HMMT 二月 2009 · 冲刺赛 · 第 27 题

HMMT February 2009 — Guts Round — Problem 27

[ 12 ] Circle Ω has radius 5. Points A and B lie on Ω such that chord AB has length 6. A unit circle ω is tangent to chord AB at point T. Given that ω is also internally tangent to Ω , find AT · BT. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th 12 HARVARD-MIT MATHEMATICS TOURNAMENT, 21 FEBRUARY 2009 — GUTS ROUND

解析

[ 12 ] Circle Ω has radius 5. Points A and B lie on Ω such that chord AB has length 6. A unit circle ω is tangent to chord AB at point T. Given that ω is also internally tangent to Ω , find AT · BT. Answer: 2 Solution: Let M be the midpoint of chord AB and let O be the center of Ω . Since AM = BM = 3 , Pythagoras on triangle AM O gives OM = 4 . Now let ω be centered at P and say that ω and Ω are tangent at Q. Because the diameter of ω exceeds 1, points P and Q lie on the same side of AB. By tangency, O, P , and Q are collinear, so that OP = OQ − P Q = 4 . Let H be the orthogonal projection 2 of P onto OM ; then OH = OM − HM = OM − P T = 3 . Pythagoras on OHP gives HP = 7 . Finally, 2 2 2 2 AT · BT = AM − M T = AM − HP = 9 − 7 = 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th 12 HARVARD-MIT MATHEMATICS TOURNAMENT, 21 FEBRUARY 2009 — GUTS ROUND