# Equidecomposability of a Cube into 6 Trirectangular Tetrahedra

In grade school we learn that a square can be equidecomposed into two congruent right isosceles triangles. Does the following three dimensional generalization hold?

Consider a trirectangular tetrahedron with vertices at $(0,0,0)$, $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$ and volume $\frac{1}{6}$. (It has three orthogonal right triangular faces and one equilateral triangular face opposite the (tri-right-angled) common vertex.) The inscribing cube of the tetrahedron has unit side length and unit volume. Is there a dissection of this cube into 6 tetrahedra of the above form?

Provided that such a dissection exists and this is the way it is to be done, it's easy to see how 4 tetrahedra can be glued together to give the boundary of the cube, but I don't quite see how the other two should fit inside.

• get a piece of cheese and cut it up – yoyo Mar 9 '11 at 21:37
• Just making sure everyone is aware: the easiest generalization of the 2-dimesional splitting of a square into two right triangles comes out of noting that you can split the square region ($0<x,y<1$) into two congruent triangular regions: $x>y$ and $y>x$. This generalizes to 3 dimensions by splitting the cubic region ($0<x,y,z<1$) into 6 congruent tetrahedral regions $x<y<z$, $y<x<z$, $x<z<y$, $z<x<y$, $y<z<x$, and $z<y<x$. This easily generalizes into higher dimensions. – Eric Nitardy Mar 10 '11 at 2:44
• You might find this page of interest: scienceu.com/geometry/activities/tetrapuzzles and also click on "answer." – Joseph Malkevitch Mar 10 '11 at 4:06
• @Joseph Malkevitch: Interesting -- but I wouldn't trust a site that speaks of a "regular cube" to correctly enumerate its decompositions into tetrahedra :-) – joriki Mar 10 '11 at 23:43

## 4 Answers

That such a decomposition doesn't exist can be shown by looking at the interior solid angles at the vertices.

The cube has eight interior solid angles of $\pi/2$ each.

Each of the six tetrahedra has one interior solid angle of $\pi/2$ and three that can be computed as shown here:

$$\Omega=\phi_{ab}+\phi_{bc}+\phi_{ac}-\pi=\pi/2 + \arccos\frac{1}{\sqrt{3}}+ \arccos\frac{1}{\sqrt{3}}-\pi=2\arccos\frac{1}{\sqrt{3}}-\pi/2\;.$$

Thus, the sum of the interior solid angles of the cube is $4\pi$, and the sum of the interior solid angles of the tetrahedra is

$$6\left(\pi/2+3\left(2\arccos\frac{1}{\sqrt{3}}-\pi/2\right)\right)=6\left(6\arccos\frac{1}{\sqrt{3}}-\pi\right)\;.$$

This is not equal to $4\pi$, since $\arccos\frac{1}{\sqrt{3}}$ is not a rational multiple of $\pi$. (Here's an elementary proof of that fact.) In fact it is almost $5\pi$.

[Edit: I just realized in the shower that all the following analysis of potential inner vertices is actually not necessary. No matter what happens at the inner vertices, we have to fill the eight vertices of the cube with vertices of the tetrahedra, and the six vertices with interior solid angle $\pi/2$ are not enough to do it, so we have to use at least one of the others, but then we can't make the sum come out to $4\pi$, a rational multiple of $\pi$.]

This establishes that you can't fill the cube with the six tetrahedra by letting all their vertices coincide with the cube's vertices. It seems geometrically obvious that you can't do it with any of the vertices inside the cube, either, but this, too, can be proved rigorously using the solid angles. If there were a vertex inside the cube, the entire solid angle of $4\pi$ around that inner vertex would have to be filled. The candidates for filling it are a face of a tetrahedron, which subtends a solid angle of $2\pi$, an edge of a tetrahedron, which subtends a solid angle of twice the dihedral angle of the intersecting planes, i.e. either $\pi$ or $2\arccos\frac{1}{\sqrt{3}}$, or an interior solid angle of a vertex of a tetrahedron. Thus, adding up all the solid angles at the eight outer vertices and at $v$ inner vertices, the following equation would have to have solutions with non-negative integer values of $j$, $k$, $l$, $m$, $n$ and $v$:

$$6\left(6\arccos\frac{1}{\sqrt{3}}-\pi\right)+j(2\pi)+k(\pi)+l\left(2\arccos\frac{1}{\sqrt{3}}\right)+m(\pi/2)+n\left(\arccos\frac{1}{\sqrt{3}}-\pi/2\right)$$

$$=4\pi+v(4\pi)\;.$$

This is impossible, since the coefficient in front of $\arccos\frac{1}{\sqrt{3}}$ is non-zero, and thus the equation would imply that this is a rational multiple of $\pi$, which it isn't (see above).

