Packaging of tetrahedra in $\mathbb Z^3$

Question

We are allowed to draw tetrahedra whose vertices lie on lattice points in $\mathbb Z^3$, the three-dimensional lattice of unit length. We can only draw tetrahedra of volume $1/6$, and we can only glue them along their faces, to produce a convex body $N$. Such $N$ cannot contain lattice points that are not the vertex of some tetrahedron.

Is it possible to produce a body $N$ that contains at least one tetrahedron whose 4 faces are all attached to other tetrahedra?

If that's the case, can one even find a body $N$ that contains at least one tetrahedron whose 4 faces and 4 vertices do not lie on the boundary of $N$?

Answer 1

Consider the six tetrahedra specified by the following lists of vertices:

\begin{align} (0,0,0),(1,0,0),(1,1,0),(1,0,1);\\ (0,0,0),(1,1,1),(1,1,0),(1,0,1);\\ (0,0,0),(0,1,0),(1,1,0),(0,1,1);\\ (0,0,0),(1,1,1),(1,1,0),(0,1,1);\\ (0,0,0),(0,0,1),(0,1,1),(1,0,1);\\ (0,0,0),(1,1,1),(0,1,1),(1,0,1). \end{align}

The idea is we divide the three faces of the unit cube about $(1,1,1)$ each into two triangles separated by the diagonals $((1,1,0),(1,0,1)),$ $((1,1,0),(0,1,1)),$ and $((1,0,1),(0,1,1)),$ and then we construct a tetrahedron on each of these triangular bases by adding the vertex $(0,0,0).$

These six tetrahedra meet all the requirements for individual tetrahedra, and together they fill the unit cube, which is convex. Now reflect this arrangement across any of the faces of the cube. This produces a convex object in which two tetrahedra are each surrounded by four others.

Further, you can construct as large an object as you want by repeated reflection so that all tetrahedra meet the requirements and most of the tetrahedra also are each surrounded by four others. In fact, you can fill all of $\mathbb Z^3$ in this way.

Answer 2

Is it possible to produce a body that contains at least one tetrahedron whose 4 faces are all attached to other tetrahedra?

Yes, pretty straightforwardly. Consider the 'unit' tetrahedron $\Delta=\{\langle0,0,0\rangle,\langle1,0,0\rangle,\langle0,1,0\rangle,\langle0,0,1\rangle\}$. Then along all of the faces which the origin we can simply glue reflected copies of this tetrahedron; the only challenge is to glue onto the face $\{\langle1,0,0\rangle,\langle0,1,0\rangle,\langle0,0,1\rangle\}$. But the tetrahedron with those three vertices and $\langle2,0,0\rangle$ as its fourth vertex also has volume $1/6$, so we can simply glue that one on.