1989 IMO 第 1 题

(AUS 2) ${ }^{\mathrm{IMO} 2}$ Let $A B C$ be a triangle. The bisector of angle $A$ meets the circumcircle of triangle $A B C$ in $A_{1}$. Points $B_{1}$ and $C_{1}$ are defined similarly. Let $A A_{1}$ meet the lines that bisect the two external angles at $B$ and $C$ in point $A^{0}$. Define $B^{0}$ and $C^{0}$ similarly. If $S_{X_{1} X_{2} \ldots X_{n}}$ denotes the area of the polygon $X_{1} X_{2} \ldots X_{n}$, prove that $$S_{A^{0} B^{0} C^{0}}=2 S_{A C_{1} B A_{1} C B_{1}} \geq 4 S_{A B C} .$$

(AUS 2) ${ }^{\mathrm{IMO} 2}$ 设 $A B C$ 为三角形。角 $A$ 的平分线与 $A_{1}$ 中三角形 $A B C$ 的外接圆相交。点 $B_{1}$ 和 $C_{1}$ 的定义类似。令$A A_{1}$ 与平分两个外角的线在$B$ 和$C$ 处相交于点$A^{0}$。类似地定义 $B^{0}$ 和 $C^{0}$。若 $S_{X_{1} X_{2} \ldots X_{n}}$ 表示多边形 $X_{1} X_{2} \ldots X_{n}$ 的面积，则证明 $$S_{A^{0} B^{0} C^{0}}=2 S_{A C_{1} B A_{1} C B_{1}} \geq 4 S_{A B C} 。$$