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.
 A: The octahedron hole inside the cube:
$\hspace{3cm}$
A: 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.
A: The hole will be the polyhedron with vertices at the centers of the faces, an  octahedron.
A: There is a "hole."  Consider any daigonal of the cube.  It will be cut into $\frac 13$'s by the various planes dissecting this cube.
There will be one plane associated with each vertex of the cube that cuts one side of the hole.  There are 8 vertexes to the cube.  The hole is a regular octohedron.
The 4 tetrahedra using with vertexes at $1,4,6,7$ are disjoint.
This leaves a remainder that is a regular tetrahedron.  The side length of this tetrahedron is $\sqrt 2$  
The volume of this remainder is $1 - 4\cdot \frac 16 = \frac 13$ 
We can rearrange the 4 tetrahedra into $\frac 12$ a regular octohedron.  The volume of this figure is $\frac {2}{3}$  The edge lenght is $\sqrt 2$  The volume of a regular ocothedron with edge length $\sqrt 2 = \frac 43$
The volume of an octohedron is $4\times$ the volume of a tetrahedron with the same edge length.
Back to our tetrahedron with volume $\frac 13$  We are going to cut the 4 vertexes off of this to leave a remainder that is a tetrahedron.
The volume of this remainder equals the volume of the 4 tetrahedra that have been cut off.
The volume of this octohedron is $\frac 16$
A: 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.
A: Intuitive understanding of the existence of the hole, and that it's an octahedron:
Fix the cube so there's a "top" and a "bottom" face. The 8 tetrahedra at the corners can be split into two groups of 4; 4 with bases on the top face, 4 with bases on the bottom face.
Focus on the bottom face. 2 tetrahedra share an edge on one diagonal of the square; the other 2 tetrahedra share an edge on the other diagonal of the square.
So consider moving $\varepsilon$ above the center of the square -- this is just above the shared edges of the 4 "bottom" tetrahedra.
It's clear this isn't a part of any of the bottom tetrahedra, by construction -- hopefully it's also clear that it's not a part of the top tetrahedra either. So the center of the face represents the outer limit of the interior hole.
The same happens at all 6 faces, and the linearity of the whole setup suggests we can connect these with lines, resulting in an octahedron.
It might also help to think of the cube as a "cave" with tetrahedral "stalagmites" on the bottom face and "stalactites" on the top face.
A: Another way is to think of this problem computationally by considering cross-sections along the $z$ axis (arbitrarily chosen).
In any cross-section (say at $z = t$), there are 8 (potentially overlapping or degenerate) line segment, one for each corner's tetrahedron, specifically:
\begin{align}
y_1 &= t   &+ x \\
y_2 &= -t  &+ x \\
y_3 &= 1-t &- x \\
y_4 &= 1+t &- x \\
y_5 &= t   &- x \\
y_6 &= 2-t &- x \\
y_7 &= -1+t &+ x \\
y_8 &= 1-t &+ x \\
\end{align}
(Just try $t=0$ and $t=1$ to convince yourself these are right)
With this in hand, we can turn to the computer to draw all of these for us for any $z$. Here's an implementation in R showing 25 cross-sections, proceeding right and down:
xx = seq(0, 1, length.out = 100L)
col = c('#e6194b', '#3cb44b', '#ffe119', '#0082c8',
        '#f58231', '#911eb4', '#46f0f0', '#f032e6')
par(mfrow = c(5, 5), mar = c(0, 0, 0, 0), oma = c(0, 0, 0, 0))
t = seq(0, 1, length.out = prod(par('mfrow')))
coef = data.frame(
  t_a = c(0, 0, 1, 1, 0, 2, -1, 1),
  t_b = c(1, -1, -1, 1, 1, -1, 1, -1),
  b = c(1, 1, -1, -1, -1, -1, 1, 1)
)
for (t_i in t) {
  plot(NA, xlim = c(0, 1), ylim = c(0, 1), asp = 1,
       xaxt = 'n', yaxt = 'n', ylab = '', xlab = '', bty = 'n')
  segments(c(0, 0, 1, 1), c(0, 1, 1, 0), c(0, 1, 1, 0), c(1, 1, 0, 0))
  del_t = abs(.5 - t_i)
  polygon(c(.5, del_t, .5, 1 - del_t), c(del_t, .5, 1 - del_t, .5),
          col = '#e6beff', border = NA)
  mtext(side = 3L, sprintf('t = %.2f', t_i), line = -2)
  for (ii in seq_len(nrow(coef))) {
    with(coef[ii, ], {
      yy = t_a + t_b * t_i + b * xx
      idx = yy >= 0 & yy <= 1
      lines(xx[idx], yy[idx], col = col[ii])
    })
  }
}

With output:

The shaded square in the middle is your hole. Each of the 8 colors tracks the internal face of each tetrahedron. 
We can also use saveGIF from the animation library to turn this into a gif:

A: Split your cube:


*

*1 octahedron in the middle





*

*8 regular tetrahedra on each face of the octahedron to obtain a stellated octahedron:





*

*12 tetrahedra (green) between adjacent pairs of regular tetrahedra (blue) to obtain a cube:



Split your tetrahedra:


*

*Notice that the tetrahedron you describe is the union of one blue tetrahedron and 3 green tetrahedra.

*Each blue tetrahedron is used once.

*Each green tetrahedron is used twice.
A: Taking the body diagonal of the cube, you see that the bases of the 2 corner simplices would divide that distance into 3 equal parts. Thus yes, there is a hole.
That hole will have its vertices at the face centers of the cube. Moreover it is defined by the base planes of the 8 corner simplices. Thus it has 8 faces. Therefore it is the inscribed octahedron.
--- rk
