# Hole inside cube with tetrahedrons at corners?

Here's a nearly-trivial proof that there is in fact a 'hole', or at least a piece left over when all the tetrahedra are removed: consider the tetrahedron with right vertex at the origin and its other three vertices at $(1,0,0)$, $(0, 1, 0)$, and $(0, 0, 1)$. Then the portion of the cube within this tetrahedron is the volume $x+y+z\lt 1$ intersected with the cube. But the center of the cube, $(\frac12, \frac12, \frac12)$ doesn't satisfy this inequality so it's not part of the tetrahedron. By symmetry it can't be part of any of the corner tetrahedra, and therefore it 'survives'; in fact, by a simple extension of this argument you can see that a ball centered at the center of the cube of radius $\sqrt{\frac1{12}}$ — i.e., passing through the point $(\frac13, \frac13, \frac13)$ — must be outside of all the tetrahedra, and so inside the left-over shape.

In fact, you can go further than this: since the leftover shape is the intersection of these half-spaces, it must have a face for each corner of the cube (You can show this by showing that the point $(\frac13, \frac13, \frac13)$ at the center of the equilateral face of one tetrahedron belongs to none of the other tetrahedra) and exactly those faces. Thus, it must be a dual polyhedron to the cube, namely the octahedron, and by careful consideration of the intersection of these half-spaces you can find the vertices of the octahedron.

In each tetrahedron, there is a distinguished corner. In your example above, the distinguished corner is the corner labeled $1$ as that connects along cube edges to each of the three other vertices.

Due to symmetry, there are only three types of interactions between two tetrahedra. These can be can be broken down into the case where $2$ is the distinguished vertex, $4$ is the distinguished vertex, and $8$ is the distinguished vertex.

Suppose $8$ is the distinguished vertex: This tetrahedron is disjoint from the given one because the two tetrahedra are separated by the plane passing through points $2$, $3$, $6$, and $7$. All points of one tetrahedron are on one side of the plane and all points of the other tetrahedron are on the other side.

Suppose $4$ is the distinguished vertex (also $6$ or $7$ is the distinguished vertex): This tetrahedron is disjoint from the given one because the two tetrahedra are separated by the plane passing through points $2$, $3$, $6$, and $7$. All points of one tetrahedron are on one side of the plane and all points of the other tetrahedron are on the other side.

Suppose $2$ is the distinguished vertex (also $3$ or $5$ is the distinguished vertex): In this case, there is some overlap between the two tetrahedra. The intersection is below the planes passing between points $2$, $3$, and $5$ as well as the plane passing between the points $6$, $1$, and $4$. It is also above the base and in front of the back wall. This is a tetrahedron ($4$ faces), so we can calculate its volume. It's height is the height of the midpoint of the back wall (i.e., where the lines between $2$ and $5$ and $6$ and $1$ intersect). This has height $\frac{1}{2}$. The base is a triangle with vertices $1$, $2$, and the midpoint of the base square (where the lines between $2$ and $3$ and $1$ and $4$ intersect). The base has area $\frac{1}{4}$. Then, the volume of the intersection is $\frac{1}{24}$.

Finally, if we note that any triple of tetrahedra do not have a common intersection (since you can't have a triple of the third type of distinguished vertex), we can calculate the volume of the union of the tetrahedra. In particular:

Sum of the volumes of the tetrahedra: $\frac{8}{6}$.

Overlap: Each edge of the cube corresponds to a single overlap, there are $12$ edges, so the overlap is $\frac{1}{2}$.

Putting this together, the volume of the union is $\frac{5}{6}$, so, yes, there is a hole.

The hole will be the polyhedron with vertices at the centers of the faces, an octahedron.