Is it possible to put an equilateral triangle onto a square grid so that all the vertices are in corners? In the following collection of problems - arXiv:1110.1556v2 [math.HO] - the following question is posed:

Is it possible to put an equilateral triangle onto a square grid so that all the vertices are in corners?

The first approach that springs to mind is to use Pick's Theorem (e.g. http://www.geometer.org/mathcircles/pick.pdf) assuming that all vertices are on lattice points. It turns out that it is not possible (by Pythagoras, the area of an equilateral triangle with two vertices on lattice points is a rational multiple of $\sqrt3$). 
My question then is - how can one establish the impossibility of such a placement without resorting to Pick's Theorem? 

Theorem: A triangle is embeddable in $\mathbb{Z}^2$ if and only if all its angles have rational tangents. 
But $\tan(\pi/3)=\sqrt{3},$ so an equilateral triangle is not embeddable.  
Reference: Triangles with Vertices on Lattice Points
Michael J. Beeson,
The American Mathematical Monthly,
Vol. 99, No. 3 (Mar., 1992), pp. 243-252
