HMMT 十一月 2010 · 团队赛 · 第 6 题

英文解析

6. [ 6 ] AB is a diameter of circle O . X is a point on AB such that AX = 3 BX . Distinct circles ω andω are tangent to O at T and T and to AB at X . The lines T X and T X intersect O again at S1
   2 1 2 1 2 1
   T T
   1 2
   and S . What is the ratio ?
   S S2
   1 2
   Answer: Since the problem only deals with ratios, we can assume that the radius of O is 1. As 3
   we have proven in Problem 5, points S and S are midpoints of arc AB . Since AB is a diameter,5
   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
   °