HMMT 十一月 2010 · 团队赛 · 第 6 题

HMMT November 2010 — Team Round — Problem 6

[ 6 ] AB is a diameter of circle O . X is a point on AB such that AX = 3 BX . Distinct circles ω and 1 ω are tangent to O at T and T and to AB at X . The lines T X and T X intersect O again at S 2 1 2 1 2 1 T T 1 2 and S . What is the ratio ? 2 S S 1 2 ◦

解析

[ 6 ] AB is a diameter of circle O . X is a point on AB such that AX = 3 BX . Distinct circles ω and 1 ω are tangent to O at T and T and to AB at X . The lines T X and T X intersect O again at S 2 1 2 1 2 1 T T 1 2 and S . What is the ratio ? 2 S S 1 2 3 Answer: Since the problem only deals with ratios, we can assume that the radius of O is 1. As 5 we have proven in Problem 5, points S and S are midpoints of arc AB . Since AB is a diameter, 1 2 S S is also a diameter, and thus S S = 2. 1 2 1 2 Let O , O , and P denote the center of circles ω , ω , and O . Since ω is tangent to O , we have 1 2 1 2 1 2 2 2 P O + O X = 1. But O X ⊥ AB . So 4 P O X is a right triangle, and O X + XP = O P . Thus, 1 1 1 1 1 1 3 5 2 2 O X + 1 / 4 = (1 − O X ) , which means O X = and O P = . 1 1 1 1 8 8 ( ) 6 / 5 P T P T T T 3 1 6 3 1 1 1 2 Since T T ‖ O O , we have T T = O O · = 2 O X · = 2 = . Thus = = . 1 2 1 2 1 2 1 2 1 P O P O 8 5 / 8 5 S S 2 5 1 1 1 2 Team Round ◦