Puzzling Asked on September 27, 2021
As a pyramid with a triangular base, the volume of a tetrahedron, like all pyramids, is $(1/3)*BH$, where $B$ is the base area and $H$ is the height.
If one had $3$ square $45$ degree pyramids (square base), it is not too hard to see how one could slice two of the pyramids with a single slice (each) and then reassemble the $5$ pieces into a rectangular block that has the same square base and a height $H$.
So, if one had $3$ equal size regular tetrahedra, how many slices are needed to create the $N$ pieces needed to reassemble them into a triangular prism of height $H$, where $H$ is the height of the initial tetrahedra?
NOTE: One might answer: "$N-2$ slices", but I need a bit more information than that.
I believe the answer is
The question is an instance of Hilbert's Third Problem.
Correct answer by Jaap Scherphuis on September 27, 2021
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