Empty lattice simplex or White's theorem White has proved (White, G. K. Lattice tetrahedra -- Canad. J. Math. 16 1964 389–396.) the following theorem:
If $T$ is a closed tetrahedron and $\Lambda$ is a lattice which contains the vertices of $T$, then the two following conditions are equivalent: 
(1) The only points of $\Lambda$ in $T$ other than the vertices lie on a pair of opposite edges of $T$; 
(2) There is a pair of parallel lattice planes of $\Lambda$ through a pair of opposite edges of $T$ such that no points of $\Lambda$ lie between these planes. 
His proof is rather complicated. Has anybody proved this theorem in a simpler way? What proof is the simplest one?
 A: A simple illustration of this interesting theorem:
          
A: I have not looked at the literature, so the following may fail for some geometric reason of which I am unaware; this reflects my thinking, and I do not yet see how else to get the results.
It should be clear that (2) implies (1), so let's think about not (2) implying not (1).  Then for each pair of parallel planes going through opposite edges of T, if they are lattice planes then there is another lattice plane between 
them that goes through T and has "as many" points of the lattice as do the parallel planes.  I am hoping there is enough regularity so that the middles plane looks like one of the parallel planes shifted; perhaps I am wrong.  Now we have for each pair of opposite edges of T a lattice plane going through T between the pair.
Following Sidney Harris, "And then a miracle occurs...", which means I do not know how to take the three planes intersecting the middle of T and produce a point lying in T and on the three planes, and make it a lattice point.  If I were looking for a simple 
(and natural) proof of the result, however, this is what I would try.  Perhaps I would need a map to pull the argument into Z^3 and use some analytic geometry to finish things, but I hope not.
Gerhard "Not Copying Someone Else's Catchphrase" Paseman, 2012.08.14
