HMMT 二月 2008 · 冲刺赛 · 第 30 题

HMMT February 2008 — Guts Round — Problem 30

[ 15 ] Triangle ABC obeys AB = 2 AC and ∠ BAC = 120 . Points P and Q lie on segment BC such that 2 2 AB + BC · CP = BC 2 2 3 AC + 2 BC · CQ = BC Find ∠ P AQ in degrees. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th 11 HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUND 2 2

解析

[ 15 ] Triangle ABC obeys AB = 2 AC and ∠ BAC = 120 . Points P and Q lie on segment BC such that 2 2 AB + BC · CP = BC 2 2 3 AC + 2 BC · CQ = BC Find ∠ P AQ in degrees. ◦ 2 Answer: 40 We have AB = BC ( BC − CP ) = BC · BP, so triangle ABC is similar to triangle 2 2 2 2 2 2 2 P BA. Also, AB = BC ( BC − 2 CQ )+ AC = ( BC − CQ ) − CQ + AC , which rewrites as AB + CQ = 2 2 ◦ ◦ BQ + AC . We deduce that Q is the foot of the altitude from A. Thus, ∠ P AQ = 90 − ∠ QP A = 90 − ◦ ◦ ◦ ∠ ABP − ∠ BAP . Using the similar triangles, ∠ P AQ = 90 − ∠ ABC − ∠ BCA = ∠ BAC − 90 = 40 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th 11 HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUND 2 2