2. Let $p_{1}, p_{2}, \cdots, p_{n}$ be $n$ points in the plane, no three of which are collinear. Connect $m$ line segments between them such that among any 4 points, there are 3 points that form a triangle, i.e., each pair of these 3 points is connected by a line segment. Find the minimum value of $m$.
(28th IMO Shortlist)
2. Connect any two points among $p_{1}, p_{2}, \cdots, p_{n-1}$ with a line segment, resulting in a total of $C_{n-1}^{2}$ line segments, and this method clearly satisfies the problem's requirements. On the other hand, suppose there is no line segment between points $p_{1}, p_{2}$. Then, any two points among the remaining $n-2$ points must be connected by a line segment. Otherwise, there must exist $3 \leqslant i<j \leqslant n$ such that points $p_{i}$ and $p_{j}$ are not connected, and thus the 4 points $\left\{p_{1}, p_{2}, p_{i}, p_{j}\right\}$ do not satisfy the problem's requirements. Additionally, for any $i(3 \leqslant i \leqslant n)$, point $p_{i}$ must be connected to at least one of $p_{1}, p_{2}$, otherwise, it will again lead to 4 points not satisfying the problem's requirements. Thus, the number of line segments in a set that meets the problem's requirements is no less than $C_{n-2}^{2}+n-2=C_{n-1}^{2}$. In summary, the minimum value of $m$ is $C_{n-1}^{2}$.