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HMMT 十一月 2012 · 团队赛 · 第 10 题

HMMT November 2012 — Team Round — Problem 10

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. [ 6 ] Triangle ABC has AB = 4, BC = 5, and CA = 6. Points A , B , C are such that B C is tangent
    ′ ′ ′ ′
    to the circumcircle of △ ABC at A , C A is tangent to the circumcircle at B , and A B is tangent to
    ′ ′
    the circumcircle at C . Find the length B C .
解析

英文解析

  1. [ 6 ] Triangle ABC has AB = 4, BC = 5, and CA = 6. Points A , B , C are such that B C is tangent
    ′ ′ ′ ′
    to the circumcircle of △ ABC at A , C A is tangent to the circumcircle at B , and A B is tangent to
    ′ ′
    the circumcircle at C . Find the length B C .
    ′ ′ ′ ′ ′ ′80
    Answer: Note that by equal tangents, B A = B C , C A = C B , and A B = A C . Moreover,
    ′ ′ ′ ′ ′ ′3
    since the line segments A B , B C , and C A are tangent to the circumcircle of ABC at C , A , and
    ′ ′ ′ ′ ′
    B respectively, we have that ∠ A BC = ∠ A CB = ∠ A , ∠ B AC = ∠ B CA = ∠ B , and ∠ C BA =
    ′ ′ ′
    ∠ C AB = ∠ C . By drawing the altitudes of the isosceles triangles BC A and AC B , we therefore have
    ′ ′
    that C A = 2 / cos C and B A = 3 / cos B .
    Now, by the Law of Cosines, we have that
    2 2 2
    a + c − b 25 + 16 − 36 1
    cos B = = =
    2 ac 2(5)(4) 8
    2 2 2
    a + b − c 25 + 36 − 16 3
    cos C = = = .
    2 ab 2(5)(6) 4
    Therefore,
    ( )
    4 80
    ′ ′ ′ ′
    B C = C A + B A = 2 + 3(8) = .
    3 3
    Team Round