Philippine Mathematical Olympiad

Solution:

Fix $m \geq 2$. First note that $x, y \geq m$, and that $M \geq 2$, implying $\sqrt{M(M+1)} \geq 2$. Thus, $f(x, y) \leq 2$ necessarily implies $|x-m|+|y-m| \leq 1$, and so $|x-m|,|y-m| \leq 1$. This implies $x \in [m-1, m+1]$, and so $x \in [m, m+1]$. Similarly, $y \in [m, m+1]$. Now if $x=m+1$, then $1 \leq |x-m|+|y-m| \leq \frac{2}{\sqrt{M(M+1)}} < 1$, contradiction. Using the same argument for $y$, it follows that $x, y \in [m, m+1)$. Thus, $\lfloor x\rfloor=\lfloor y\rfloor=m$ always, and so $M=m$.

The inequality is then equivalent to $|x-m|+|y-m| \leq \frac{2}{\sqrt{m(m+1)}}$. Let $r \leq 1$ be the upper bound in the inequality. Keeping in mind the fact that $x, y \geq m$, this region is a right-triangle with vertices $(m, m)$, $(m+r, m)$ and $(m, m+r)$, which then has area $\frac{r^{2}}{2}=\frac{2}{m(m+1)}$. This region is within the set for $2 \leq m \leq 2021$, so the desired sum is

$$\sum_{m=2}^{2021} \frac{2}{m(m+1)}=2 \sum_{m=2}^{2021}\left(\frac{1}{m}-\frac{1}{m+1}\right)=2\left(\frac{1}{2}-\frac{1}{2022}\right)=\frac{1010}{1011}$$

It then follows that $a+b=2021$.