The 48th question: Let $\mathrm{S}$ be the set of points inside and on the boundary of a regular hexagon with side length 1. Find the smallest $\mathrm{r}$ such that there exists a way to color the points in $S$ with three colors, so that the distance between any two points of the same color is less than $r$.

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Question 48:
Solution: On the one hand, construct the following coloring: Let the center of the regular hexagon $\mathrm{ABCDEF}$ be $\mathrm{O}$, and draw perpendiculars from $\mathrm{O}$ to $\mathrm{AF}$, $\mathrm{BC}$, and $\mathrm{DE}$, with the feet of the perpendiculars being $\mathrm{X}$, $\mathrm{Y}$, and $\mathrm{Z}$, respectively. Color the three regions marked in the figure with the 1st, 2nd, and 3rd colors, respectively. Specifically, points $\mathrm{X}$, $\mathrm{Y}$, and $\mathrm{Z}$ are colored with the 1st, 2nd, and 3rd colors, respectively (the points at the intersections of other regions are colored with the color of either side). In this way, the maximum distance between any two points in any region is $\mathrm{XY}=\mathrm{YZ}=\mathrm{ZX}=\frac{3}{2}$, and the two points that achieve this maximum distance do not belong to the same region. Therefore, the distance between any two points of the same color is less than $\frac{3}{2}$, so $\frac{3}{2}$ is feasible.
Next, we prove that for any $\mathrm{r} < \frac{3}{2}$, there exists an $\varepsilon > 0$ such that for any $r < r + \varepsilon < \frac{3}{2}$, there are two points $\mathrm{A}$ and $\mathrm{B}$ with $\mathrm{AB} = r + \varepsilon$ that must be colored differently. Assume the contrary, i.e., there exists an $\mathrm{r} < \frac{3}{2}$ such that for any $\varepsilon > 0$, there are two points $\mathrm{A}$ and $\mathrm{B}$ with $\mathrm{AB} = r + \varepsilon$ that can be colored the same. Consider a point $\mathrm{C}$ such that $\mathrm{AC} = r$ and $\mathrm{BC} > r$. Since $\mathrm{B}$ and $\mathrm{C}$ are colored differently, $\mathrm{A}$ must be colored the same as $\mathrm{A'}$. Therefore, for any two points with a distance less than $\varepsilon$, they must be colored the same, which implies that the entire circle will be one color, leading to a contradiction.
Thus, the value of $\mathrm{r}$ that satisfies the condition is not less than $\frac{3}{2}$.