Croatian Mathematical Olympiad

Let $a_n$ be the maximum possible number of sections of a set of $n$ points in the plane. Let $S$ be a set of $n+1$ points in the plane and let us consider one of these points, call it $T$, on the convex hull of that set. Note that each section of $S$ restricts to a section of the set $S\setminus \{T\}$. Let $\{A,B\}$ and $\{C,D\}$ be two different sections of the set $S$ that restrict to the same section of the set $S\setminus \{T\}$. This means that $\{A\setminus \{T\},B\setminus \{T\}\}=\{C\setminus \{T\},D\setminus \{T\}\}$. Furthermore, it means, without loss of generality, that the sets $A$ and $C$ are the same and the sets $B$ and $D$ are the same, up to the point $T$. Hence, for the section $\{A\setminus \{T\},B\setminus \{T\}\}$ of the set $S\setminus \{T\}$ we can take a line passing through the point $T$ such that all the points of the set $A\setminus \{T\}$ are on one side of the line, while all the points of the set $B\setminus \{T\}$ are on the other. Let us consider all the lines passing through the point $T$ and none of the points of the set $S\setminus \{T\}$. Two such lines could give two different sections of the set $S\setminus \{T\}$ only if there is at least one point of the set $S\setminus \{T\}$ between them. Hence, the number of different sections of the set $S\setminus \{T\}$ corresponding to a line passing through $T$ is at most $n$, i.e. as many as there are points in the set $S\setminus \{T\}$. Finally, we note that $$a_{n+1}\le n+a_n\le n+(n-1)+a_{n-1}\le \cdots \le n+(n-1)+\cdots +2+1+a_1=\left(\genfrac{}{}{0ex}{}{n+1}{2}\right)+1.$$ So, the set of $n$ points in the plane can have at most $\left(\genfrac{}{}{0ex}{}{n}{2}\right)+1$ sections. That number is achievable, e.g. in the case of a regular $n$-gon.