Source: TSTST-2019-P8
Category: Counting

Problem:
Let $\mathcal{S}$ be a set of $16$ points in the plane, no three collinear. Let $\chi(\mathcal{S})$ denote the number of ways to draw $8$ line segments with endpoints in $\mathcal{S}$, such that no two drawn segments intersect, even at endpoints. Find the smallest possible value of $\chi(\mathcal{S})$ across all such $\mathcal{S}$.

Solution:
The answer is $1430$. 

In general, we prove that for a set $\mathcal{S}$ of $2n$ points, the minimum possible value of $\chi(\mathcal{S})$ is the $n$-th Catalan number $C_n = \frac{1}{n+1}\binom{2n}{n}$. For $16$ points, this gives $C_8 = \frac{1}{9}\binom{16}{8} = 1430$.

First, it is a well-known result that if the points of $\mathcal{S}$ are in convex position (forming the vertices of a convex $2n$-gon), then $\chi(\mathcal{S}) = C_n$.

It remains to prove the lower bound $\chi(\mathcal{S}) \ge C_n$ for any configuration. We proceed by strong induction on $n$, with the base cases $n = 0$ and $n = 1$ being clear. Suppose the statement is proven for $0, 1, \dots, n$, and consider a set $\mathcal{S}$ with $2(n+1)$ points.

Let $P$ be a point on the convex hull of $\mathcal{S}$, and label the other $2n+1$ points $A_1, \dots, A_{2n+1}$ in order of angle from $P$.

Consider drawing a segment $\overline{PA_{2k+1}}$ for some $k \in \{0, 1, \dots, n\}$. This splits the $2n$ remaining points into two sets $\mathcal{U} = \{A_1, \dots, A_{2k}\}$ and $\mathcal{V} = \{A_{2k+2}, \dots, A_{2n+1}\}$, with $2k$ and $2(n-k)$ points respectively.

Note that by the choice of $P$ on the convex hull, the ray from $P$ through $A_{2k+1}$ separates $\mathcal{U}$ and $\mathcal{V}$. Thus, no segment with endpoints in $\mathcal{U}$ can intersect a segment with endpoints in $\mathcal{V}$. By the inductive hypothesis, $\chi(\mathcal{U}) \ge C_k$ and $\chi(\mathcal{V}) \ge C_{n-k}$. Thus, if we draw $\overline{PA_{2k+1}}$, we have at least $C_k C_{n-k}$ ways to complete the drawing by matching the points within $\mathcal{U}$ and matching the points within $\mathcal{V}$. 

Since matchings resulting from different choices of $k$ are disjoint (as the point $P$ is matched to a different point $A_{2k+1}$ in each case), we can sum over all choices of $k$ to obtain
\[ \chi(\mathcal{S}) \ge \sum_{k=0}^n C_k C_{n-k} = C_0 C_n + C_1 C_{n-1} + \dots + C_n C_0 = C_{n+1} \]
as desired. This completes the induction.

Answer: 1430

Grading guidelines:
(Partial) 1. Correctly identified the extremal configuration as 16 points in convex position and stated that the number of noncrossing perfect matchings then equals the Catalan number C8=1430 (with a correct counting argument or citation). 2. Proved a valid decomposition/recurrence for noncrossing matchings by choosing a hull point P and pairing it with an “odd-indexed” point in the angular order, obtaining a sum of products of smaller subproblems (a Catalan-type recurrence), even if not fully justified for all configurations. 3. Established a correct inductive lower-bound step in some nontrivial special case (e.g., when all points are in convex position, or when the chosen separating segment PR guarantees the remaining points split into two noninteracting sets), but did not finish the full general proof. (Almost) 1. Gave the full induction lower bound χ(S)≥Cn for 2n points using a convex-hull point P, but left a localized gap in justifying the separation claim (that points on the two sides of ray/segment PA_{2k+1} cannot produce crossing segments across the split). 2. Completed the Catalan recurrence and summation ∑_{k=0}^n CkC_{n-k}=C_{n+1} and concluded the minimum is C8, but did not clearly argue that the matchings counted for different k are disjoint (i.e., that P is matched to different points). 3. Solution is almost complete, but made minor mistakes which are not negligible.