Hole inside cube with tetrahedrons at corners?

Given is a unit cube with a tetrahedron at each corner, as shown here for one corner out of the $8$ :

It is noticed that the tetrahedrons are not disjoint. Because I cannot look through the cube, I have great difficulty imagining whether there is a hole left inside or not. If there is a hole, what then is the shape of that hole? And what then is the volume of that hole?
The volume of one tetrahedron is $1/6$ . This would make a total of $\,8/6\,$ if they were disjoint, but - as I've said - they are not. Apart from the facts some sort of proof would be nice.

Solution 1:

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.

Solution 2:

The octahedron hole inside the cube:

$\hspace{3cm}$