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The diagram is the outline of the surface of a 3D object. Several objects, like the one created from this given surface, may be used to create a cube.

enter image description here

Let me know if you need more clarifications to solve this one

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  • $\begingroup$ Is this the unwrapped surface? I first read this as saying that this is the projection of the 3D object onto a 2D surface. $\endgroup$ – Acccumulation Jun 10 at 22:21
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Here's a silly looking animation with

3

copies of the piece:

enter image description here

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I think we can do it with

3

copies of the given 3D object.


Note: I don't have a mathematical background, so this is mostly an intuitive "solve"and may not be very rigorous (or even correct). It also assumes that lengths that are visually equal, are (e.g. FD = EB = EJ, FA = EA = AB = BC = CD = DA = BH = JH).

Firstly fold the object on the red lines, like so:

folds

Assuming (as stated above), edges FA = AB = AD, etc, then the resulting polyhedron is

a skewed square-based pyramid, which fits neatly inside a cube of side length FA such that the base ABCD is one face, and the point E/F is one of the top corners. Thus, FAD and EBA are both half faces of the cube, meaning we have two totally covered faces of the cube (so even intuitively we can see that three copies should cover all six faces).

If we call the remaining points of the cube, X (above B), Y (above C), and Z (above D), then we can

take another copy and place its base square on the DCYZ face and its apex at E. This covers the second half of the AEZD face and half the top (EXYZ). Then we take a third copy and slide it in such that its vertex is again at E, but with its base covering the whole of the XYCB face. This covers the second half of the EXYZ "top" and the second half of the EXBA face.

Another way to think of this is to take the original object and

duplicate it three times, each time rotating through the diagonal EC axis, moving the base to a new face of the cube each rotation.

As I said at the top, I'm not sure how to approach a rigorous proof, but it's hopefully intuitive that

because all copies share the central EC axis and each internal face is at 45°, the copies can't be overlapping each other. Similarly, the large internal 45° triangles they must be completely face to face, meaning we can't be left with any gaps, so we've filled the cube entirely.

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