_0(P_2)$. Is it then true that $P_1$ and $P_2$ are congruent (related by an orthogonal transformation)?

Answer 1

Multiple polytopes can have the same data, as pictured below.

Take a pyramidal frustum, and twist it slightly clockwise or slightly counterclockwise. Make one polytope by gluing two identical versions, and make another polytope by gluing two opposite versions.

These will have the same combinatorial type, the same edge lengths, and the same distances from the origin to the edges, but they are not orthogonally equivalent.

The images show polytopes with vertices at \begin{align} &(\cos (k+\frac15)\alpha,&\sin (k+\frac15)\alpha, &\ \ \ \ +1)\\ &(\ \ \ \ 3\cos k\alpha,&3\sin k\alpha,\ \ \ \ &\ \ \ \ \ \ \ \ \ 0)\\ &(\cos (k\pm\frac15)\alpha,&\sin (k\pm\frac15)\alpha, &\ \ \ \ -1) \end{align} with $\alpha=\pi/2$, and $+$ for one polytope, $-$ for the other.

Answer 2

Not an answer, just an additional illustration for the existing answer by @MattF. (which I find exhaustive).

Answer 3

A 2-dimensional counterexample

In the image below, the white dot represents the origin and is located outside the polygon. And it has to be: if the origin were inside, the shape would be unique, as shown here.

One can imagine to build higher dimensional counterexamples from this, e.g. prisms over these shapes.

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Proof of a special case

Suppose that for each 2-face $\sigma\in\mathcal F(P)$ the (perpendicular) projection of the origin onto $\mathrm{aff}(\sigma)$ ends up in the relative interior of $\sigma$. Then the polytope is uniquely determined by its edge-lengths and vertex-origin-distances.

Proof.

Let $P$ be a $d$-polytope.

Each 2-face of $P$ and the origin form a (possibly degenerate) pyramid, in which all edge-length are known, and the apex projects to the interior of the base. This case is discussed in this question where it is also proven that the base face of the pyramid is uniquely determined.

If $d=2$, we are done. If $d\ge 3$, we can apply Cauchy's rigidity theorem in the 2-face version, that is, the last version mentioned over here, to obtain that $P$ is uniquely determined.

$\square$

In fact, using much stronger tool it is possible to show that it suffices that the origin lies strictly inside the polytope (or in a facet). Then the combinatorial type, edge lengths and vertex-origin distances determine the polytope up to isometry. A proof of this is given in

• M. Winter, "Rigidity, Tensegrity, and Reconstruction of Polytopes Under Metric Constraints", International Mathematics Research Notices, https://doi.org/10.1093/imrn/rnad298.