P.S.: I just realized I forgot to mention an essential part of the proof:

$$\cos 2\phi = 2 \cos^2 \phi - 1\;,$$

$$\arccos x = \frac{1}{2}\arccos(2x^2-1)\;,$$

$$\arccos \frac{1}{\sqrt{3}}=\frac{1}{2}\arccos(-\frac{1}{3})\;,$$

so $\arccos \frac{1}{\sqrt{3}}$ is rational iff $\arccos(-\frac{1}{3})$ is rational; then we can apply the theorem I linked to above.

• Someone just removed an upvote from this answer -- would be nice to know what made you change your mind :-) – joriki Mar 11 '11 at 13:53
• +1 for the write up. However, I'm not sure I agree with the following: "we have to fill the eight vertices of the cube with vertices of the tetrahedra, and the six vertices with interior solid angle π/2 are not enough to do it". You can use four trirectangular tetrahedra (by gluing only along an edge of each pair of tetrahedra to make a square face) to fill all of the vertices of the cube. What is left over is an internal regular tetrahedral void and two unused trirectangular tetrahedra. – user02138 Mar 11 '11 at 13:54
• @user02138: You can glue together two edges to make a square face, but that doesn't mean you've filled all the vertices of the cube. By "filling", I mean taking up the entire solid angle of the vertex. The "internal regular tetrahedral void" that you describe reaches up to the cube's vertices, so they aren't filled. And what's left of them is such that you can't fill it with any of the vertices of the two remaining tetrahedra. – joriki Mar 11 '11 at 13:59
• Agreed. Well done. – user02138 Mar 11 '11 at 14:04

I'm pretty sure there isn't. The most straightforward way of proving this is probably to (a) show that any embedding of the trirectangular tetrahedron in the cube is 'extremal' (that is, that its right-triangle faces are of necessity aligned with the faces of the cube) and then (b) show that any possible extremal embedding of a trirectangular tetrahedron misses some point in the cube. (a) is a little bit tricky, but once you've got that, (b) is easy (what point is most likely to be missed?).

Here is what I've discovered so far. Four trirectangular tetrahedra glued into a cubical shape leaves a regular tetrahedral void (a 3-simplex actually) within the cube. There is no way to get two more trirectangular tetrahedra into that space without further dissection (a proof follows by considering angles).

This, however, doesn't settle the question because I've looked at only one gluing.

The slickest proof of the above fact is that the Dehn invariant of the trirectangular tetrahedron is non-zero, while that of the cube is zero. Since the Dehn invariant is additive in equidecomposition, the result follows.

• As I said, I think the key is to prove that every means of placing a trirectangular tetrahedron in the cube forces it to be aligned into a corner with its three rectilinear sides along the three sides of the cube; once you've done that, the rest is easy, but I don't think looking at specific gluings of multiple tetrahedra will get you far in and of itself. – Steven Stadnicki Mar 9 '11 at 22:38
• @Steven: Agreed. – user02138 Mar 9 '11 at 22:39

You could also consider decomposition into 'orthoschemes'. One of these tetrahedra has vertices (1,0,0), (0,0,0), (0,1,0), (0,1,1). A cube will decompose into three shapes directly isometric to this, and another three to the reflection of this shape.

This decomposition is somewhat similar to the decomposition into trirectangular tetrahedra you asked about: all the vertices in this decomposition are vertices of the original cube (as you might have hoped would be the case with the trirectangular tetrahedra).

There is an illustration of this on Wikipedia: https://en.wikipedia.org/wiki/Schl%C3%A4fli_orthoscheme.