Philippine Mathematical Olympiad

Solution:

Let $r$ be the inradius, $r = 3$. Let $S$ be the area, $S = 2022$. Let $a = BC$.

Recall that $S = r \cdot s$, where $s$ is the semiperimeter. So:
$$s = \frac{S}{r} = \frac{2022}{3} = 674$$

Let $AI$ be the distance from $A$ to the incenter $I$. There is a formula:
$$AI^2 = \frac{bc}{(b + c)^2} \left[ (b + c)^2 - a^2 \right] + r^2$$
But this is complicated. Alternatively, recall that:
$$AI^2 = \frac{r^2 + (s - a)^2}{1}$$
So:
$$AI^2 = r^2 + (s - a)^2$$
Given $AI = 5$, $r = 3$, $s = 674$:
$$5^2 = 3^2 + (674 - a)^2$$
$25 = 9 + (674 - a)^2$
$16 = (674 - a)^2$
$674 - a = \pm 4$
So $a = 674 \pm 4 = 678$ or $670$

But $a$ must be less than $s$ (since $a$ is a side, $s$ is the semiperimeter), and $a$ must be positive. Both $678$ and $670$ are possible, but let's check which is correct.

If $a = 678$, then $s - a = -4$, which is not possible (since $s - a$ is the sum of the other two sides divided by $2$ and must be positive). So $a = 670$.

Thus, the answer is $\boxed{670}$.