# Concerning Tetrahedra

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.

• Are we dealing with regular tetrahedra, i.e. with equilateral triangles as faces? – Florian F Jul 27 '20 at 21:22
• @FlorianF Yes. Edited. – Jiminion Jul 27 '20 at 21:26
• Looks impossible to me. – Florian F Jul 27 '20 at 21:30

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.