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Question: Which Euclidean tetrahedra are scissor congruent to cubes?

Consider a Euclidean tetrahedron $T$ in $\mathbb{R}^3$ with edge lengths $l_1,\ldots, l_6$ and dihedral angles $\alpha_1,\ldots, \alpha_6.$ According to Sydler's theorem this is equivalent to vanishing of the Dehn invariant $$ D(T)=\sum l_i \otimes \alpha_i \in \mathbb{R}\otimes \mathbb{R}/2\pi \mathbb{Z}. $$ Denote by $r(T)$ the dimension of the $\mathbb{Q}-$span $\langle \alpha_1,\ldots,\alpha_6, \pi \rangle.$ It is easy to see that $1 \leq r(T)\leq 6.$

A list of partial results:

  1. If $r(T)=1$ then all angles of $T$ are rational multiples of $\pi,$ such tetrahedra were classified by Coxeter (this is explained here).

  2. Some examples of $1-$dimensional families of tetrahedra with $r(T)=2,3$ are presented here.

  3. A similar question about right-angled pyramid appeared on MOF recently and was beautifully solved. It is easy to see that such pyramid is
    scissor congruent to a centrally symmetric Schläfli orthoscheme.

It might be simpler to restrict to a certain family of examples, like Schläfli orthoschemes.

Simpler questions:

  1. What examples of Euclidean tetrahedra scissor congruent to cubes are known?

  2. What is the biggest possible dimension of an algebraic family of tetrahedra scissor congruent to cubes?

  3. Which tetrahedra are known not to be scissor congruent to cubes?

  4. Can a partial answer be obtained for certain values of $r(T)?$

F. C.
  • 3,507
Daniil Rudenko
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  • There's a guy in my department who knows all about the tetrahedrons; I might ask him. – user145307 Sep 04 '19 at 12:50
  • Oh, who is that? – Daniil Rudenko Sep 04 '19 at 12:56
  • Imagine someone who you could not surprise with any fact about tetrahedrons --- that is the guy. Possibly you already know him but for my own anonymity reasons I will demur to name him. – user145307 Sep 04 '19 at 14:51
  • Maybe, I can suggest a bet? If I tell a fact about tetrahedron, which will surprise your colleague, you will tell their name:) – Daniil Rudenko Sep 04 '19 at 16:13
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    You have a bet! I should say that classes do not begin for several weeks so you have to give me a month. If this fact is unknown to my colleague, then you will get an email from him personally. If not, then you owe me a coffee, which I am allowed to claim if we get a chance to meet in person and I reveal my identity. (Or, if you prefer, you can send me $0.00035162733$ of a bitcoin and I can buy one for myself.) – user145307 Sep 04 '19 at 17:52
  • Classes start this week and I await your interesting Tetrahedral fact! – user145307 Sep 21 '19 at 14:50
  • What about this one: https://mathoverflow.net/questions/336464/a-curious-relation-between-angles-and-lengths-of-edges-of-a-tetrahedron/336753#336753 – Daniil Rudenko Sep 21 '19 at 15:33
  • I determined that my colleague already knows this fact. Indeed, I got the impression that he thought he came up with it himself... – user145307 Oct 01 '19 at 14:40
  • Then you definitely should suggest him to write me ASAP, since I am going to submit the paper quite soon. – Daniil Rudenko Oct 01 '19 at 14:42
  • I promise to talk to him tomorrow at tea and then will get back to you then – user145307 Oct 01 '19 at 14:43
  • According to our conversation today, I think that we can safely say you lost the bet :) – user145307 Oct 03 '19 at 01:46
  • I totally agree with you, coffee is on me! – Daniil Rudenko Oct 03 '19 at 01:57
  • Are you sure that Coxeter classified the rationally angled tetrahedra? This recent preprint is claimed to be the first one which achieves this. Do you maybe mean tetrahedra with angles $\pi/n$? – M. Winter Feb 28 '21 at 10:12
  • You are right, I should correct myself. I saw this fact mentioned in some paper and did not check. – Daniil Rudenko Feb 28 '21 at 10:13

0 Answers0