2019 IMO Shortlist C6

Let $n>1$ be an integer. Suppose we are given $2 n$ points in a plane such that no three of them are collinear. The points are to be labelled $A_{1}, A_{2}, \ldots, A_{2 n}$ in some order. We then consider the $2 n$ angles $\angle A_{1} A_{2} A_{3}, \angle A_{2} A_{3} A_{4}, \ldots, \angle A_{2 n-2} A_{2 n-1} A_{2 n}, \angle A_{2 n-1} A_{2 n} A_{1}$, $\angle A_{2 n} A_{1} A_{2}$. We measure each angle in the way that gives the smallest positive value (i.e. between $0^{\circ}$ and $180^{\circ}$ ). Prove that there exists an ordering of the given points such that the resulting $2 n$ angles can be separated into two groups with the sum of one group of angles equal to the sum of the other group.

令 $n>1$ 为整数。假设我们在平面上有 $2 n$ 个点，其中没有三个点共线。这些点将以某种顺序标记为 $A_{1}、A_{2}、\ldots、A_{2 n}$。然后我们考虑$2 n$角$\angle A_{1} A_{2} A_{3}, \angle A_{2} A_{3} A_{4}, \ldots, \angle A_{2 n-2} A_{2 n-1} A_{2 n}, \angle A_{2 n-1} A_{2 n} A_{1}$, $\angle A_{2 n} A_{1} A_{2}$。我们以给出最小正值的方式测量每个角度(即在 $0^{\circ}$ 和 $180^{\circ}$ 之间)。证明给定点存在排序，使得所得的 $2 n$ 角度可以分为两组，其中一组角度的总和等于另一组角度的总和。