2009 APMO 第 3 题

Let three circles $\Gamma_{1}, \Gamma_{2}, \Gamma_{3}$, which are non-overlapping and mutually external, be given in the plane. For each point $P$ in the plane, outside the three circles, construct six points $A_{1}, B_{1}, A_{2}, B_{2}, A_{3}, B_{3}$ as follows: For each $i=1,2,3, A_{i}, B_{i}$ are distinct points on the circle $\Gamma_{i}$ such that the lines $P A_{i}$ and $P B_{i}$ are both tangents to $\Gamma_{i}$. Call the point $P$ exceptional if, from the construction, three lines $A_{1} B_{1}, A_{2} B_{2}, A_{3} B_{3}$ are concurrent. Show that every exceptional point of the plane, if exists, lies on the same circle.

设平面内不重叠且互为外部的三个圆$\Gamma_{1}、\Gamma_{2}、\Gamma_{3}$。对于平面上三个圆之外的每个点 $P$，构造六个点 $A_{1}、B_{1}、A_{2}、B_{2}、A_{3}、B_{3}$，如下所示： 对于每个 $i=1,2,3，A_{i}、B_{i}$ 是圆 $\Gamma_{i}$ 上的不同点，使得线 $P A_{i}$ 和 $P B_{i}$ 均与 $\Gamma_{i}$ 相切。如果从构造来看，三行 $A_{1} B_{1}、A_{2} B_{2}、A_{3} B_{3}$ 是并发的，则称点 $P$ 为异常。证明平面上的每个异常点(如果存在)都位于同一个圆上。