97. a) It is clear that any plane broken line of length 1 can be enclosed in a circle of radius 1: for this, it is sufficient for the center $O$ of the circle to coincide with one of the ends of the broken line. What is the radius of the smallest circle in which any broken line of length $l$ can be enclosed?

b) It is clear that any plane closed broken line of perimeter 1 can be enclosed in a circle of radius $\leqslant \frac{1}{2}$: for this, it is sufficient for the center $O$ of the circle to coincide with any point of the broken line. (In this case, for any point $M$ of the broken line, the length of one of the two segments $O M$ of the broken line will be $\leqslant \frac{1}{2}$, and therefore the distance $O M$ will be $\leqslant \frac{1}{2}$.) What is the radius of the smallest circle inside which any closed broken line of length 1 can be enclosed?
97. a) Let $S$ be the midpoint of the broken line $AB$, i.e., such a point that the broken lines $AS$ and $SB$ have the same length $1/2$. In this case, each point $M$ of the broken line is such that the length of the broken line $SM$ does not exceed $1/2$, and even more so, the distance $SM \leqslant 1/2$, from which it follows that a circle of radius $1/2$ with center $S$ completely covers the broken line. Thus, every broken line of length 1 can be covered by a circle of radius $1/2$, and the number $1/2$ cannot be reduced, for it is obvious that a segment of length 1 cannot be covered by a circle of radius $<1/2$.

![](https://cdn.mathpix.com/cropped/2024_05_21_d4ffc236376258137647g-314.jpg?height=566&width=578&top_left_y=883&top_left_x=310)

Fig. 176.

b) We will prove that any closed plane broken line of perimeter 1 can be enclosed within a circle of radius 1/4. Let $A$ be an arbitrary point of our broken line, and $B$ be such a point on the broken line that both parts of the broken line connecting points $A$ and $B$ have the same length $1/2$, and let $Q$ be the midpoint of the segment $AB$ (Fig. 176). Draw a circle of radius $1/4$ with center at point $Q$. We claim that the entire broken line will be enclosed within this circle, i.e., that the distance from any point $M$ of the broken line to point $Q$ is no more than 1/4. Indeed, let $M'$ be the point symmetric to $M$ with respect to point $Q$. Connect $M$ and $M'$ with points $A$ and $B$. Since $AM$ is no greater than the part of the broken line enclosed between points $A$ and $M$, and $BM$ is no greater than the part of the broken line enclosed between points $B$ and $M$, then $AM + BM$ is no greater than the part of the broken line enclosed between points $A$ and $B$, i.e., no greater than $1/2$. But from Fig. 176, it is not difficult to see that $AM = BM', \quad BM = AM', \quad QM = QM'$. From the consideration of $\triangle AMM'$ it follows that

$$
MM' = 2QM \leqslant AM + AM' = AM + BM \leqslant 1/2
$$

from which

$$
QM \leqslant 1/4
$$

which is what we needed to prove.

On the other hand, the diameter of a broken line of perimeter 1 can be arbitrarily close to $1/2$ (this will be the case, for example, if the broken line represents a rhombus with side $1/4$ and a very small acute angle). Since the diameter of a circle containing the broken line is obviously not less than the diameter of the broken line, no circle of radius less than $1/4$ can cover any closed plane broken line of perimeter 1. Therefore, the required value of the radius of the circle is $1/4$.