42*. It is obvious that any system of points with a diameter of 1 can be enclosed in a circle of radius 1: for this, it is sufficient for the center of the circle to coincide with any point of the figure. What is the radius of the smallest circle that can enclose any system of points with a diameter of 1?
42. If our system of points coincides with the vertices of an equilateral triangle with side 1, then it cannot be enclosed in a circle of radius less than $\frac{\sqrt{3}}{3}$ ( $\frac{\sqrt{3}}{3}$ is the radius of the circle circumscribed around an equilateral triangle with side 1; see the solution to problem 18). Now let us consider any system of points with diameter 1. It is not difficult to show,
![](https://cdn.mathpix.com/cropped/2024_05_21_962b75a91b7185035ab9g-149.jpg?height=738&width=372&top_left_y=530&top_left_x=146)

Fig. 122. that any three points $A, B$, and $C$ of the system can be enclosed in a circle of radius $\frac{\sqrt{3}}{3}$. Indeed, if triangle $A B C$ is obtuse, then it fits entirely within the circle described on the largest side as the diameter (Fig. $122, a$); but since the diameter of the system of points does not exceed 1, the radius of this circle is no more than $1 / 2$, i.e., even less than $\frac{\sqrt{3}}{3}$. If, however, triangle $A B C$ is acute (Fig. $122, \sigma$), then at least one of its angles is not less than $60^{\circ}$ (since the sum of all angles is $180^{\circ}$); let this be, for example, angle $B$. Then, by the sine theorem, we have

$$
R=\frac{A C}{2 \sin B} \leqslant \frac{1}{2 \sin 60^{\circ}}=\frac{\sqrt{3}}{3},
$$

where $R$ is the radius of the circle circumscribed around triangle $A B C$. From the fact that any three of the points under consideration can be enclosed in a circle of radius $\frac{\sqrt{3}}{3}$, it follows that all the points can be enclosed in a circle of radius $\frac{\sqrt{3}}{3}$ (see problem 18). Thus, the smallest circle in which any system of points with diameter 1 can be enclosed has a radius of $\frac{\sqrt{3}}{3}$.