10. [70] A convex 2019-gon $A_{1} A_{2} \ldots A_{2019}$ is cut into smaller pieces along its 2019 diagonals of the form $A_{i} A_{i+3}$ for $1 \leq i \leq 2019$, where $A_{2020}=A_{1}, A_{2021}=A_{2}$, and $A_{2022}=A_{3}$. What is the least possible number of resulting pieces?
Answer: 5049
Each time we draw in a diagonal, we create one new region, plus one new region for each intersection on that diagonal. So, the number of regions will be
$$
1+\text { (number of diagonals) }+ \text { (number of intersections), }
$$
where (number of intersections) counts an intersection of three diagonals twice. Since no four diagonals can pass through a point, the only nonconstant term in our expression is the last one. To minimize this term, we want to maximize the number of triples of diagonals passing through the same point. Consider the set $S$ of triples of diagonals $A_{n} A_{n+3}$ that intersect at a single point. Each triple in $S$ must come from three consecutive diagonals, and two different triples can only have one diagonal in common, so $S$ has at most $\left\lfloor\frac{2019}{2}\right\rfloor=1009$ triples. Hence the number of resulting pieces is at least
$$
1+(2019)+(2 \cdot 2019-1009)=5049 \text {. }
$$

To show that 5049 is attainable, we use the following construction. Let $B_{1} \ldots B_{1010}$ be a regular 1010and $A_{2 n+1}=\ell_{n} \cap \overleftrightarrow{B_{n+1} B_{n+2}}$. It follows that, for all $n=0, \ldots, 1008, \overline{A_{2 n-1} A_{2 n+2}}, \overline{A_{2 n} A_{2 n+3}}$, and $\overline{A_{2 n+1} A_{2 n+4}}$ intersect at $B_{n+1}$.