59. (USS 6) On the circle with center $O$ and radius 1 the point $A_{0}$ is fixed and points $A_{1}, A_{2}, \ldots, A_{999}, A_{1000}$ are distributed in such a way that $\angle A_{0} O A_{k}=k$ (in radians). Cut the circle at points $A_{0}, A_{1}, \ldots, A_{1000}$ How many arcs with different lengths are obtained?
59. By the arc $A B$ we shall always mean the positive arc $A B$. We denote by $|A B|$ the length of are $A B$. Let a basic arc be one of the $n+1$ arcs into which the circle is partitioned by the points $A_{0}, A_{1}, \ldots, A_{n}$, where $n \in \mathbb{N}$. Suppose that $A_{p} A_{0}$ and $A_{0} A_{q}$ are the basic arcs with an endpoint at $A_{0}$, and that $x_{n}, y_{n}$ are their lengths, respectively. We show by induction on $n$ that for each $n$ the length of a basic arc is equal to $x_{n}, y_{n}$ or $x_{n}+y_{n}$. The statement is trivial for $n=1$. Assume that it holds for $n$, and let $A_{i} A_{n+1}, A_{n+1} A_{j}$ be basic arcs. We shall prove that these two arcs have lengths $x_{n}, y_{n}$, or $x_{n}+y_{n}$. If $i, j$ are both strictly positive, then $\left|A_{i}, A_{n+1}\right|=$ $\left|A_{i-1} A_{n}\right|$ and $\left|A_{n+1} A_{j}\right|=\left|A_{n} A_{j-1}\right|$ are equal to $x_{n}, y_{n}$ or $x_{n}+y_{n}$ by the inductive hypothesis.
Let us assume now that $i=0$, i.e, that $A_{p} A_{n+1}$ and $A_{n+1} A_{0}$ are basic arcs. Then $\left|A_{p} A_{n+1}\right|=\left|A_{0} A_{n+1-p}\right| \geq\left|A_{0} A_{q}\right|=y_{n}$ and similarly $\left|A_{n+1} A_{q}\right| \geq x_{n}$, but $\left|A_{p} A_{q}\right|=x_{n}+y_{n}$, from which it follows that $\left|A_{p} A_{n+1}\right|=\left|A_{0} A_{q}\right|=y_{n}$ and consequently $n+1=p+q$. Also, $x_{n+1}=\left|A_{n+1} A_{0}\right|=y_{n}-x_{n}$ and $y_{n+1}=y_{n}$. Now, all basic ares have lengths $y_{n}-x_{n}, x_{n}, y_{n}, x_{n}+y_{n}$. A presence of a basic arc of length $x_{n}+y_{n}$ would spoil our inductive step. However, if any basic arc $A_{k} A_{k}$ has length $x_{n}+y_{n}$, then we must have $l-q=k-p$ because $2 \pi$ is irrational, and therefore the arc $A_{k} A_{l}$ contains either the point $A_{k-p}$ (if $k \geq p$ ) or the point $A_{k+q}$ (if $k1, \\
\left(p_{k} p_{k}+q_{k}\right), \text { if }\left\{p_{k} /(2 \pi)\right\}+\left\{q_{k} /(2 \pi)\right\}<1 .
\end{array}\right.
$$

It is now "easy" to calculate that $p_{19}=p_{20}=333, q_{19}=377, q_{20}=710$, and thus $n_{19}=709<1000<1042=n_{20}$. It follows that the lengths of the basic arcs for $n=1000$ take exactly three different values.