Third question Given ten points in space, where no four points lie on the same plane. Connect some of the points with line segments. If the resulting figure contains no triangles and no spatial quadrilaterals, determine the maximum number of line segments that can be drawn.
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    Solution 1 The maximum number of line segments that can be connected is 15, an
example is shown in Figure 5.
        It is only necessary to prove that 16 line segments are not possible.
        If 16 line segments are connected, it will produce 32 endpoints.
    By the pigeonhole principle, there is a point with a degree of at least 4.
        First, we prove a lemma by contradiction.
        Lemma: If there are 16 line segments that meet the conditions and
    the maximum degree of each point is 4, then the number of points with a degree of 4
    does not exceed two.
        Proof: Assume there are three points with a degree of 4, let
    these three points with a degree of 4 be }A, B, C\mathrm{. In this case, there are
    two scenarios.
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(1) If points $A$ and $B$ are not connected, this can be further divided into two scenarios: the four points connected to $A$ and $B$ are not repeated; the four points connected to $A$ and $B$ have one repetition (if the points connected to $A$ and $B$ have two repetitions, it will form a quadrilateral).

By the pigeonhole principle, among the four points connected to $C$, there must be two points in one of the above two groups, forming a triangle or a quadrilateral, which is a contradiction.
(2) If points $A$ and $B$ are connected, assume point $A$ is connected to $A_{1}$, $A_{2}$, $A_{3}$, and point $B$ is connected to $B_{1}$, $B_{2}$, $B_{3}$, as shown in Figure 6. Similarly, by the pigeonhole principle, a contradiction is derived. The lemma is proved.
For the original problem, we discuss two scenarios:
(1) The maximum degree of all points is 4.
By the lemma, there are two points with a degree of 4, and the rest of the points have a degree of 3. Assume these two points are $A$ and $B$. At this time, the degree of the points other than $A$ and $B$ is 3.
(i) If points $A$ and $B$ are not connected, this can be further divided into two scenarios: the four points connected to $A$ and $B$ are not repeated; the four points connected to $A$ and $B$ have one repetition (if the points connected to $A$ and $B$ have two repetitions, it will form a quadrilateral). Additionally, there is at most one point $D$.

But each point in the above groups needs to draw two more edges (not to the points in the same group, but to the other group, at most one edge). Therefore, the edges drawn to point $D$ are more than 4, which is a contradiction.
(ii) If points $A$ and $B$ are connected, assume point $A$ is connected to $A_{1}$, $A_{2}$, $A_{3}$, and point $B$ is connected to $B_{1}$, $B_{2}$, $B_{3}$, as shown in Figure 6. The other two points are $C$ and $D$.

Similarly, $A_{i}$ and $B_{i}$ ($i=1,2,3$) each draw one edge to point $C$ or $D$. Therefore, one of the points $C$ or $D$ draws three edges, forming a quadrilateral, which is a contradiction.
(2) The maximum degree of a point is not less than 5, assume the maximum degree is $i$ ($i \geqslant 5$).

Assume the point with a degree of $i$ is $A$, and the $i$ points connected to it are $A_{1}$, $A_{2}$, ..., $A_{i}$; the other $9-i$ points are $B_{1}$, $B_{2}$, ..., $B_{9-i}$. Then, point $B_{j}$ ($j=1,2, \cdots, 9-i$) can only be connected to $A_{1}$, $A_{2}$, ..., $A_{i}$. Among them, $B_{j}$ cannot be connected to two points at the same time (otherwise, it will form a quadrilateral).

Therefore, $B_{1}$, $B_{2}$, ..., $B_{9-i}$ can connect at most $9-i$ line segments to $A_{1}$, $A_{2}$, ..., $A_{i}$.

Now consider how many line segments can be connected between $B_{1}$, $B_{2}$, ..., $B_{9-i}$.

According to the assumption, we need to connect $16-(9-i)-i=7$ more line segments.
The maximum number of line segments that can be connected between $9-i$ points is
$$
C_{9-i}^{2}=\frac{(9-i)(8-i)}{2} \text {. }
$$

By $i \geqslant 5$, we know $C_{9-i}^{2} \leqslant 6<7$, which is a contradiction, so this scenario is excluded.
In summary, the maximum number of line segments that can be connected is not 16. Therefore, the original proposition holds, and the maximum value is 15.
(Du Yulin, Experimental Class 4, Tianjin Yaohua Middle School, 300040)

Solution 2 Let the ten points be $A_{1}$, $A_{2}$, ..., $A_{10}$, and the edges between them form a graph $G(V, E)$. Then, the graph $G$ has exactly $\sum_{i=1}^{10} C_{d(1)}^{2}$ angles.

According to the conditions, in any $\angle A_{i} A_{k} A_{j}$ in graph $G$, $A_{i}$ and $A_{j}$ cannot be connected; and in any two different $\angle A_{i_{1}} A_{k_{1}} A_{j_{1}}$ and $\angle A_{i_{2}} A_{k_{2}} A_{j_{2}}$ in graph $G$, we have
$$
\left\{A_{i_{1}}, A_{j_{1}}\right\}=\left\{A_{i_{2}}, A_{j_{2}}\right\} .
$$

Thus, the complement graph of $G(V, E)$ has at least $\sum_{i=1}^{10} C_{d(A, i)}^{2}$ edges.
$$
\begin{array}{l}
\text { Hence } C_{10}^{2} \geqslant \sum_{i=1}^{10} C_{d(A, i)}^{2}+|E| \\
=\frac{\sum_{i=1}^{10} d^{2}(A_{i})-\sum_{i=1}^{10} d(A_{i})}{2}+|E| \\
\geqslant \frac{\frac{1}{10}\left(\sum_{i=1}^{10} d(A_{i})\right)^{2}-\sum_{i=1}^{10} d(A_{i})}{2}+|E| \\
=\frac{|E|^{2}}{5} .
\end{array}
$$
Solving this, we get $|E| \leqslant 15$.
Figure 7 is a graph with 15 edges.
In summary, the maximum number of edges that can be connected is 15.
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