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.

  • $\begingroup$ Are we dealing with regular tetrahedra, i.e. with equilateral triangles as faces? $\endgroup$ – Florian F Jul 27 at 21:22
  • $\begingroup$ @FlorianF Yes. Edited. $\endgroup$ – Jiminion Jul 27 at 21:26
  • $\begingroup$ Looks impossible to me. $\endgroup$ – Florian F Jul 27 at 21:30

I believe the answer is

that it is impossible.

The question is an instance of Hilbert's Third Problem.

Any triangular prism has a zero Dehn invariant because you can pair up the top and bottom edges and their contributions to the Dehn invariant cancel (the angles add to $\pi$ radians), and similarly the three vertical edges together cancel.

The set of three tetrahedra has a non-zero Dehn invariant. The angle between its faces is not a nice fraction of $\pi$ radians, so cannot cancel.

Therefore they cannot be transformed into each other through dissection.

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