# Intersection of the cube tessellation and the plane $x+y+z=0$

Consider the solid tessellation of cubes with centers at each $(x,y,z)$ point s.t. $x,y,z$ are integers and the plane $x+y+z=0$ cutting it.

The plain intersects with each cube in a hexagon, my question is this:

If I look at one specific cube with center at $(x,y,z)$, is the center of the hexagon (the intersection between this specific cube and the plane) is the same center of the cube ($(x,y,z)$)?

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Yes. Take the cube $\max(|x|,|y|,|z|)=\frac12$ centered at $O=(0,0,0)$. Now the plane $x+y+z=0$ passes through the origin, and the hexagon has its center there as well. –  bgins Apr 9 '12 at 18:32

If we parametrize the plane $P:x+y+z=0$ (bijectively) using $$\matrix{ u&=&\frac{x+y}{\sqrt2}\\ v&=&\frac{x-y}{\sqrt2} } \qquad \text{as} \qquad \matrix{ x&=&\frac{u+v}{\sqrt2}\\ y&=&\frac{u-v}{\sqrt2}\\ z&=&-\sqrt2\,u }$$ then the orthogonal projection of the hexagon $H$ at the intersection of the cube $C$ $$\max\Bigl(|x|,|y|,|z|\Bigr)=\frac12$$ onto $P$ can be found by simply describing $(u,v)$ along $H$. (In fact $H$ is already in the plane $P$.) In particular, if a symmetry about $(0,0)$ can be observed, then we will have demonstrated that the centers of $C$ and $H$ coincide. Using the given parametrization actually already guarantees that $(u,v)$ that $(x,y,z)$ lies in the plane $P$, so we need only transform the equation for $C$:
$$\max\left( \left|\frac{u+v}{\sqrt2}\right|, \left|\frac{u-v}{\sqrt2}\right|, \left|\sqrt2\,u\right| \right)=\frac12$$
$$\max\left( \left|u+v\right|, \left|u-v\right|, \left|2\,u\right| \right)=\frac1{\sqrt2}$$
This can be visualized in the $uv$-plane as the hexagonal curve with the origin at its interior connecting the finite line segments of the six lines with equations $$\matrix{ u+v&=&\pm s \\\\ u-v&=&\pm s \\\\ u&=&\pm\frac{s}{2} }$$ for $s=\frac1{\sqrt2}$. It's pretty easy to see that its center is $(0,0)$. The three pairs of equations above are plotted below in red, green and blue, respectively, and the projections of the portions of the $xzy$-coordinate axes interior to $C$ (between the centers of opposite faces) is shown dotted in gray.