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HMMT 二月 2008 · 冲刺赛 · 第 30 题

HMMT February 2008 — Guts Round — Problem 30

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

  1. [ 15 ] Triangle ABC obeys AB = 2 AC and ∠ BAC = 120 . Points P and Q lie on segment BC suchthat
    2 2
    AB + BC · CP = BC
    2 2
    3 AC + 2 BC · CQ = BC
    Find ∠ P AQ in degrees.
    11 HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUNDth
    2 2

英文原题

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

解析

英文解析

  1. [ 15 ] Triangle ABC obeys AB = 2 AC and ∠ BAC = 120 . Points P and Q lie on segment BC suchthat
    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 .
    11 HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUNDth
    2 2