Minimum difference of angles between points on square lattice

Question

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?

Answer 1

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.

Answer 2

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).