Shirov V.

On a circle, $2 N$ points are marked ($N$ is a natural number). It is known that through any point inside the circle, no more than two chords with endpoints at the marked points pass. We will call a matching a set of $N$ chords with endpoints at the marked points such that each marked point is the endpoint of exactly one of these chords. We will call a matching even if the number of points where its chords intersect is even, and odd otherwise. Find the difference between the number of even and odd matchings.
Let the marked points be $A_{1}, A_{2}, \ldots, A_{2 N}$ in the order of a clockwise traversal of the circle. We will prove by induction on $N$ that the number of even matchings is one more than the number of odd matchings. For $N=1$, the statement is obvious: there is only one matching, and it is even.

Inductive step. First method. Suppose in the matching, the chord $A_{1} A_{i}$ is involved and it is intersected by exactly $k$ chords. Consider the points $A_{2}, \ldots, A_{i-1}$; exactly $k$ of them are endpoints of chords intersecting $A_{1} A_{i}$. The remaining $i-2-k$ points are paired by chords that do not intersect $A_{1} A_{i}$. Thus, the number $i-2-k$ is even, meaning that $i$ and $k$ have the same parity.

We divide all matchings into $2 N-1$ groups $\Pi_{2}, \ldots, \Pi_{2 N}$: in group $\Pi_{i}$, the point $A_{1}$ is connected to $A_{i}$. Remove the chord $A_{1} A_{i}$ from each matching in $\Pi_{i}$; we obtain all possible matchings on the remaining $2 N-2$ points. By the inductive hypothesis, among them, there is one more even matching than odd. If $i$ is even, the parity of the matching does not change upon removal, and if $i$ is odd, the parity changes. Therefore, in each of the $N$ sets $\Pi_{2}, \ldots, \Pi_{2 N}$, there is one more even matching than odd, and in each of the $N-1$ sets $\Pi_{3}, \ldots, \Pi_{2 N-1}$, there is one more odd matching than even. In total, the number of even matchings is greater than the number of odd matchings by $N-(N-1)=1$.

Second method. Consider all matchings in which $A_{2 N-1}$ and $A_{2 N}$ are connected by a chord. This chord does not intersect any other. Thus, by removing it from each of the considered matchings, we obtain all matchings on the points $A_{1}, \ldots, A_{2 N-2}$, and the parity of each of them is preserved. By the inductive hypothesis, among our matchings, there is one more even matching than odd.

It remains to prove that among the remaining matchings, there are equally many even and odd. Consider any of them; suppose it contains chords $A_{2 N-1} A_{i}$ and $A_{2 N} A_{k}$. Replace these chords with $A_{2 N} A_{i}$ and $A_{2 N-1} A_{k}$. If the original chord intersected with some other chord, the new chord will also intersect with it. On the other hand, if the chords $A_{2 N-1} A_{i}$ and $A_{2 N} A_{k}$ did not intersect, the new chords will intersect, and vice versa. Thus, we have divided the remaining matchings into pairs of different parity.

## Answer

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