Example 4.1.5 $n$ points $(n \geqslant 5)$ in the plane, no three of which are collinear. Among the triangles formed with these points as vertices, at most $70 \%$ are acute triangles.
Solution 1: First, discuss the case when $n=5$. At this time, there are $\mathrm{C}_{5}^{3}=10$ triangles. We only need to prove that there are at least three non-acute triangles, and discuss according to different situations of the convex hull of the five-point set $P_{1}, P_{2}, P_{3}, P_{4}, P_{5}$.
(1) The convex hull is a convex pentagon $P_{1} P_{2} P_{3} P_{4} P_{5}$. A convex pentagon has at least two obtuse angles, which may be adjacent or not. If $\angle P_{1}, \angle P_{2}$ are obtuse angles, then $\triangle P_{5} P_{1} P_{2}$, $\triangle P_{1} P_{2} P_{3}$ are obtuse triangles. In the quadrilateral $P_{1} P_{3} P_{4} P_{5}$, at least one of the four interior angles is non-acute, so there are at least three non-acute triangles. If $\angle P_{1}, \angle P_{3}$ are obtuse angles, the same proof applies.
(2) The convex hull is a convex quadrilateral $P_{1} P_{2} P_{3} P_{4}$, and $P_{5}$ is inside $P_{1} P_{2} P_{3} P_{4}$. Any three of the five points are not collinear. $P_{5}$ must be in one of $\triangle P_{1} P_{2} P_{3}$ or $\triangle P_{1} P_{3} P_{4}$. If $P_{5} \in \triangle P_{1} P_{2} P_{3}$, then $\angle P_{1} P_{5} P_{2}$, $\angle P_{2} P_{5} P_{3}$, $\angle P_{3} P_{5} P_{1}$ have at least two non-acute angles. Also, the four interior angles of the quadrilateral $P_{1} P_{2} P_{3} P_{4}$ have at least one non-acute angle, so the conclusion holds.
(3) The convex hull is $\triangle P_{1} P_{2} P_{3}$, and $P_{4}, P_{5} \in \triangle P_{1} P_{2} P_{3}$. Then, $\angle P_{1} P_{4} P_{2}$, $\angle P_{2} P_{4} P_{3}$, $\angle P_{3} P_{4} P_{1}$ have at least two non-acute angles; $\angle P_{1} P_{5} P_{2}$, $\angle P_{2} P_{5} P_{3}$, $\angle P_{3} P_{5} P_{1}$ also have at least two non-acute angles, so the conclusion holds.

Now, discuss the $n$ points $P_{1}, P_{2}, \cdots, P_{n}$. Take any five points $P_{i_{1}}, P_{i_{2}}, P_{i_{3}}, P_{i_{4}}, P_{i_{5}}$, among which there are at least three non-acute triangles, and at most 7 acute triangles. There are $\mathrm{C}_{n}^{5}$ groups of five points in $P_{1}, P_{2}, \cdots, P_{n}$, so there are at most $7 \times \mathrm{C}_{n}^{5}$ acute triangles. In the above calculation, there is repetition. Each acute triangle is contained in $\mathrm{C}_{n-3}^{2}$ groups of five points, so each acute triangle is counted $\mathrm{C}_{n-3}^{2}$ times. Therefore, the total number of acute triangles is at most $\frac{7 \times \mathrm{C}_{n}^{5}}{\mathrm{C}_{n-3}^{2}}$. There are $\mathrm{C}_{n}^{3}$ triangles in total, so the proportion of acute triangles does not exceed
$$
\frac{7 \times \mathrm{C}_{n}^{5}}{\mathrm{C}_{n-3}^{2} \times \mathrm{C}_{n}^{3}}=\frac{7}{10}=70 \%.
$$

Discussion: We can prove by another method that when $n \geqslant 16$, the proportion of acute triangles does not exceed $67 \%$.

Solution 2: Let $A=\left\{P_{i}\right\}_{i=1}^{n}, A_{i}=A \backslash\left\{P_{i}\right\} \quad(i=1,2, \cdots, n)$. $\alpha_{n}(A)$ is the number of acute triangles with vertices in $A$, and $\alpha_{n-1}\left(A_{i}\right)$ has a similar meaning. $\alpha_{n}=\operatorname{Sup}_{A} \alpha_{n}(A)$. Then we have $\alpha_{n}(A)=\frac{1}{n-3}\left[\alpha_{n-1}\left(A_{1}\right)+\alpha_{n-1}\left(A_{2}\right)+\cdots+\alpha_{n-1}\left(A_{n}\right)\right]$ (since each acute triangle in $A$ is counted $n-3$ times in $\sum_{i=1}^{n} \alpha_{n-1}\left(A_{i}\right)$).
$$
\alpha_{n}(A) \leqslant \frac{n}{n-3} \alpha_{n-1}.
$$

Since $A$ is any set of $n$ points where no three points are collinear, we have
$$
\begin{array}{l}
\alpha_{n} \leqslant \frac{n}{n-3} \alpha_{n-1}, \\
\frac{\alpha_{n}}{\mathrm{C}_{n}^{3}} \leqslant \frac{n}{n-3} \cdot \frac{\alpha_{n-1}}{\mathrm{C}_{n}^{3}}=\frac{\alpha_{n-1}}{\mathrm{C}_{n-1}^{3}},
\end{array}
$$

i.e., $\frac{\alpha_{n}}{\mathrm{C}_{n}^{3}}$ is monotonically decreasing, $\frac{\alpha_{n}}{\mathrm{C}_{n}^{3}} \leqslant \frac{\alpha_{5}}{\mathrm{C}_{5}^{3}} \leqslant 70 \%$.
Since $\alpha_{n} \in \mathbf{N}$, from $\alpha_{n} \leqslant \frac{n}{n-3} \alpha_{n-1}$, we can deduce
$\alpha_{n} \leqslant\left[\frac{n}{n-3} \alpha_{n-1}\right]$, from which we can calculate $\frac{\alpha_{16}}{\mathrm{C}_{16}^{3}} \leqslant \frac{375}{560}=67 \%$,
When $n \geqslant 16$, $\frac{\alpha_{n}}{\mathrm{C}_{n}^{3}} \leqslant 67 \%$.
Discussion: $\frac{\alpha_{n}}{\mathrm{C}_{n}^{3}}$ is monotonically decreasing, so $\lim _{n \rightarrow \infty} \frac{\alpha_{n}}{\mathrm{C}_{n}^{3}}$ exists. Studying this limit is an interesting but challenging problem.

Combinatorial geometry problems are flexible and varied, often without a fixed method. It is necessary to innovate "techniques" according to specific situations. Below, we will present some other methods for solving finite point set problems, hoping that readers will carefully appreciate the techniques involved.