On a circle of radius 1, a point $O$ is marked, and from it, a notch is made to the right with a radius of $l$. From the resulting point $O_{1}$, another notch is made in the same direction with the same radius, and this is repeated 1968 times. After this, the circle is cut at all 1968 notches, resulting in 1968 arcs. How many different lengths of arcs can be obtained in this way?
We will prove by induction on $n$ that the number of different arcs after $n$ cuts does not exceed 3. For $n=2$, this is obvious. Let $A_{k}$ denote the cut with number $k$. Suppose $n$ cuts have been made and the point $A_{n}$ falls on the ARC $A_{k} A_{l}$. Then the point $A_{n-1}$ falls on the arc $A_{k-1} A_{l-1}$. Therefore, if $k, l \neq 1$, no new lengths will appear, and the required statement is proved. Now suppose, for example, that $l=1$. We will prove that then the length of any arc $A_{p} A_{q}$ between neighboring cuts is equal to the length of one of the arcs $A_{k} A_{1}, A_{n} A_{1}, A_{1} A_{s}$, where $A_{s}$ is the nearest cut to $A_{1}$, different from $A_{n}$. Indeed, if $p, q \neq 1$, the length of the arc $A_{p} A_{q}$ is equal to the length of the arc $A_{p-1} A_{q-1}$, and there are no other cuts between the cuts $A_{p-1}$ and $A_{q-1}$ except possibly $A_{n}$. Thus, in a finite number of steps, we arrive at one of the three considered arcs.

## Answer

no more than three.