13 Let $S$ be the set of points inside and on the boundary of a regular hexagon with side length 1. Find the smallest $r$ such that there exists a way to three-color the points in $S$ so that the distance between any two points of the same color is less than $r$.
Solution 1: $r_{\min }=\frac{3}{2}$.
On one hand, construct the following coloring (as shown in Figure 1): Let the center of the regular hexagon $A B C D E F$ be $O$, and draw perpendiculars from $O$ to $A F$, $B C$, and $D E$, with the feet of the perpendiculars being $X$, $Y$, and $Z$ respectively. Color the three regions marked in the figure with colors I, II, and III respectively. Specifically, points $X$, $Y$, and $Z$ are colored with colors I, II, and III respectively (the points at the boundaries of the regions can be colored with the color of either adjacent region). In this way, the maximum distance between any two points in the same region is $X Y = Y Z = Z X = \frac{3}{2}$, and the maximum value is not achieved by two points in the same region. Therefore, the distance between any two points of the same color is less than $\frac{3}{2}$, i.e., $r = \frac{3}{2}$ is feasible.

Now consider the points on the boundary of the hexagon. For any two points $A^{\prime}$ and $B^{\prime}$ on the boundary, the minimum length required to travel from $A^{\prime}$ to $B^{\prime}$ along the boundary is called the boundary distance between $A^{\prime}$ and $B^{\prime}$ (for example, the boundary distance between $A$ and $E$ is $A F + F E = 2$). It can be proven that for $0 < \varepsilon < 1$, there exist points $M$ and $N$ on the boundary such that the boundary distance between $M$ and $N$ is $2 - \varepsilon$. Since $M$ and $N$ cannot be on the same side, we consider the following three cases:
$1^{\circ}$ If $M$ and $N$ are on adjacent sides (as shown in Figure 2), then
$$
\begin{array}{c}
A M + A N \geqslant 2 - \varepsilon, \\
M N = \sqrt{A M^{2} + A N^{2} + A M \cdot A N} \geqslant \frac{\sqrt{3}}{2}(A M + A N) \\
\geqslant \frac{\sqrt{3}}{2}(2 - \varepsilon) = \sqrt{3} - \frac{\sqrt{3}}{2} \varepsilon > \frac{3}{2} - \frac{\sqrt{3}}{2} \varepsilon.
\end{array}
$$
$2^{\circ}$ If $M$ and $N$ are on opposite sides (as shown in Figure 3), then
$$
A M + A B + B N \geqslant 2 - \varepsilon, \text{ i.e., } A M + B N \geqslant 1 - \varepsilon.
$$

Draw perpendiculars from $M$ and $N$ to $A B$, with the feet of the perpendiculars being $M_{1}$ and $N_{1}$, then
$$
\begin{aligned}
M N & \geqslant M_{1} N_{1} = M_{1} A + A B + B N_{1} = \frac{A M}{2} + A B + \frac{B N}{2} \\
& \geqslant 1 + \frac{1 - \varepsilon}{2} = \frac{3}{2} - \frac{\varepsilon}{2} > \frac{3}{2} - \frac{\sqrt{3}}{2} \varepsilon.
\end{aligned}
$$
$3^{\circ}$ If $M$ and $N$ are on opposite sides, then
$$
M N \geqslant \sqrt{3} > \frac{3}{2} - \frac{\sqrt{3}}{2} \varepsilon.
$$

In summary, from $1^{\circ}$ to $3^{\circ}$, there exist points $M$ and $N$ such that $M N > \frac{3}{2} - \frac{\sqrt{3}}{2} \varepsilon$, i.e., $r > \frac{3}{2} - \frac{\sqrt{3}}{2} \varepsilon \Rightarrow \lim _{\varepsilon \rightarrow 0+} r \geqslant \lim _{\varepsilon \rightarrow 0+}\left(\frac{3}{2} - \frac{\sqrt{3}}{2} \varepsilon\right)$, i.e., $r \geqslant \frac{3}{2}$.
In summary, $r_{\min } = \frac{3}{2}$.
Solution 2: When $r = \frac{3}{2}$, it is feasible as shown in Solution 1. Now we prove that $r \geqslant \frac{3}{2}$.

Assume $r < \frac{3}{2}$. Then there exists $\varepsilon > 0$ such that $r + \varepsilon < \frac{3}{2}$. Consider the regular hexagon $A B C D E F$ with side length 1. For any point $A^{\prime}$ on the boundary of the hexagon, there exists a point $B$ on the boundary such that the boundary distance between $A^{\prime}$ and $B$ is $2 - \varepsilon$. Since $A^{\prime} B > r$, $A^{\prime}$ and $B$ must be colored differently. Similarly, there exists a point $C$ on the boundary such that the boundary distance between $B$ and $C$ is $2 - \varepsilon$. Since $B C > r$, $B$ and $C$ must be colored differently. However, $A^{\prime} C > r$, but $B$ and $C$ are colored differently, so $A^{\prime}$ must be colored the same as $A$.

Therefore, for any two points with a distance less than $\varepsilon$, they must be colored the same, which means the entire circle will be one color, a contradiction.
Therefore, $r$ is not less than $\frac{3}{2}$.