Example 20 Given $n$ points $(n \geqslant 5)$ in a plane, where no four points are coplanar, find the smallest positive integer $m$ such that after connecting $m$ lines between the $n$ points, there must exist two triangles that share exactly one common point. (This problem is essentially a 2003 Chinese National Training Team practice problem, with only a slight change in the wording)
First, divide $n$ points into two disjoint sets $M_{1}$ and $M_{2}$, where $M_{1}$ contains $\left[\frac{n+1}{2}\right]$ points, and $M_{2}$ contains $\left[\frac{n}{2}\right]$ points. Connect each point in $M_{1}$ to each point in $M_{2}$, and connect each pair of points in $M_{1}$. The total number of line segments is $\left[\frac{n+1}{2}\right]\left[\frac{n}{2}\right]+1=\left[\frac{n^{2}}{4}\right]+1$. However, there do not exist two triangles with exactly one common vertex.
Therefore, the smallest positive integer $m \geqslant\left[\frac{n^{2}}{4}\right]+2$.
Next, we use mathematical induction to prove that if $n(\geqslant 5)$ points are connected by $\left[\frac{n^{2}}{4}\right]+2$ line segments, there must exist two triangles with exactly one common vertex.

When $n=5$, the maximum number of line segments that can be drawn between 5 points is $C_{5}^{2}=10$. Now, $\left[\frac{5^{2}}{4}\right]+2=8$ line segments are drawn, so only two pairs of points are not connected. Therefore, there must be a point $A$ that is connected to the other 4 points $B, C, D, E$. Among the three pairs of opposite edges (edges without a common endpoint) formed by $B, C, D, E$, at least one pair of opposite edges must exist. Suppose $B$ is connected to $C$ and $D$ is connected to $E$. Thus, there exist two triangles with a common vertex: $\triangle A B C$ and $\triangle A D E$. Therefore, the conclusion holds for $n=5$.

Assume the conclusion holds for $n-1$, and consider the case for $n \geqslant 6$. In this case, the total number of line segments from $n$ points is $2\left[\frac{n^{2}}{4}\right]+4$, so there must be a point $A$ from which the number of line segments is at most $\left[\frac{2\left[\frac{n^{2}}{4}\right]+4}{n}\right]$. When $n=$
$$
\begin{array}{l}
2 k \geqslant 6 \text {, }\left[\frac{2\left[\frac{n^{2}}{4}\right]+4}{n}\right]=\left[\frac{2 k^{2}+4}{2 k}\right]=\left[k+\frac{2}{k}\right]=k=\left[\frac{n}{2}\right] ; \text { when } n=2 k+1 \geqslant 7 \text {, } \\
{\left[\frac{2\left[\frac{n^{2}}{4}\right]+4}{n}\right]=\left[\frac{2 k(k+1)+4}{2 k+1}\right]=\left[k+\frac{k+4}{2 k+1}\right]=\left\{\begin{array}{l}
k+1=\left[\frac{n}{2}\right]+1(n=7) \\
k=\left[\frac{n}{2}\right](n \geqslant 9)
\end{array} \text {, so if } n=6\right.}
\end{array}
$$

or $n \geqslant 8$, removing $A$ and the line segments from $A$ leaves $n-1$ points, and the number of line segments among them is at least
$$
\left[\frac{n^{2}}{4}\right]+2-\left[\frac{n}{2}\right]=\left[\frac{n}{2}\right]\left(\left[\frac{n+1}{2}\right]-1\right)+2=\left[\frac{n}{2}\right]\left[\frac{n-1}{2}\right]+2=\left[\frac{(n-1)^{2}}{4}\right]+2 \text {. }
$$

By the induction hypothesis, there must exist two triangles with exactly one common vertex.
Therefore, we only need to consider the case $n=7$. In this case, 7 points are connected by $\left[\frac{7^{2}}{4}\right]+2=14$ line segments. If the number of line segments from any point is at most $\left[\frac{n}{2}\right]=3$, then the problem can be reduced to the case with $n-1=6$ points, so we only need to consider the case where each point has exactly 4 line segments.

Consider the 4 points $B, C, D, E$ connected to $A$. Since each point has 4 line segments and there are only 7 points, each of $B, C, D, E$ must be connected to at least one of the other 3 points. Suppose $B$ is connected to $C$. If $D$ is connected to $E$, the conclusion holds. Therefore, assume $D$ and $E$ are not connected. If $D$ and $E$ are connected to $B$ and $C$ (or $C$ and $B$), the conclusion also holds. Therefore, we only need to consider the case where $D$ and $E$ are both connected to $B$ (or both to $C$). Suppose $D$ and $E$ are both connected to $B$, and $C, D, E$ are not connected to each other. Consider the other two points $F, G$. Since $C, D, E$ each have 4 line segments, $C, D, E$ are all connected to $F$ and $G$. Since $F$ and $G$ each have 4 line segments, $F$ and $G$ are connected (since $F$ and $G$ are not connected to $A$ or $B$), thus we get the graph 4-16, where $\triangle A D B$ and $\triangle F G D$ are two triangles with exactly one common vertex.
In summary, $m=\left[\frac{n^{2}}{4}\right]+2$.