Question 40: Given any $n$ points $\mathrm{P}_{1}, \mathrm{P}_{2}, \ldots, \mathrm{P}_{\mathrm{n}}$ on a plane, where no three points are collinear. Each point $\mathrm{P}_{\mathrm{i}}(1 \leq i \leq n)$ is arbitrarily colored either red or blue. Let $S$ be the set of triangles with vertex set $\left\{P_{1}, P_{2}, \ldots, P_{n}\right\}$, and has the property: for any two line segments $\mathrm{P}_{\mathrm{i}} \mathrm{P}_{\mathrm{i}}$ and $\mathrm{P}_{\mathrm{u}} \mathrm{P}_{\mathrm{v}},$ the number of triangles in $S$ that have $\mathrm{P}_{\mathrm{i}} \mathrm{P}_{\mathrm{j}}$ as an edge is the same as the number of triangles in $S$ that have $\mathrm{P}_{\mathrm{u}} \mathrm{P}_{\mathrm{v}}$ as an edge. Find the smallest $n$ such that in $S$ there are always two triangles, each of which has vertices of the same color.
Question 40:
Solution: Let the number of triangles in $\mathrm{S}$ with $\mathrm{P}_{\mathrm{i}} \mathrm{P}_{\mathrm{j}}$ as an edge be $\mathrm{k}$, then $\mathrm{k}$ is a positive integer and is independent of $\mathrm{i}$ and $\mathrm{j}$. Since there are $C_{n}^{2}$ edges $\mathrm{P}_{\mathrm{i}} \mathrm{P}_{\mathrm{j}}$, the total number of edges (counted with repetition) produced by all triangles in $\mathrm{S}$ is $\mathrm{k} \cdot \mathrm{C}_{\mathrm{n}}^{2}$. Since each triangle has three edges, $|\mathrm{S}| = \frac{k C_{n}^{2}}{3}$ (i.e., there are $\frac{k C_{n}^{2}}{3}$ triangles in total).

Let there be $x$ triangles in $S$ with vertices of the same color, then the number of triangles in $S$ with vertices of different colors is $\frac{\mathrm{kC}_{2}^{2}}{3}-x$. At the same time, each triangle with vertices of different colors produces two edges with endpoints of different colors, so the total number of edges with endpoints of different colors in $S$ is $2\left(\frac{\mathrm{kC}_{2}^{2}}{3}-\mathrm{x}\right)$ (counted with repetition).

On the other hand, suppose among $P_{1}, P_{2}, \ldots, P_{n}$, there are $n_{1}$ points colored red and $n_{2}$ points colored blue, with $\mathrm{n}_{1}+\mathrm{n}_{2}=\mathrm{n}$. By assumption, each edge with endpoints of different colors appears $\mathrm{k}$ times in the set of all triangles in $\mathrm{S}$, so there are $\mathrm{kn}_{1} \mathrm{n}_{2}$ such edges. Therefore:
$$
\begin{array}{l}
2\left(\frac{k C_{2}^{2}}{3}-x\right)=k n_{1} n_{2} \\
\Rightarrow x=\frac{k\left(2 C_{n}^{2}-3 n_{1} n_{2}\right)}{6}=\frac{k\left(n^{2}-n-3 n_{1} n_{2}\right)}{6} \geq \frac{k\left[n^{2}-n \frac{\left.3\left(n_{1}+n\right)_{2}\right)^{2}}{4}\right]}{6}=\frac{k n\left(\frac{n}{4}-1\right)}{6}=\frac{k n(n-4)}{24}
\end{array}
$$

Thus, when $n \geq 8$, $x \geq \frac{8(8-4)^{6}}{24}=\frac{4}{3}>1$, hence $x \geq 2$. That is, there are at least two triangles, each with vertices of the same color.

When $n=7$, the conclusion does not necessarily hold. For example, if 1, 2, 4 are colored red and 3, 5, 6, 7 are colored blue, then the set of triangles $\{1,2,4\}, \{2,3,5\}, \{3,4,6\}, \{4,5,7\}, \{5,6,1\}, \{6,7,2\}, \{7,1,3\}$ meets the requirement (each edge $\mathrm{ij}$ appears in one triangle), but there are no two triangles with vertices of the same color.