Question 47: In a Cartesian coordinate system, take $\mathrm{n}$ points such that no three points are collinear, and color each point in one of three colors: red, green, or yellow, so that the following conditions are met:
(1) In any triangle with all vertices red, there is at least one green point inside;
(2) In any triangle with all vertices green, there is at least one yellow point inside;
(3) In any triangle with all vertices yellow, there is at least one red point inside. Find the maximum value of the positive integer $\mathrm{n}$.
Question 47:
Solution: Let there be $\mathrm{a}$ red points, and their convex hull is a convex $\mathrm{r}$-gon; inside the convex $\mathrm{r}$-gon, there are $\mathrm{b}$ green points, and their convex hull is a convex $s$-gon; inside the convex $s$-gon, there are $c$ yellow points, and their convex hull is a convex $t$-gon; inside the convex $t$-gon, there are $a^{\circ}$ red points.
If the three vertices of a triangle are all of the same color, then the triangle is called a "red, green, or yellow triangle."
Since the convex hull of all red points is exactly divided into $2 a-r-2$ non-overlapping red triangles, we have $b \geq 2 a-r-2$.
Similarly, $c \geq 2 b-s-2 \geq b-2 ; a=2 c-t-2 \geq c-2$.
Also, because the $\mathrm{r}$ vertices of the red convex hull are not counted among the $\mathrm{a}^{\prime}$ red points, hence
$$
a \geq r+a^{\prime} \geq r+(c-2) \geq r+b-4 \geq r+(2 a-r-2)-4=2 a-6
$$

Solving this, we get $a \leq 6$. Similarly, $b 、 c \leq 6$. Therefore, $n=a+b+c \leq 18$.
Below is an example with $n=18$.
First, consider the construction of six red points.
According to the previous discussion, there should be four green points inside the convex hull of these red points, placed as shown in the figure below, where $R_{i}(i=1 、 2 、 \ldots 、 6)$ are the six red points, and $G_{i}(i=1 、 2 、 3 、 4)$ are the four green points. All points are symmetrically placed about point $\mathrm{O}$, and the aspect ratio of the rectangles $\mathrm{R}_{2} \mathrm{G}_{5} \mathrm{R}_{5} \mathrm{G}_{2}$ and $\mathrm{R}_{2} \mathrm{G}_{6} \mathrm{R}_{6} \mathrm{G}_{3}$ is $\sqrt{3}$. Note that each red triangle contains exactly one green point. Then, place the two green points $\mathrm{G}_{1} 、 \mathrm{G}_{4}$ at the positions obtained by rotating the red points $\mathrm{R}_{1} 、 \mathrm{R}_{4}$ counterclockwise by $60^{\circ}$ about point $\mathrm{O}$. Thus, the 6 green points are obtained by rotating the 6 red points counterclockwise by $60^{\circ}$ about point $O$. Then, rotate the 6 green points counterclockwise by $60^{\circ}$ about point $O$ to get 6 yellow points. By symmetry, this construction of 18 points satisfies the conditions.