A regular octahedron has a side length of 1. What is the distance between two opposite faces?

I've tried to find this from drawing a square within, and seeing if the lengths of the individual triangles could be used in anyway. I tried other cross sections too, but none of them got me anywhere.

Any ideas? Any and all help is appreciated.

Solution 1:

Let the vertices be $A=(0,0,0), B=(1,0,0),C=(1,1,0), D=(0,1,0), E=(a,b,c), F=(a,b,-c)$ with $c>0$. By L-R symmetry we have $a=b=1/2.$ The center of the octohedron is $G=(1/2,1/2,0)$. And $\angle AGE=90^o.$ So $1=|AE|^2=|AG|^2+|GE|^2=1/2+|GE|^2=1/2+c^2.$ So $c=1\sqrt 2.$ So $E=(1/2,1/2,1/\sqrt 2).$

Reflection thru the point $G$ maps $ABE$ to $DCF$. So the planes $ABE$ and $DCF$ are parallel and their distance apart is twice the distance from $G$ to either of them.

In $\Bbb R^2$ if $p,q$ are not both $0$ then the distance from $(x,y)$ to the line $\{(u,v): pu+qv=0\}$ is $\dfrac {|px+qy|}{\sqrt {p^2+q^2}}.$

Similarly in $\Bbb R^3$ if $p,q,r$ are not all $0$ then the distance from $(x,y,z)$ to the plane $\{(u,v,w):pu+qv+rw=0\}$ is $$\frac {|px+qy+rz|}{p^2+q^2+r^2}.$$

So find $p,q,r,$ not all $0,$ such that $pu+qv+rw=0$ when $(u,v,w)\in\{A,B,E)$. And compute the distance from $(x,y,z)=G=(1/2,1/2,0)$ to the plane $ABE=\{(u,v,w):pu+qr+rw=0\}$, which is half the distance between opposite sides of the octohedron.