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I am reading an article and the article claims (without proof or any explnasion) the following:

Consider the solid tessellation of cubes with a cube center at each integer point $X=(x,y,z)$, and a plane $x+y+z=0$ cutting it. This plane intersects each cube in a hexagon and $x+y+z=\pm 1$ in a triangle".

From what I understand if we look at the projection to $\mathbb{R}^2$ then this looks like in this link

My question is how can I prove or at least understand why the shapes are hexagon\triangles ?

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Is +-1 supposed to be $\pm 1$? – Arturo Magidin Mar 23 '12 at 21:03
@ArturoMagidin - yes. again, sorry for the writing – Belgi Mar 23 '12 at 21:04
I think the triangle is not hard to see. Imagine cutting off a small corner of a cube with a plane perpendicular to the body diagonal. The cut surface is an equilateral triangle. You can even try it on a stick of butter. The hexagon is harder to imagine. – Ross Millikan Mar 23 '12 at 21:17
up vote 3 down vote accepted

I recommend you get some thin cardboard or stiff paper. The idea is to make the two halves resulting from cutting a cube along a plane through its center, and orthogonal to a major diagonal.

Pick a unit of measurement. Draw three squares, each two units on a side. For each, mark the midpoints of two consecutive edges. Draw the line segment between the two edges. Now you have three squares, each of which has a corner triangle. Cut out the squares, then further cut off the triangle corner pieces, Now you have three right triangles, edges $(1,1,\sqrt2),$ and three pentagons, edges $(2,2,1,\sqrt2,1).$ Finally, draw one regular hexagon, all edges $\sqrt 2.$ Cut that out as well.

Tape together the pentagons at their right angles, so you have part of a cube. Tape in the right triangles in the remaining three corners that have right angles. Finally, tape the hexagon where it fits.

Make another one.

Place them with the hexagons matching up, see what happens.

Oh, well, pictures at . I still recommend you construct these yourself.

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