Promotion 1: Given $n$ points in a plane, where no three points are collinear, connect some points arbitrarily with line segments (these line segments are called edges). The condition to ensure that the graph contains a triangle with the given points as vertices is that the number of edges $x \geqslant \frac{2 \mathrm{C}_{n}^{3}+1}{\mathrm{C}_{n-2}^{1}}$, i.e., $x \geqslant \frac{n(n-1)(n-2)+3}{3(n-2)}$.
Proof: Construct drawers: each drawer contains three distinct points, yielding a total of $\mathrm{C}_{n}^{3}$ drawers. Since the same edge will appear in $\mathrm{C}_{n-2}^{1}$ drawers, according to the Dirichlet drawer principle, when $x \cdot \mathrm{C}_{n-2}^{1} \geqslant 2 \mathrm{C}_{n}^{3}+1$, it ensures that there is one drawer containing 3 edges, and these 3 edges form a triangle with three distinct non-collinear points.

This means that the condition for ensuring the appearance of a triangle in the graph is that the number of edges $x \geqslant \frac{2 \mathrm{C}_{n}^{3}+1}{\mathrm{C}_{n-2}^{1}}$, i.e., $x \geqslant$ $\frac{n(n-1)(n-2)+3}{3(n-2)}$