Inf of a particular subset of vectors Let $N\geq2$ an even natural number and $A_N=\{k=(k_1,k_2) \in \mathbb{N} \times \mathbb{N}:1<k_2<k_1\leq N\}$. Let $U_N=\{(k,l): k \in A_N, l \in A_N, \angle(k,l)>0\}$, where $\angle(k,l)$ is the angle between two vectors $k$ and $l$ in the plane. I have to study:
\begin{equation} \alpha_N=\inf_{(k.l)\in U_N}\angle(k,l) \ \ \ \text{and} \ \ \ \lim_{N \to \infty} N\alpha_N. \end{equation}
I tried in this way: $k \cdot l = |k||l|\cos(\angle(k,l))$, so, since $|k|=\sqrt(k_1^2+k_2^2)$ and $|l|=\sqrt(l_1^2+l_2^2)$,
\begin{equation} \cos(\angle(k,l))=\frac{k_1l_1+k_2l_2}{\sqrt(k_1^2+k_2^2)\sqrt(l_1^2+l_2^2)} \end{equation}
Now using that $k,l \in A_N$
\begin{equation} \cos(\angle(k,l))>\frac{2k_2l_2}{\sqrt(2k_1^2)\sqrt(2l_1^2)}>\frac{k_2l_2}{k_1l_1}>\frac{1}{N^2} \end{equation}
Could it be a good way to find $\alpha_N$?
 A: Here is a representation of $A_6$. In the picture it is considered the triangle with vertices $(6,5), (5,4), (0,0)$.

Recall the formula for the area of a triangle:
$$A = \frac{1}{2} |k| |l| \sin ( \alpha )$$
Note that for all $(k,l) \in U_N$ the area formed by the two vectors is the area of a triangle whose vertices have integer coordinates. This means that, by Pick's Theorem, the area of such a triangle is at least
$$A \ge \frac{1}{2}$$
Hence
$$\sin ( \angle (k,l)) = \frac{2A}{|k||l|} \ge \frac{1}{|k||l|}$$
If we want to minimize $\angle (k,l)$, we have to minimize the sine of $\angle (k,l)$: that's because $\sin: [0, \pi/2] \to \Bbb R$ is increasing.
This means that we need to minimize
$$\frac{1}{|k||l|}$$ or, equivalently, we need to maximize its inverse $|k||l|$.
It happens that the maximum is achieved for:
$$l= (N, N-1) \\ k= (N-1, N-2)$$
This is intuitive, but it is not trivial. I have to think about how to prove this.
After we have established this, we have
$$\alpha_N= \arcsin \left( \frac{1}{\sqrt{(N^2 + (N-1)^2)((N-1)^2 + (N-2)^2)}} \right)$$
Asymptotically, since $\sin ( \theta ) \approx \theta$ we have
$$\alpha_N \approx \frac{1}{\sqrt{(2N^2)(2N^2)}} = \frac{1}{2N^2}$$
This means that
$$\lim_{N \to \infty} N \alpha_N = 0$$
