The Problem:
Denote by $V_{(x, y)}$ the vertex at ordered pair $(x, y)$ in the Cartesian coordinate system. Denote by $A_{(x, y)}$ the measure of angle $\angle V_{(x, y)}V_{(0, 0)}V_{(1, 0)}$. Let $$\theta = \sum_{0 \le i, j \le 5 \ \ (i, j) \neq (0, 0)}{A_{(i, j)}}$$ Find $\theta \pmod{2\pi}$.
I understand the question and everything, but I am slightly overwhelmed about how to find an organized approach to computing the sum of angles. I got the value down to the sum $$\sum_{0 \le i, j \le 5 \ \ (i, j) \neq (0, 0)}{\arctan{\frac ij}}$$ But I don't know how to evaluate this. $\arctan(x+y)$ formula fails. I think there is a different approach I am not aware of.
Is there a nice way to generalize to $$\sum_{0 \le i, j \le k \ \ (i, j) \neq (0, 0)}{A_{(i, j)}}$$
I got it now, Note that if we represent $A_{m, n}+A_{n, m}$ in terms of complex numbers we get $\arg{(m+ni)(n+mi)}=\pi/2$. There are $k(k+1)$ of these numbers (ignnoring diagonals so we get $(k(k+1)/2)(\pi/2)$. Adding back the diagonal (ignoring $(0, 0)$) we get $k\pi/4$. $$(k(k+1)/2)(\pi/2)+k\pi/4=\frac{k(k+2)\pi}{4}$$ Which is the general solution. Substituting $k=5$ gives $35\pi/4 \equiv 3\pi/4 \pmod{2\pi}$