Lengths on the unit octahedron Consider the face of the unit octahedron, defined by:
$$O^2 = \{(x,y,z): |x|+|y|+|z|=1\}$$
Every point on the octahedron has between 0 and 3 positive coordinates. E.g, in $(0.2,-0.3,0.5)$, the $x$ and $z$ coordinates are positive. For every point $(x,y,z)$ on the octahedron, define: $P(x,y,z)$ as the sum of the positive coordinates, e.g, $P(0.2,-0.3,0.5)=0.7$.
For any $r\in [0,1]$, let $L(r)$ be the set of all points whose sum-of-positive-coordinates is $r$:
$$L(r) = \{(x,y,z)\in O^2: P(x,y,z)=r\}$$
Given $r$, what is the length of $L(r)$? 
I know that the length of $L(0)$ is infinite, as $L(0)$ is the entire bottom-west-south orthant (where all coordinates are negative). Similarly, the length of $L(1)$ is infinite. 
But, the lengths of $L(r)$ for $r \in (0,1)$ seem finite - they look like lines or circles on the octahedron. Do they all have the same length? How do you calculate their lengths? 
 A: As you have determined in finding $L(0)$ and $L(1)$,
if all three of $x,$ $y,$ and $z$ are positive then $P(x,y,z) = x + y + z = 1,$
and if all three are negative then $P(x,y,z) = 0.$
So if $P(x,y,z) \in (0,1)$ then at least one of $x,y,z$ is positive, but
not all three.
Let $P(x,y,z) = r$.
If $x$ is the only positive coordinate then $x = r$ and $y+z = r-1$
(so that $|y| + |z| = 1-r$).
So we have a line segment from $(r,r-1,0)$ to $(r,0,r-1)$,
length $(1-r)\sqrt2.$
If $x$ and $y$ are positive then $x+y = r$ and $z = r-1$
(so that $|z| = 1-r$).
So we have a line segment from $(r,0,r-1)$ to $(0,r,r-1)$,
length $r\sqrt2.$
Taking into account all six possible combinations of one or two positive
coordinates with the other coordinates being negative,
we have three segments of length $(1-r)\sqrt2$ and three of length $r\sqrt2,$
for a total of
$$3(1-r)\sqrt2 + 3r\sqrt2 = 3\sqrt2.$$
This is also the perimeter of the boundary of one triangular face,
which it should be, since in the limit as $r$ goes to $1$, $L(r)$
approaches the boundary of $L(1)$.
I noticed also that 
$$L(r) = \{(x,y,z) \mid |x|+|y|+|z|=1 \land x + y + z = 2r - 1\}.$$
That is, $L(r)$ is a hexagon 
in the plane whose equation is $x + y + z = 2r - 1.$
