Recall that the first octant in 3D is the set of points $(x,y,z)$ such that $x\geq 0$, $y\geq 0$, and $z\geq 0$. This intersects the XY, XZ, and YZ planes in quadrants. In particular:
- for XY: the set of points such that $x\geq 0$, $y\geq 0$, and $z=0$.
- for XZ: the set of points such that $x\geq 0$, $y=0$, and $z\geq 0$.
- for YZ: the set of points such that $x=0$, $y\geq0$, and $z\geq 0$.
What seems to be going on is that you're taking these three quadrants and projecting them onto the plane described by the equation $x+y+z=0$. Every point in this plane comes from the projection of exactly one point in the union of the three quadrants. The quadrants share one point in common (the origin), and every pair of quadrants shares a ray.
The projection formula onto this plane is given by the function
$$ \pi(x,y,z) = (\tfrac{2}{3}x-\tfrac{1}{3}y-\tfrac{1}{3}z,-\tfrac{1}{3}x+\tfrac{2}{3}y-\tfrac{1}{3}z,-\tfrac{1}{3}x-\tfrac{1}{3}y+\tfrac{2}{3}z)$$
All this is doing is subtracting $\tfrac{1}{3}(x+y+z)$ from each coordinate to make the sum of the resulting coordinates equal $0$.
The formula for each quadrant reduces to
\begin{align*}
\pi(x,y,0) &= (\tfrac{2}{3}x-\tfrac{1}{3}y,-\tfrac{1}{3}x+\tfrac{2}{3}y,-\tfrac{1}{3}x-\tfrac{1}{3}y) \\
\pi(x,0,z) &= (\tfrac{2}{3}x-\tfrac{1}{3}z,-\tfrac{1}{3}x-\tfrac{1}{3}z,-\tfrac{1}{3}x+\tfrac{2}{3}z) \\
\pi(0,y,z) &= (-\tfrac{1}{3}y-\tfrac{1}{3}z,\tfrac{2}{3}y-\tfrac{1}{3}z,-\tfrac{1}{3}y+\tfrac{2}{3}z)
\end{align*}
And for each ray:
\begin{align*}
\pi(x,0,0) &= (\tfrac{2}{3}x,-\tfrac{1}{3}x,-\tfrac{1}{3}x) \\
\pi(0,y,0) &= (-\tfrac{1}{3}y,\tfrac{2}{3}y,-\tfrac{1}{3}y) \\
\pi(0,0,z) &= (-\tfrac{1}{3}z,-\tfrac{1}{3}z,\tfrac{2}{3}z) \\
\end{align*}
And of course the origin:
\begin{align*}
\pi(0,0,0) &= (0,0,0)
\end{align*}
Given a point on the $x+y+z=0$ plane, can we see which point it's the projection of? The only points that project to $(x,y,z)$ are points of the form $(x+c,y+c,z+c)$. All we need to do is let $c$ be the least number such that either $x+c=0$, $y+c=0$, or $z+c=0$. From this, we get your idea:
A way to parameterize a 2-dimensional plane is the set of points $(x,y,z)$ with $x\geq 0$, $y\geq 0$, and $z\geq 0$ such that either $x=0$, $y=0$, or $z=0$. We can realize this plane as the plane with the equation $x+y+z=0$ by taking $(x,y,z)$ and subtracting the average $\tfrac{1}{3}(x+y+z)$ from each coordinate.
Notice that there is a 3-fold rotational symmetry from rotating 3D space 120 degrees about the line $x=y=z$. This corresponds to the 120 degree rotation you noticed in your system of coordinates.
This suggests a generalization to higher dimensions. For example, to work with 3D, you can think of it as the set of points in 4D satisfying $x+y+z+w=0$. Then, with your system we can instead think of the points $(x,y,z,w)$ such that $x,y,z,w\geq 0$ and at least one of these numbers is $0$. To project onto the $x+y+z+w=0$ plane, we just subtract $\tfrac{1}{4}(x+y+z+w)$ from each coordinate. This time there is tetrahedral symmetry. Rather than three axes, you'll have four axes pointing in the directions of the vertices of a regular tetrahedron.
Even more generally, what we're doing is taking a convex polyhedron (the octant in 3D for example) and a hyperplane (a space of one less dimension, the plane $x+y+z=0$ for example) with the property that the projection of the boundary of the polyhedron onto the hyperplane is a bijection. Each face of the polyhedron gives a different region of the hyperplane this way, and you can use the plane to give the faces of the polyhedron a system of coordinates.
Best Answer
Another description of the Mayan numbering system, including important historical facts explaining how limited our knowledge is, is at the site http://www-groups.dcs.st-and.ac.uk/history/HistTopics/Mayan_mathematics.html.
We have only a few surviving documents from the Mayan civilization. We have examples of large numbers written in a not-quite-base-$20$ system, in which the place value always increases by $20$ except once: the place value above $20$ is $360$ instead of $400$. But those examples are said to represent numbers of days in calendar calculations, and we think it would have been more convenient to deal with groups of $360$ days at a time and account in some other way for the extra $5.2422$ days that the Mayans knew the average solar year contains, rather than have groups of $400$ days, which would be $34.7578$ days too many per year.
We too have an irregularity in the way we tell time, in that it takes $60$ seconds to make one minute, $60$ minutes to make one hour, but only $24$ hours to make one day. Why not use factors of $60$ in all three places? This is much stranger than the Mayan system, which at least has the excuse that the day and the year are naturally-occurring units of time that we cannot redefine for our convenience; if we had all been born in a world in which the day was divided in $60$ equal parts instead of $24$, would we ever have noticed that there was anything wrong with our timekeeping?
Whether the Mayans used a true base-$20$ system when they were not writing calendars seems to be a matter of debate among historians. It seems that it is very hard to find good evidence supporting either side of that debate, due to the lack of surviving documents that would have used such numbers.