HMMT 二月 2007 · 几何 · 第 6 题

HMMT February 2007 — Geometry — Problem 6

[ 5 ] Triangle ABC has ∠ A = 90 , side BC = 25 , AB > AC , and area 150. Circle ω is inscribed in ABC , with M its point of tangency on AC . Line BM meets ω a second time at point L . Find the length of segment BL .

解析

[ 5 ] Triangle ABC has ∠ A = 90 , side BC = 25 , AB > AC , and area 150. Circle ω is inscribed in ABC , with M its point of tangency on AC . Line BM meets ω a second time at point L . Find the length of segment BL . √ Answer: 45 17 / 17 . Let D be the foot of the altitude from A to side BC . The length of AD is 2 2 · 150 / 25 = 12. Triangles ADC and BDA are similar, so CD · DB = AD = 144 ⇒ BD = 16 and CD = 9 ⇒ AB = 20 and AC = 15. Using equal tangents or the formula inradius as area divided by semiperimeter, we can find the radius of ω to be 5. Now, let N be the tangency point of ω on AB . 2 By power of a point, we have BL · BM = BN . Since the center of ω together with M, A, and N √ √ determines a square, BN = 15 and BM = 5 17, and we have BL = 45 17 / 17.