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According to several sources on the internet, the volume of a regular octahedron with unit edge lengths is approximately 0.47. I haven't seen an explanation for why this is so yet. When I tried to figure the volume myself, I obtained a unit volume of 1/(sqrt(2)), which is approximately 0.71.

Let me explain my reasoning. If you take an octahedron, you can see it contains two squares on the vertical and horizontal axis (which are the base squares of the two pyramids it is made up of, depending on the angle you look at it at.) Each of these squares has an area of 1, since every line has a measure of 1 in a unit octahedron.

If you take a cross section of this octahedron parallel to one of these main squares, you will always get a similar square, because each vertex on the main square is equidistant to the other two vertexes in the octahedron.

Where the two main squares forming the octahedron intersect, the line is equal to the diagonal of a unit square, or the square root of 2.

Since all cross sections of the octahedron parallel to either of these squares are similar, we can therefore say that the relationship between the line on the base of any cross section and the area of the cross section is sqrt(2) (base line) to 1 (area), or a factor of 1/sqrt(2).

Extending this logic, we can multiply the area of the base square by this ratio to obtain the volume, which, because the area is 1, would be 1/sqrt(2) units cubed.

Where did I mess up my math reasoning?

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  • $\begingroup$ It's implicit what is meant later in the question statement, but it might be useful to spell out that by "unit octahedron" you mean "octahedron with unit edge length" (as opposed to, say, meaning the circumscribing sphere has unit radius). $\endgroup$ – Travis Willse Sep 30 at 19:13
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The volume of a pyramid is $\frac 13 hA$ where $h$ is the height and $A$ is the area of the base. Using this for two pyramids suck together gives me $\frac{\sqrt 2}3\approx 0.47$ for a regular octagon of side $1$.

You are implicitly arguing that the volume of a pyramid is $\frac 12 hA$ - I think you are applying two dimensional intuition to three dimensions. The area of a triangle is $\frac 12 bh$ because the similar pieces diminish linearly if you draw lines parallel to the base. But the areas of the cross sections of the pyramid here are two-dimensional.

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  • $\begingroup$ I was about to say the same thing. The proof is fine up to “Extending this logic...”...because the logic doesn't extend. $\endgroup$ – Matthew Leingang Sep 30 at 19:16
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The height for half octahedron is

$$h=\sqrt{1^2-\left(\frac{\sqrt 2}2\right)^2}=\frac{\sqrt 2}2$$

then

$$V=2\cdot\frac13\cdot 1\cdot 1\cdot h=\frac23h\approx 0.47$$

Here below a sketch for the derivation

enter image description here

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