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Going over degrees of freedom, it appears that the following construction works, but I have no idea how to calculate exact positions of the vertices, or even a practical approach to approximating them well.

There's a deformed cube, meaning the vertices are all moved around a bit but the four vertices of each face are still planar, and it has the property that the length of UFL (up front left) to UBL (up back left) equals UFL to DFL equals DFL to DFR. Also UFR to UBR equals UBR to DBR equals DBR to DBL. Also the other six edges are all the same length. But not all edges are the same length. I believe there's continuum of these defined by a single parameter which can be, for example, the ratio in length between two of the edges.

Any idea how to calculate exact vertex positions, or at least dimensions of the faces? This is meant to be a 'find the symmetry' puzzle, where each face misleadingly hints at a possible symmetry of the puzzle as a whole but the actual one is something different, and it can be made either as a solid block or a fold-out on paper.

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    $\begingroup$ I will point out that the tag (geometry) is deprecated on MO, see the tag-info. So choosing some other tag might be suitable. (I am not exactly sure which tag to choose, but perhaps (mg.metric-geometry) would be reasonable.) $\endgroup$
    – Martin Sleziak
    Mar 5, 2019 at 9:22
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    $\begingroup$ One could write out all your constraints in terms of vertex coordinates, and find solutions to the systems of equations. That the faces are planar is expressed as the volume of the tetrahedron determined by the four vertices is zero. The six equal-length edges could be fixed to $1$. The length ratios can be expressed algebraically. One vertex could be placed at $(0,0,0)$ wlog. So this is a collection of polynomial equations. I don't see an impediment to writing out all the equations and attempting a solution. $\endgroup$
    – Joseph O'Rourke
    Mar 9, 2019 at 3:05
  • $\begingroup$ Slogging through the polynomials is a lot more than I for one can pull off by hand! $\endgroup$
    – Bram Cohen
    Mar 10, 2019 at 7:29

1 Answer 1

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May as well put up a diagram. (Looks like getting the handedness correct messed up the edgematching).

deformed cube

It brings up more questions:

Are the two blue-green faces the same?
Are the two red-green faces the same?
Are the red-blue-green faces the same?

Let dfl={0,0,0}, dbl={0,1,0}, ufl={0,cos(a),sin(a)} ubl={0,b cos(a/2),b sin(a/2)}

That's a start.

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  • $\begingroup$ Yes to all three. I believe that's enforced by the requirements given, but if not then it should be a consequence of the unstated requirement, which is that the object as a whole needs to be 180 degree rotationally symmetric about an axis going through the midpoints of ubl-dbl and ufr-dfr $\endgroup$
    – Bram Cohen
    Jun 28, 2019 at 22:04

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