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HMMT 二月 2011 · 冲刺赛 · 第 28 题

HMMT February 2011 — Guts Round — Problem 28

专题
Discrete Math / 离散数学
难度
L3
来源
HMMT

题目详情

  1. [ 16 ] Let ABC be a triangle, and let points P and Q lie on BC such that P is closer to B than Q is. Suppose that the radii of the incircles of triangles ABP , AP Q , and AQC are all equal to 1, and that the radii of the corresponding excircles opposite A are 3, 6, and 5, respectively. If the radius of the 3 incircle of triangle ABC is , find the radius of the excircle of triangle ABC opposite A . 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th 14 HARVARD-MIT MATHEMATICS TOURNAMENT, 12 FEBRUARY 2011 — GUTS ROUND
解析
  1. [ 16 ] Let ABC be a triangle, and let points P and Q lie on BC such that P is closer to B than Q is. Suppose that the radii of the incircles of triangles ABP , AP Q , and AQC are all equal to 1, and that the radii of the corresponding excircles opposite A are 3, 6, and 5, respectively. If the radius of the 3 incircle of triangle ABC is , find the radius of the excircle of triangle ABC opposite A . 2 Answer: 135 Let t denote the radius of the excircle of triangle 4 ABC . Lemma: Let 4 ABC be a triangle, and let r and r be the inradius and exradius opposite A . Then A r B C = tan tan . r 2 2 A Guts Round Proof. Let I and J denote the incenter and the excenter with respect to A . Let D and E be the foot of the perpendicular from I and J to BC , respectively. Then B r = ID = BI sin 2 ◦ 180 − B B r = JE = BJ sin = BJ cos A 2 2 C BI = BJ tan ∠ AJB = BY tan . 2 The last equation followed from ◦ 180 − B A C ◦ ∠ AJB = 180 − ∠ ABJ − ∠ JAB = − = . 2 2 2 Hence B sin r BI B C 2 = · = tan · tan . B r BJ 2 2 cos A 2 ∠ AP Q ∠ AQP ∠ AQC ∠ AP B Noting tan tan = tan tan = 1 and applying the lemma to 4 ABC , 4 ABP , 2 2 2 2 4 AP Q , and 4 AQC give 3 / 2 ∠ ABC ∠ ACB = tan · tan t 2 2 ( ) ( ) ( ) ∠ ABC ∠ AP B ∠ AP Q ∠ AQP ∠ AQC ∠ ACB = tan · tan · tan · tan · tan · tan 2 2 2 2 2 2 1 1 1 = · · 3 6 5 Therefore, t = 135.