Is a pyramid uniquely determined by its edge-lengths?

Question

I am looking for a nice/short proof of the following:

The shape of a pyramid with a convex polygonal base is already uniquely determined by knowing the length of all its edges.

By "knowing the length of each edge", I mean that I know the edge-graph of the pyramid, and to each edge of the graph I know its length. So we cannot freely permute the lengths among the edges.

Also, I consider mirror images as the same "shape".

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Proof of a special case

Suppose that the perpendicular projection of the apex onto the (affine hull of the) base ends up inside the base (in its relative interior).

In that case, we can mirror the pyramid on its base to obtain the corresponding double-pyramid, which again is a convex polyhedron.

All the faces of the double-pyramid are triangles, and so their shapes are uniquely determined by the edge lengths (which we know). Then the whole double-pyramid is uniquely determined by Cauchy's rigidity theorem, which then uniquely determined the original pyramid as well.

Apparently, this no longer works if the projection of the apex is outside of the base (or on its boundary), as the mirroring trick no longer gives a convex polyhedron, or might give a polyhedron with non-triangular faces.

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Discussing degrees of freedom

We can assume that the vertices of the base are contained in a fixed plane, and that only the apex is hovering above the plane. This gives the configuration $2n+3$ degrees of freedom.

Now, we also have $2n$ length constraint. The remaining three degrees of freedom are exactly translation parallel to the plane (two degrees of freedom), and rotation around an axes spanned by the normal vector of the plane (one degree of freedom).

So the configuration has no degrees of freedom to deform continuously, but this does not exclude the case of multiple rigid configurations with the same given edge-lengths.

Answer 1

I found a counterexample in the case when the apex is not "above" the base face:

Here are two non-congruent polygons with pair-wise identical edge-lengths and pair-wise identical vertex-distances to the white dot. You can imagine to raise the white dot out of the plane do create two non-congruent pyramids with identical edge lengths.

Answer 2

Here's a simple counterexample: The base is a many-sided polygon inscribed in a circle and the lengths of sides are distinct; the apex is a point not in that plane. ("Many"${}={}$at least four.)

Now interchange two adjacent edges of the base to get a different shape.