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 $V_o = 4V_t$.

Length-2 octahedron dissected into six octahedra and four tetrahedra

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"?

  • $\begingroup$ When you say "an octahedron" and "a tetrahedron" what is the implied normalization? Do you require that they both have edges of length $1$? (Anyway, I think your image is already quite a nice explanation!) $\endgroup$ – Qiaochu Yuan Sep 4 '20 at 4:36
  • $\begingroup$ @QiaochuYuan, yes, all edges of length $1$ (or equivalently, all edges of length $\ell$, I suppose). $\endgroup$ – Peter Kagey Sep 4 '20 at 5:38
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    $\begingroup$ @QiaochuYuan, I agree that the image is a slick insight and the argument is elementary and convincing, but it somehow feels a little convoluted—perhaps since it requires solving for $V_o$ in the equation. I'm hoping that another explanation will scratch whatever itch I have. $\endgroup$ – Peter Kagey Sep 4 '20 at 5:42
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    $\begingroup$ Instead of doubling an octahedron, it is a little bit simpler to double a tetrahedron. This results in one octahedron in the centre, and four tetrahedra, which together should have the same volume as eight tetrahedra. $\endgroup$ – Jaap Scherphuis Sep 4 '20 at 8:14
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    $\begingroup$ Joining midpoints of a tetrahedron's edges yields an octahedron. The four smaller tetrahedra outside the octahedron but inside the larger tetrahedron are an eighth the size, so $O+4T=8T$. (This is an alternate decomposition proof.) $\endgroup$ – runway44 Sep 17 '20 at 8:48

Join the vertices of the unit-sided octahedron with its centre. That will divide it into eight regular pyramids, having the faces of the octahedron as bases and three lateral edges with length $1/\sqrt2$.

Pythagora's theorem gives then a height of $1/\sqrt6$ for these eight pyramids, whereas the height of a regular unit-sided tetrahedron is $2/\sqrt6$. The volume of the tetrahedron is then double that of each regular pyramid in the octahedron, which explains why the volume of the octahedron is four times the volume of the tetrahedron.


Consider a cube with a tetrahedron inside it sharing four of its vertices. The cube dissects into this tetrahedron, and four identical triangular pyramids. Look at this picture of the cube standing on one vertex:

enter image description here

A body diagonal of the cube, vertical in this picture, is split into three equal parts by the heights of the vertices. This shows that the inner tetrahedron has twice the height of each small pyramid, and hence twice the volume. Eight of those small pyramids can form an orctahedron, so the tetrahedron is a quarter of the volume of the octahedron.

I don't think this can be generalised to higher dimensions in the direction you are looking for.


If you inscribe two tetrahedra in a cube, their overlap is an octahedron:

inscribed 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.

  • $\begingroup$ This is a very nice purely geometric argument! It is relatively straight forward to argue with cartesian coordinates and integration, but these would give little to no geometric insight. $\endgroup$ – M. Winter Nov 1 '20 at 10:02

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