44. It is obvious that any flat closed broken line with a perimeter of 1 can be enclosed in a circle of radius $1 / 2$: 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 $A$ of the broken line, the length of one of the two segments of the broken line with ends $O$ and $A$ does not exceed $1 / 2$ and, consequently, the distance $O A$ is also no more than $1 / 2$). What is the radius of the smallest circle that can enclose every flat closed broken line with a perimeter of 1?
44. We will prove that any closed broken line of perimeter 1 can be enclosed within a circle of radius $1 / 4$. Let $A$ be an arbitrary point on our broken line, and $B$ be a point on the broken line such that both parts of the broken line connecting points $A$ and $B$ have the same length $1 / 2$. Let $O$ be the midpoint of segment $A B$ (Fig. 124). Draw a circle of radius $1 / 4$ with center at point $O$. We claim that the entire broken line will be contained within this circle, i.e., that the distance from any point $M$ on the broken line to point $O$ is no more than $1 / 4$. Indeed, let $M$ be an arbitrary point on the broken line, and $M^{\prime}$ be the point symmetric to $M$ with respect to point $O$. Connect points $M$ and $M^{\prime}$ with points $A$ and $B$. Since $A M$ is no greater than the part of the broken line between points $A$ and $M$, and $B M$ is no greater than the part of the broken line between points $B$ and $M$, then $A M + B M$ is no greater than the part of the broken line between points $A$ and $B$, i.e., no greater than $1 / 2$. But from Fig. 124, it is easy to see that $A M = B M^{\prime}, \quad B M = A M^{\prime}, O M = O M^{\prime}$. From the consideration of triangle $A M M^{\prime}$, it is not difficult to derive that

$$
M M^{\prime} = 2 O M \leqslant A M + A M^{\prime} = A M + B M \leqslant \frac{1}{2}, O M \leqslant \frac{1}{4}
$$

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 is a rhombus with side $1 / 4$ and a very small acute angle). Since the diameter of the circle containing the broken line is obviously not less than the diameter of the broken line, no circle with a radius less than $1 / 4$ can cover any planar closed broken line of perimeter 1. Therefore, the required radius of the circle is $1 / 4$.