Inradius of an octahedron Max here. I would like to know an alternative answer to the following question:
What is the inradius of the octahedron with sidelength $a$? 
Now, I have found a solution by considering two scenarios below. There is, however, trigonometric relations involved, which I would like to avoid. Can someone come up with a simpler proof? (Perhaps using duality, which I was unable to do.)


Let $C$ be the center of the octahedron. (Spoiler: If you are curious, first try to find it out on your own.)
1) Determine the height of one of two constituent square pyramids by considering a right triangle, using the fact that the height of the equilateral triangle is $a \sqrt{3}/2$. (Incidentally, by this you obtain the circumradius of the Octahedron.)
2) Now cut the (solid) octahedron along 4 of these latter heights, i.e. across two non-adjacent vertices and two midpoints of parallel sides of the "square base". (Figure 1)

Consider the resultant rhombic polygon, whose sidelength is $a\sqrt{3}/2$. Choose a convenient pair of adjacent corners $A$ and $B$, s.t. the angle in $A$ is obtuse. Then the insphere touches $AB$, say in $M$.
I now use the fact that $C$ has to be the center of the insphere. (Why?) 
Now we can extract information using trigonometric relations by considering two right triangles, 
one whose catheti are $BC$ and $AC$ [$BC$ is the height calculated in 1)],
another whose catheti are $CM$ and $AM$.
Obtain a fraction as the sine of $\angle CAB$ first, then use it in the second right-triangle. You're done.

This avoids any explicit calculation of sine/cosine/tangent/arcus-functions, since you can use the sine of the angle $\angle CAB$ twice. However, I would like to know if there is a way that could avoid even trigonometric functions, a really geometric way in the original sense of the word.
Edit: I am aware that one can avoid the sine in this case by making explicit calculations, using only the Pythagorean-Theorem. So the question becomes: Is there a still simpler proof?
Thanks! 
Yours,
Max
 A: An analytic solution: If we center the octahedron on the origin, this problem amounts to finding the sphere centered at the origin that’s tangent to all of the faces. By symmetry, we only need to consider one face. A plane with equation $Ax+By+Cz+D=0$ is tangent to the sphere $x^2+y^2+z^2=r^2$ iff the coefficients satisfy the dual quadric equation $A^2+B^2+C^2=D^2/r^2$. W.l.o.g. we can place the three vertices at $(a/2,a/2,0)$, $(-a/2,a/2,0)$ and $(0,0,a/\sqrt2)$. The plane through these points has equation $$\begin{vmatrix}x&y&z&1 \\ \frac a2&\frac a2&0&1\\-\frac a2&\frac a2&0&1 \\ 0&0&\frac a{\sqrt2}&1\end{vmatrix} = 0,$$ which simplifies to $2y+\sqrt2z=a$. Substituting the coefficients of this plane equation into the dual quadric equation produces $a^2/r^2=6$, therefore $r=a/\sqrt6$.
A: Sketch of idea.  I tend to think of it as follows.  Consider an octahedron with points on the $x$, $y$, and $z$ axes, at coordinates $\pm 1$.  Then by symmetry, the inradius is the distance from the origin to the point $\left(\frac13, \frac13, \frac13\right)$, which is $\sqrt\frac13$.  By similarity, the inradius of an octahedron with side length $a$ is $\frac{a}{\sqrt6}$.
(I'm aware this is by no means a complete notion; I'll come back to flesh it out.)
A: Consider this full elementar geometrical derivation:
Inscribe into the octahedron its dual cube. Then the searched for inradius of the octahedron would be the circumradius of that cube. Let the edgelength of the octahedron be $a$ and that of the inscribed dual cube be $c$.
Next consider a variation of that dual cube as to be an accordingly oriented square prism of varying height, still with vertices on the faces of the octahedron. That prism would have base sides of size
$$b=\frac1{\sqrt2}a\cdot t$$
where $t$ is some variation parameter, starting with $t=0$ at the top and ending with $t=1$ at the equatorial section of the octahedron. The corresponding height of that prism then is
$$h=\sqrt2 a(1-t)$$
For the prism to become a cube we have $b=h$ or 
$$\frac1{\sqrt2}a\cdot t=\sqrt2 a(1-t)$$
Solving for $t$ results in $t=2/3$ or
$$c=b|_{t=\frac23}=\frac1{\sqrt2}a\cdot \frac23=\frac{\sqrt2}3a$$
Now the inradius $r$ of the $a$-sized octahedron is the circumradius of the $c$-sized cube and thus clearly
$$r=\frac{\sqrt3}2c=\frac{\sqrt3}2\cdot \frac{\sqrt2}3a=\frac1{\sqrt6}a$$
--- rk
