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PUMaC 2010 · 几何（A 组） · 第 3 题

PUMaC 2010 — Geometry (Division A) — Problem 3

专题
Discrete Math / 离散数学
难度
L3
来源
PUMaC

题目详情

  1. Triangle ABC has AB = 4, AC = 5, and BC = 6. An angle bisector is drawn from angle A , AM and meets BC at M . What is the nearest integer to 100 ? CM
解析
  1. Triangle ABC has AB = 4, AC = 5, and BC = 6. An angle bisector is drawn from angle A , AM and meets BC at M . What is the nearest integer to 100 ? CM [Answer] 100 [Solution] By Angle-Bisector Theorem, BM : CM = AB : AC = 4 : 5, hence BM = 8 / 3, √ CM = 10 / 3. By Angle-Bisector Length Formula, AM = AB · AC − BM · CM = 10 / 3. Hence AM/CM = 1.