Minimum difference of angles between points on square lattice I have integer grid of size $N \times N$.
If I calculate angles between all point triples - is it possible analytically find minimal non-zero difference between those angles?
 A: By brute force, for $N = 4, 5, 6$ the smallest non-zero angle is formed between the segments from $(N-1,N-2)$ to $(0,0)$ and from $(N-1,N-2)$ to $(1,1)$, if the coordinates have range $0$ to $N - 1$ inclusive.
The first segment makes an angle
$\theta_1 = \tan^{-1}\left(\frac{N-2}{N-1}\right)$
with the edge of the lattice, and the second segment makes an angle
$\theta_2 = \tan^{-1}\left(\frac{N-3}{N-2}\right)$
with that edge.
The magnitude of the resulting angle between these segments is 
$$| \,\theta_1 - \theta_2| = 
 \tan^{-1}\left(\frac{N-2}{N-1}\right) - \tan^{-1}\left(\frac{N-3}{N-2}\right).
$$
(Since $0 < \theta_2 < \theta_1 < \frac{\pi}{4}$, the right-hand side of this equation is positive.)
While this exercise suggests that this might be the formula for larger $N$ as well, it is hardly a proof. I too am curious about the general result now.
A: Isn't the minimum angle just the one who's arms are along the diagonal (1,1) to (N,N) the back down to (1,2)? This gives the minimum angle as pi/4 - arctan((N-1)/N).
