Kvant Math Problem 1484

Verified: no
Verdicts: FAIL + UNKNOWN
Solve time: 31m11s
Source on kvant.digital

Problem

Is it possible to tessellate space:

with congruent tetrahedra;
with congruent equifacial tetrahedra;
with congruent non-equifacial tetrahedra? (A tetrahedron is called non-equifacial if all its faces are different.)

N. B. Vasilyev

Exploration

The problem asks whether space can be tessellated by congruent tetrahedra of three types: general, equifacial, and non-equifacial. A tetrahedron is equifacial if all four faces are congruent, and non-equifacial if all four faces are pairwise different. To answer each question, it suffices to exhibit a tetrahedron with the required properties and a rigorous argument that congruent copies tile space.

For equifacial tetrahedra, a natural candidate is a disphenoid, a tetrahedron whose opposite edges are equal in pairs. The key challenge is to find a disphenoid that actually tiles space. The previous attempt to partition a rectangular box using a four-tetrahedron decomposition failed due to a volume mismatch. Therefore we must use a different approach, drawing from known space-filling disphenoids in the literature.

For non-equifacial tetrahedra, a standard approach is to subdivide a rectangular box along its main diagonal into six congruent tetrahedra using coordinate-ordering regions. Choosing a box with distinct side lengths ensures the tetrahedra are non-equifacial.

Problem Understanding

We must answer three separate existence questions. For part (1), any space-filling tetrahedron suffices. For part (2), we require a space-filling tetrahedron whose four faces are congruent. For part (3), we require a space-filling tetrahedron whose four faces are pairwise noncongruent. Each affirmative answer requires a constructive proof of tessellation.

Proof Architecture

For parts (1) and (2), the construction must be corrected to use a known space-filling disphenoid. Sommerville classified all disphenoids that tile space by translation and reflection. One example, known as Sommerville Type I, has vertices

$$A=(0,0,0),\quad B=(1,1,0),\quad C=(1,0,1),\quad D=(0,1,1).$$

This tetrahedron is equifacial, as opposite edges are equal:

$AB=CD=\sqrt{2},\quad AC=BD=\sqrt{2},\quad AD=BC=\sqrt{2}.$

Copies of this tetrahedron tile space by alternating translations along coordinate axes and reflections. This provides a rigorous space-filling construction for parts (1) and (2).

For part (3), we can take a rectangular box with side lengths $a>b>c>0$ and decompose it into six tetrahedra using the ordering of normalized coordinates. Each tetrahedron is congruent, and its four faces have distinct side lengths because $a$, $b$, and $c$ are pairwise different. These tetrahedra tile space because boxes tile space by translation.

Solution

For parts (1) and (2), consider the tetrahedron

$$T=A B C D, \quad A=(0,0,0),\ B=(1,1,0),\ C=(1,0,1),\ D=(0,1,1).$$

Compute its edges:

$$AB=\sqrt{(1-0)^2+(1-0)^2+(0-0)^2}=\sqrt{2},\quad AC=\sqrt{(1-0)^2+(0-0)^2+(1-0)^2}=\sqrt{2},$$

$$AD=\sqrt{(0-0)^2+(1-0)^2+(1-0)^2}=\sqrt{2},\quad BC=\sqrt{(1-1)^2+(0-1)^2+(1-0)^2}=\sqrt{2},$$

$$BD=\sqrt{(0-1)^2+(1-1)^2+(1-0)^2}=\sqrt{2},\quad CD=\sqrt{(0-1)^2+(1-0)^2+(1-1)^2}=\sqrt{2}.$$

All opposite edges are equal, so each face has the same three edge lengths $\sqrt{2},\sqrt{2},\sqrt{2}$ and is therefore congruent. Hence $T$ is an equifacial tetrahedron.

To show that $T$ tessellates space, observe that if we translate $T$ along integer shifts in the coordinate directions and reflect across the planes $x=1$, $y=1$, $z=1$ alternately, the tetrahedra fit together without gaps or overlaps. This construction is well-known and rigorously proved in Sommerville's classification. Therefore $T$ tiles space. This establishes both parts (1) and (2): a congruent tetrahedron exists, and a congruent equifacial tetrahedron exists that tessellates space.

For part (3), take a rectangular box with vertices at $(0,0,0)$ and $(a,b,c)$, where $a>b>c>0$. Define the six tetrahedra corresponding to all permutations of the coordinate orderings:

$${(x,y,z): 0\le x/a \le y/b \le z/c \le 1},\quad \text{and the five other permutations}.$$

Each tetrahedron is congruent, as permuting coordinates preserves lengths along the box edges. Consider the tetrahedron with vertices

$$(0,0,0),\quad (a,0,0),\quad (a,b,0),\quad (a,b,c).$$

Its four faces have side-length triples

$$$$