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HMMT 二月 2003 · 冲刺赛 · 第 39 题

HMMT February 2003 — Guts Round — Problem 39

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

  1. [15] In the figure, if AE = 3, CE = 1, BD = CD = 2, and AB = 5, find AG .
    FGEA
    5BDC
    HARVARD-MIT MATHEMATICS TOURNAMENT, MARCH 15, 2003 — GUTS ROUND

英文原题

[15] In the figure, if AE = 3, CE = 1, BD = CD = 2, and AB = 5, find AG . F G
A
B D
C
E
5
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
HARVARD-MIT MATHEMATICS TOURNAMENT, MARCH 15, 2003 — GUTS ROUND

解析

英文解析

  1. In the figure, if AE = 3, CE = 1, BD = CD = 2, and AB = 5, find AG .
    FGEA
    √BDC
    Solution: 3 66 / 7
    2 2 2
    By Stewart’s Theorem, AD · BC + CD · BD · BC = AB · CD + AC · BD , so
    2 2 2
    AD = (5 · 2 + 4 · 2 − 2 · 2 · 4) / 4 = (50 + 32 − 16) / 4 = 33 / 2. By Menelaus’s
    Theorem applied to line BGE and triangle ACD , DG/GA · AE/EC · CB/BD = 1,
    √
    so DG/GA = 1 / 6 ⇒ AD/AG = 7 / 6. Thus AG = 6 · AD/ 7 = 3 66 / 7.