Belarus2022

$2021\cdot 1009=2039189$. Let's solve the problem for an arbitrary odd number $n$ of marked points and drawn segments. We enumerate the marked points on the circle by the numbers from $1$ to $n$ in the order of going around the circle. For each $i=1,\overline{\{}n\}$ we denote by $a_i$ the number of drawn segments for which $i$ is an endpoint. Among all $\left(\genfrac{}{}{0ex}{}{\{}{n}\right)\}\{2\}$ pairs of segments, at least ${\left(\genfrac{}{}{0ex}{}{\{}{a}\right)}_i\}\{2\}$ pairs do not intersect, since they have a common endpoint $i$. For different $i=1,\overline{\{}n\}$ these pairs do not coincide, hence the number of intersection points does not exceed $$M=\left(\genfrac{}{}{0ex}{}{\{}{n}\right)\}\{2\}-\left({\left(\genfrac{}{}{0ex}{}{\{}{a}\right)}_1\}\{2\}+{\left(\genfrac{}{}{0ex}{}{\{}{a}\right)}_2\}\{2\}+\cdots +{\left(\genfrac{}{}{0ex}{}{\{}{a}\right)}_n\}\{2\}\right).$$ Let's determine the largest possible value of $M$, for which we find the smallest possible value of the sum in brackets. It is easy to see, by opening the brackets, that the inequality $\left(\genfrac{}{}{0ex}{}{\{}{b}\right)\}\{2\}+(\genfrac{}{}{0ex}{}{\{}{a})\}\{2\}>\left(\genfrac{}{}{0ex}{}{\{}{b}\right)-1\}\{2\}+(\genfrac{}{}{0ex}{}{\{}{a})+1\}\{2\}$ is equivalent to the inequality $b-1>a$. Therefore, the numbers $a_i,i=1,\overline{\{}n\}$, must differ at most by one. Given that $a_1+a_2+\cdots +a_n=2n$, it's only possible if $a_1=a_2=\cdots =a_n=2$. Hence, the maximum possible value of $M$ is equal to $\frac{\{}{n}(n-1)\}\{2\}-n=\frac{\{}{n}(n-3)\}\{2\}$. To make sure that this value is the answer for the problem, we provide an example. Let's arrange points at the vertices of a regular $n$-gon and sequentially perform the following operation with each of them: choose a point $i$ and consider the set $p_i$ of intersection points of all possible segments with vertices at the remaining points. The set $p_i$ is finite, and the set of points on the arc containing the point $i$ is infinite. Therefore, we can choose such a new location for it that no segment connecting the point $i$ with any other point passes through the points of the set $p_i$. As a result of such operations, we obtain an arrangement of points such that no three segments with vertices at these points have a common point inside the circle. Now it suffices to draw segments connecting the points with numbers $i$ and $i+[n/2]$, $i=1,\overline{\{}n\}$, (we assume that the numbers are cyclic modulo $n$, i.e. $n+k=k$ if $k=1,\overline{\{}n\}$). In this case, any two segments that don't have a common endpoint, intersect and all $a_i$ are equal to two. Substituting the value $n=2021$ into the found formula $\frac{\{}{n}(n-3)\}\{2\}$, we find the answer $2021\cdot 1009=2039189$.