47. What is the radius of the smallest circle in which $n$ points ( $n=2,3,4, \ldots, 10,11$ ) can be placed, one of which coincides with the center of the circle, so that the distance between any two points is at least 1?
47. Let's denote the radius of the smallest "circle in which $n$ points can be placed, one of which coincides with the center of the circle and the distances between each pair are at least 1" by $R_{n}$. Our task is to determine the values of $R_{n}$ for the first few values of $n$.

$1^{\circ}$ It is completely obvious that

$$
R_{2}=R_{3}=R_{4}=R_{5}=R_{6}=R_{7}=1
$$

(a circle of radius 1 can contain 7 points, the distances between each pair of which are at least 1; it is sufficient to take the center of the circle and the six vertices of a regular inscribed hexagon; see Fig. 129, a).

$2^{\circ}$ Suppose inside a circle of radius $R$ there are $n$ points such that one of the points coincides with the center of the circle and the distance between each pair of points is at least 1. In this case, $n-1$ of these points are located within the annulus centered at the $n$-th point $A_{0}$, formed by circles of radii 1 and $R$. The distance between each pair of these $n-1$ points is at least 1, and at least one of the angles under which the segments connecting these points are seen from the center of the annulus does not exceed $\frac{360^{\circ}}{n-1}$. This observation indicates a connection between the present problem and the previous one.

If $n=8, n-1=7$, then our problem reduces to finding the smallest number $R$ such that within the annulus centered at $A_{0}$, formed by circles of radii 1 and $R$, two points $A_{1}$ and $A_{2}$ can be placed, the distance between which is at least 1 and the angle $A_{1} A_{0} A_{2}$ does not exceed $\frac{360^{\circ}}{7}$.
![](https://cdn.mathpix.com/cropped/2024_05_21_962b75a91b7185035ab9g-157.jpg?height=1232&width=838&top_left_y=310&top_left_x=266)

Fig. 129.

By the result of problem 46, this requires that $2 \arcsin \frac{1}{2 R} \geqslant \frac{360^{\circ}}{7}$; from this it follows that $R_{8}=\frac{1}{2 \sin \frac{180^{\circ}}{7}}=1.15 \ldots$ (Fig. 129, b).

In exactly the same way, we find

$$
\begin{gathered}
R_{9}=\frac{1}{2 \sin \frac{180^{\circ}}{8}}=1.30 \ldots, \quad R_{10}=\frac{1}{2 \sin \frac{180^{\circ}}{9}}=1.46 \ldots \\
R_{11}=\frac{1}{2 \sin \frac{180^{\circ}}{10}}=1.61 \ldots
\end{gathered}
$$

Remark. Note that for $n>11$ we will have $R_{n}-\frac{1}{R_{n}}>1$, and accordingly, the most favorable arrangement of $n$ points will no longer resemble Fig. 129, $\sigma$ (see the solution to problem 46). For example, using the second formula from the solution to problem 46, it is not difficult to derive that $\arccos \frac{R_{13}}{2}=\frac{360^{\circ}}{12}$, from which $R_{13}=2 \cos 30^{\circ}=$ $=1.73 \ldots$; the arrangement of 13 points in a circle of radius $\sqrt{3}$, satisfying the condition of the problem, will look like the one depicted in Fig. 129, v.