Question 4 Let $S$ be the set of all points inside and on the boundary of a regular hexagon with side length 1. Find the smallest $r>0$, such that there exists a 3-coloring of $S$ where the distance between any two points of the same color is less than $r$.
(2003, China National Training Team Test)
Solve: Let the vertices of the regular hexagon be labeled in counterclockwise order as $A, B, C, D, E, F$, and let $P, Q, R$ be the midpoints of segments $A F$, $B C$, and $D E$ respectively, with $O$ being the center.

The polygons $A B Q O P$, $O Q C D R$, and $O R E F P$ are colored red, yellow, and blue respectively, including their boundaries. The segment $O P$ is colored red, the segment $O Q$ (excluding point $O$) is colored yellow, and the segment $O R$ (excluding point $O$) is colored blue.

Clearly, the distance between any two points of the same color is less than $\frac{3}{2}$. Therefore, the smallest $r \leqslant \frac{3}{2}$.
Next, we prove that the smallest value of $r$ is $\frac{3}{2}$.
If not, then the smallest value $r_{0} < \frac{3}{2}$.
Take a sufficiently large natural number $n$ such that $\frac{1}{n} < \frac{3}{2} - r_{0}$.
Consider the set of points $\left\{M_{k}, M_{2 n+k}, M_{4 n+k}\right\}$. These three points are pairwise of different colors.
Now consider the set of points $\left\{M_{k+1}, M_{2 n+k}, M_{4 n+k}\right\}$. At this point,
$$
\begin{array}{l}
d\left(M_{k+1}, M_{2 n+k}\right) \\
\geqslant d\left(M_{k}, M_{2 n+k}\right) - d\left(M_{k+1}, M_{k}\right) \\
\geqslant \frac{3}{2} - \varepsilon = \frac{3}{2} - \frac{1}{2}\left(\frac{3}{2} - r_{0}\right) \\
= \frac{1}{2}\left(\frac{3}{2} + r_{0}\right) > r_{0}.
\end{array}
$$

Similarly, $d\left(M_{k+1}, M_{4 n+k}\right) > r_{0}$.
Thus, these three points are also of different colors.
Therefore, the points $M_{k}$ and $M_{k+1}$ are of the same color.
By the arbitrariness of $k$, it follows that these $6 n$ points are of the same color, which is a contradiction.
In conclusion, the smallest $r$ is $\frac{3}{2}$.