mole trapped in a snowball In Famous Puzzles of Great Mathematicians by Miodrag S. Petković there is this problem:
A large avalanche in the Alps traps an unhappy mole. When the avalanche stops,  it turns out that the poor mole has been buried somewhere inside a snowball with an ellipsoidal shape with a volume of 500 cubic meters. The mole can dig a hole through the snow advancing at one meter per minute, but he only has the strenght and breath for 24 minutes. Can the mole reach the surface of the snowball and save his life?
In the provided answer, Petković states that the mole should dig a tunnel through the points $A$, $B$, $C$, $D$ in the picture below, which are vertices of a cube with sides 8 meters. Then he says: «Indeed, if all four points $A$, $B$, $C$, $D$ would lie inside the snowball, then all interior points of the cube constructed from the perpendicular segments $AB$, $BC$ and $CD$ would belong to the interior of the snowball.» (After this, the result follows immediately, of course)
Trouble is, I don't know how to prove that this must be the case. Of course if all eight vertices lied in the snowball this would be true, but here I may only say that the convex hull of the four point are within the snowball. Am I forgetting something obvious?

 A: Let us pick the coordinates for the points as
$$
A:(1,0,0),\quad B:(0,0,0),\quad C:(0,1,0),\quad D:(0,1,1).
$$
It is easy to verify that the tetrahedron with the vertices
$$
A':(1,0,0),\quad B':(0,0,0),\quad C':\left(\frac12,\frac{\sqrt{3}}{2},0\right),\quad D':\left(\frac12,\frac{1}{2\sqrt{3}},\frac{\sqrt{2}}{\sqrt{3}}\right)
$$
is regular. We can map the original tetrahedron ABCD to the regular one $A'B'C'D'$ by a linear map:
$$
BA\mapsto B'A',\quad BC\mapsto B'C',\quad BD\mapsto B'D'.
$$
The standard constructions gives the matrix of the map as
$$
M=\begin{bmatrix}B'A' & B'C' & B'D'-B'C'\end{bmatrix}=
\begin{bmatrix}1 & \frac12 & 0\\0 & \frac{\sqrt{3}}{2} & -\frac{1}{\sqrt{3}}\\ 0 & 0 & \frac{\sqrt{2}}{\sqrt{3}}\end{bmatrix}
$$
with $\det(M)=\frac{1}{\sqrt{2}}$. It is known that the determinant is the volume scaling under linear mapping. It is also known (and clear from the symmetry) that the minimal volume ellipsoid around the regular tetrahedron is the circumscribed ball. The ball radius can be found as $R=\frac{\sqrt{3}}{2\sqrt{2}}$, so the ball volume is
$$
V'=\frac{4\pi}{3}R^3=\frac{\pi\sqrt{3}}{4\sqrt{2}}.
$$
Thus the original minimum volume ellipsoid has the volume
$$
V=\det(M^{-1})V'=\sqrt{2}\frac{\pi\sqrt{3}}{4\sqrt{2}}=\frac{\pi\sqrt{3}}{4}\approx 1.3603
$$
which is larger than the volume of the unit cube. Hence, an ellipsoid of a volume that is smaller than that of the cube cannot include all the points $A,B,C,D$.
