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HMMT 二月 2003 · 几何 · 第 1 题

HMMT February 2003 — Geometry — Problem 1

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. AD and BC are both perpendicular to AB , and CD is perpendicular to AC . If AB = 4
    and BC = 3, find CD .
    A BCD
解析

英文解析

  1. AD and BC are both perpendicular to AB , and CD is perpendicular to AC . If AB = 4
    and BC = 3, find CD .
    A BCD
    Solution: 20 / 3
    °
    6 6 6
    By Pythagoras in 4 ABC , AC = 5. But CAD = 90 − BAC = ACB , so righttriangles CAD, BCA are similar, and CD/AC = BA/CB = 4 / 3 ⇒ CD = 20 / 3.