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.
Let me know if you need more clarifications to solve this one
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I think we can do it with
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:
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
FAsuch that the base
ABCDis one face, and the point
E/Fis one of the top corners. Thus,
EBAare 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,
D), then we can
take another copy and place its base square on the
DCYZface and its apex at
E. This covers the second half of the
AEZDface 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
XYCBface. This covers the second half of the
EXYZ"top" and the second half of the
Another way to think of this is to take the original object and
duplicate it three times, each time rotating through the diagonal
ECaxis, 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
ECaxis 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.