Given ten points in space, where no four points lie on the same plane. Some points are connected by line segments. If the resulting figure contains no triangles and no spatial quadrilaterals, determine the maximum number of line segments that can be drawn. ${ }^{[1]}$
(2016, National High School Mathematics Joint Competition)
Proof Let $v$ be a vertex in graph $G$, and $N_{i}(v)$ denote the set of points at a distance $i$ from $v$. For example,
$$
N_{0}(v)=\{v\},
$$
$N_{1}(v)=\{u \mid u$ is adjacent to $v\}$,
$N_{2}(v)=\{w \mid w$ is adjacent to $u$, not adjacent to $v$, and $u$ is adjacent to $v\}$.

First, we prove a lemma.
Lemma Let $G=(V, E)$ be a graph with girth $g(G) \geqslant 5$, average degree, number of vertices, and number of edges be $d$, $n$, and $e$ respectively. Then there exists a vertex $v$ such that
$$
\left|N_{2}(v)\right| \geqslant d^{2}-\operatorname{deg} v.
$$

Proof Assume the conclusion does not hold. Then for any vertex $v$, we have
$$
\left|N_{2}(v)\right|<d^{2}-\operatorname{deg} v.
$$

Since $g(G) \geqslant 5$, for any vertex $v$, if $w \in N_{2}(v)$, then $w$ is adjacent to exactly one vertex in $N_{1}(v)$. Because $w \in N_{2}(v)$, there is at least one vertex in $N_{1}(v)$ that is adjacent to it. If there were two vertices adjacent to $w$, say $u_{1}$ and $u_{2}$, then $v, u_{1}, w, u_{2}$ would form a cycle of length 4, contradicting $g(G) \geqslant 5$.
Therefore, for any vertex $v$, we have
$\left|N_{2}(v)\right|=\sum_{u \in N_{1}(v)}(\operatorname{deg} u-1)$.
Thus, $\sum_{u \in N_{1}(v)}(\operatorname{deg} u-1)<d^{2}-\operatorname{deg} v$.
Summing over all $v$ gives
$\sum_{v \in V} \sum_{u \in N_{1}(v)}(\operatorname{deg} u-1)<\sum_{v \in V}\left(d^{2}-\operatorname{deg} v\right) = n d^{2}-2 e$.
The left-hand side is $\sum_{v \in V}(\operatorname{deg} v)(\operatorname{deg} v-1)$, so
$$
\sum_{v \in V} \operatorname{deg}^{2} v < n d^{2} = n\left(\frac{\sum_{v \in V} \operatorname{deg} v}{n}\right)^{2}.
$$

Let $\boldsymbol{a}=(1,1, \cdots, 1)$,
$\boldsymbol{b}=\left(\operatorname{deg} v_{1}, \operatorname{deg} v_{2}, \cdots, \operatorname{deg} v_{n}\right)$.
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
|\boldsymbol{a} \cdot \boldsymbol{b}|^{2}=\left(\sum_{v \in V} \operatorname{deg} v\right)^{2} \\
\leqslant|\boldsymbol{a}|^{2}|\boldsymbol{b}|^{2}=n \sum_{v \in V} \operatorname{deg}^{2} v \\
\Rightarrow \sum_{v \in V} \operatorname{deg}^{2} v \geqslant n\left(\frac{\sum_{v \in V} \operatorname{deg} v}{n}\right)^{2},
\end{array}
$$

which is a contradiction.
The lemma is proved.
Since for any vertex $v$, we have
$$
n \geqslant\left|N_{0}(v)\right|+\left|N_{1}(v)\right|+\left|N_{2}(v)\right|,
$$

by the lemma, we get
$n \geqslant 1+\operatorname{deg} v+d^{2}-\operatorname{deg} v=1+d^{2}$
$\Rightarrow d \leqslant \sqrt{n-1}$.
Since $d=\frac{1}{|V|} \sum_{v \in V} \operatorname{deg} v=\frac{2 e}{n}$, we have
$e \leqslant \frac{n \sqrt{n-1}}{2}$.
Another proof: Construct a simple 10-vertex graph $G$ with the ten points as vertices and the connecting line segments as edges.
Since graph $G$ contains no triangles or quadrilaterals, $g(G) \geqslant 5$.
By the theorem, the number of edges is at most
$\frac{10 \times \sqrt{10-1}}{2}=15$.
Additionally, the Petersen graph satisfies the conditions, so the maximum number of edges is 15.