HongKong 2022-23 IMO Selection Tests

Let the lengths of $BD$ and $DC$ be $x$ and $y$ respectively. By the angle bisector theorem, we have $\frac{x}{y}=\frac{AB}{AC}=\frac{EB}{EC}$, i.e. $$\frac{x}{y}=\frac{2022}{2022+x+y}$$



$$y=\frac{x^2+2022x}{2022-x}=-x-4044+\frac{2022\cdot 4044}{2022-x}$$ Hence we must choose positive integers $x$ for which $2022-x$ is a positive integer dividing $2022\cdot 4044$ (note that every such $x$ will result in a positive integer value of $y$ and the existence of such a figure satisfying all the conditions). As $2022\cdot 4044=2^3\cdot 3^2\cdot 337^2$, its only positive factors that are less than $2022$ are the 12 positive factors of $2^3\cdot 3^2$ as well as $337,337\cdot 2,337\cdot 3$ and $337\cdot 4$. Hence there are altogether 16 such values of $2022-x$, and they correspond to 16 possible values of $x$.