2004 IMO Shortlist S16

Let $A_1A_2A_3\ldots A_n$ be a regular $n$ -gon. Let $B_1$ and $B_{n-1}$ be the midpoints of its sides $A_1A_2$ and $A_{n-1}A_n$ . Also, for every $i\in\left\{2,3,4,\ldots ,n-2\right\}$ . Let $S$ be the point of intersection of the lines $A_1A_{i+1}$ and $A_nA_i$ , and let $B_i$ be the point of intersection of the angle bisector bisector of the angle $\measuredangle A_iSA_{i+1}$ with the segment $A_iA_{i+1}$ .

Prove that $\sum_{i=1}^{n-1} \measuredangle A_1B_iA_n=180^{\circ}$ .

*Proposed by Dusan Dukic, Serbia and Montenegro*

设 $A_1A_2A_3\ldots A_n$ 为正则 $n$ 边形。令 $B_1$ 和 $B_{n-1}$ 为其边 $A_1A_2$ 和 $A_{n-1}A_n$ 的中点。另外，对于每个 $i\in\left\{2,3,4,\ldots ,n-2\right\}$ 。令 $S$ 为线 $A_1A_{i+1}$ 和 $A_nA_i$ 的交点，并令 $B_i$ 为角 $\measuredangle A_iSA_{i+1}$ 的角平分线平分线与线段 $A_iA_{i+1}$ 的交点。

证明 $\sum_{i=1}^{n-1} \measuredangle A_1B_iA_n=180^{\circ}$ 。

*由塞尔维亚和黑山杜桑·杜基奇提议*