Cube to tetrahedron
Puzzle
Cut cube in two pieces from which tetrahedron can be
assembled.
Solution
The third Hilbert problem states "Can solid tetrahedron be cut into finite
number of pieces from which cube can be assembled?". The answer to this problem
was found by Max Dehn a few months after the problem was stated (in fact, this
was the first solved Hilbert problem) and this answer was negative. Dehn
proved that it is impossible.
This puzzle does not ask you to solve impossible problem. Instead it asks if
surface of solid cube can be cut in to pieces, which then can be
unfolded into plane, rearranged and folded back into tetrahedron.
Cut cube along some edges and unfold it into the following figure.
(It can be folded back into a cube along black dashed lines.)
If you cut it along solid blue line and rearrange parts as this
the resulting parallelogram can be folded into tetrahedron along blue dashed
lines.
Improvements
Can number of pieces be further reduced? Down to 1? At first it may sound
crazy, but maybe, it is possible to come up with unfolding that can be folded
into both cube and tetrahedron. There certainly exists a polygon that can be
folded both into cube and irregular tetrahedron. If you find common
unfolding for cube and regular tetrahedron, please drop me a note.
Possible variants
If we allow immeasurable sets as pieces this porblem might have a solution for
solid cube and tetrahedron. After all, when you can cut a solid sphere into 5
pieces from which you can assemble 2 solid spheres of the same radius,
then merely turning cube into tetrahedron does not seem implausible. I haven't
tried this path, quite possibly, there is a solution there too.
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