Problem 2440

Proof. For the edge $BC$, a partial order $\prec_A$ can be defined on the set $P = P_1, \cdots, P_n$ as follows:
$$P_i \prec_A P_j \iff \begin{aligned}[t] & P_i \text{ is an interior point of the } \triangle P_j BC \\ & \text{the ray } P_j P_i \text{ intersects with the segment } BC. \end{aligned}$$
Similarly, a partial order $\prec_B$ can be defined on $P$ using the edge $CA$. If two distinct points $P_i, P_j$ are incomparable under $\prec_A$, then the line $P_i P_j$ does not intersect with the segment $BC$. Since the line $P_i P_j$ intersects with the boundary of $\triangle ABC$ at two points, it follows that $P_i, P_j$ are comparable under $\prec_B$. Thus, the anti-chains in the order $\prec_A$ are chains in the order $\prec_B$.
By using the Dilworth theorem, under the partial order $\prec_A$, there either exists a chain of length at least $\sqrt{n}$, or there exists an anti-chain of length at least $\sqrt{n}$. Combining with the above-mentioned property that an anti-chain in the order $\prec_A$ is a chain in the order $\prec_B$, we know that there either exists a chain of length at least $\sqrt{n}$ under the partial order $\prec_A$, or there exists a chain of length at least $\sqrt{n}$ under the partial order $\prec_B$.
Without loss of generality, suppose there exists a chain of length at least $\sqrt{n}$ under the partial order $\prec_A$, denoted as $P_1 \prec_A \cdots \prec_A P_t$, where $t \ge \sqrt{n}$. Connect the segments
$$BP_i, CP_i (1 \le i \le t), P_1 P_2, \cdots, P_{t-1} P_t, P_t A,$$
which partition $\triangle ABC$ into $2t + 1$ triangles
$$(*) \quad \triangle BCP_1, \triangle BP_i P_{i+1}, \triangle CP_i P_{i+1} (1 \le i \le t-1), \triangle BP_t A, \triangle CP_t A.$$
Let the number of points from set $P$ in the interiors of $\triangle BCP_1, \triangle BP_1 P_2, \cdots, \triangle BP_t A$ be $k_0, k_1, \cdots, k_t$, respectively. For each such triangle $\triangle$, denoted as $\triangle BQR$, suppose the interior points from set $P$ in $\triangle$ are $U_j (1 \le j \le k)$, where $\angle QBU_j (1 \le j \le k)$ are arranged in increasing order. Connect the segments $BU_j (1 \le j \le k)$ and $QU_1, U_1 U_2, \cdots, U_k R$, yielding $k + 1$ small triangles that all include $B$, and divide the polygon $QU_1 \cdots U_k R$ into a union of $k$ small triangles arbitrarily (as it is well-known that any polygon can be triangulated). This way, $\triangle$ is partitioned into the union of $2k + 1$ small triangles, $k + 1$ of which include $B$ as a vertex. Similarly, suppose the number of points from set $P$ in the interiors of $\triangle CP_1 P_2, \cdots, \triangle CP_t A$ are $\ell_1, \cdots, \ell_t$, respectively. For each such triangle $\triangle'$, with $\ell$ points from set $P$ in its interior, $\triangle'$ can be partitioned into a union of $2\ell + 1$ small triangles, $\ell + 1$ of which include $C$ as a vertex.
This way, we have constructed a triangulation of $\triangle ABC$, where at least
$$T = \sum_{i=0}^{t} (k_i + 1) + \sum_{i=1}^{t} (\ell_i + 1)$$
of the small triangles include a point from $\{B, C\}$ as a vertex. Noting that $k_0 + k_1 + \cdots + k_t + \ell_1 + \cdots + \ell_t = n - t$, we get $T = (n - t) + (2t + 1) = n + t + 1 \ge \sqrt{n} + 1$, which proves the conclusion of the theorem. $\square$