2013 IMO 第 6 题

Let $n\geq 3$ be an integer, and consider a circle with $n + 1$ equally spaced points marked on it. Consider all labellings of these points with the numbers $0,1,\ldots ,n$ such that each label is used exactly once; two such labellings are considered to be the same if one can be obtained from the other by a rotation of the circle. A labelling is called beautiful if, for any four labels $a< b< c< d$ with $a + d = b + c$ , the chord joining the points labelled $a$ and $d$ does not intersect the chord joining the points labelled $b$ and $c$ . Let $M$ be the number of beautiful labellings, and let $N$ be the number of ordered pairs $(x,y)$ of positive integers such that $x + y\leq n$ and $\gcd (x,y) = 1$ . Prove that $M = N + 1$ .

令 $n\geq 3$ 为整数，并考虑一个圆，其上标记有 $n + 1$ 等距点。考虑这些点的所有标签都带有数字 $0,1,\ldots ,n$ ，以便每个标签仅使用一次；如果可以通过旋转圆圈从另一个获得一个这样的标签，则认为两个这样的标签是相同的。如果对于任何四个标签 $a< b< c< d$ 和 $a + d = b + c$ ，连接标记为 $a$ 和 $d$ 的点的弦不与连接标记为 $b$ 和 $c$ 的点的弦相交，则该标记被称为漂亮。令 $M$ 为漂亮标签的数量，令 $N$ 为正整数有序对 $(x,y)$ 的数量，使得 $x + y\leq n$ 和 $\gcd (x,y) = 1$ 。证明 $M = N + 1$ 。