TITLE: Conceptual reason for why the volume of an ocahedron is four times the volume of a tetrahedron
QUESTION [7 upvotes]: The image below shows that a regular octahedron can be scaled by a factor of $2$ (resulting in a $2^3$ factor in volume) and decomposed as six octahedra and eight tetrahedra.
If $V_o$ and $V_t$ respectively represent the volumes of a regular octahedron and a regular tetrahedron with the same edge lengths, then $$
  2^3V_o = 6V_o + 8V_t,
$$ and solving for $V_o$ yields $V_o = 4V_t$.

Image from Wikipedia

Is there a conceptual reason why the volume of an octahedron is $4$ times the volume of a tetrahedron that doesn't rely on a decomposition like this? For example, is there a way that you can chop up four tetrahedra to fit them into an octahedron?
Equally useful, is there some nice way to see that a square-based pyramid has twice the volume of a tetrahedron? Perhaps integrating as slices of equilateral triangles vs slices of squares?
A higher dimensional analog.
A "nice to have" quality of the answer would be if it generalizes to the higher dimensional case. If $V_o^{(n)}$ and $V_t^{(n)}$ denote the (hyper)volumes of the $n$-dimensional cross-polytope and $n$-dimensional simplex respectively, then
$$
  V_o^{(n)} = \frac{\sqrt{2^n}}{n!} \text{ and } 
  V_t^{(n)} = \frac{\sqrt{n+1}}{n!\sqrt{2^n}} \text { with ratio } 
  \frac{V_o^{(n)}}{V_t^{(n)}} = \frac{2^n}{\sqrt{n+1}}.
$$
Is there a conceptual reason why this relationship is "nice"?

REPLY [3 votes]: If you inscribe two tetrahedra in a cube, their overlap is an octahedron:
 
Equivalently, the octahedron may be constructed by joining midpoints of a tetrahedron's edges. Notice that within this (say, red) tetrahedron, outside the octahedron $O$ there are four smaller tetrahedra $T$. The side lengths of these smaller $T$s are half the side-length of the original, red tetrahedra, so the red one has eight times the volume, and so $8T=4T+O$.
This is another decomposition proof, but it's more direct. 3D only though.