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HMMT 二月 2007 · 冲刺赛 · 第 36 题

HMMT February 2007 — Guts Round — Problem 36

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. [ 25 ] The Marathon. Let ω denote the incircle of triangle ABC . The segments BC, CA , and AB aretangent to ω at D , E , and F , respectively. Point P lies on EF such that segment P D is perpendicularto BC . The line AP intersects BC at Q . The circles ω and ω pass through B and C , respectively,
    1 2
    and are tangent to AQ at Q ; the former meets AB again at X , and the latter meets AC again at Y .
    The line XY intersects BC at Z . Given that AB = 15 , BC = 14, and CA = 13, find b XZ · Y Z c . 5
解析

英文解析

  1. [ 25 ] The Marathon. Let ω denote the incircle of triangle ABC . The segments BC, CA , and AB aretangent to ω at D , E , and F , respectively. Point P lies on EF such that segment P D is perpendicularto BC . The line AP intersects BC at Q . The circles ω and ω pass through B and C , respectively,
    1 2
    and are tangent to AQ at Q ; the former meets AB again at X , and the latter meets AC again at Y .
    The line XY intersects BC at Z . Given that AB = 15 , BC = 14, and CA = 13, find b XZ · Y Z c .
    ′ ′ ′
    Answer: 101 . Construct D diametrically opposed to D , so that ∠ DF D and ∠ DED are right, and
    ′ ′ ′
    note that P lies on DD . By standard angle chasing, m ∠ F DD = β (half angle B ) and m ∠ D DE = γ .
    ′ ° ′ ° ′ ′
    Thus, m ∠ DD F = 90 − β and m ∠ ED D = 90 − γ . Then by the law of sines, DE : ED : D F :
    ′ ′
    F D = cos( γ ) : sin( γ ) : sin( β ) : sin( γ ). Now using 4 DEP ∼ 4 F D P and 4 DF P ∼ 4 ED P , we have
    ′
    EP ED · ED sin( γ ) cos( γ ) c = = = .
    ′
    P F F D · F D sin( β ) sin( β ) b
    ′ ′
    Let the dilation centered at A sending E to C map P and F to P and F , respectively. Note that
    ′ ′ ′ ′
    AF = AC as AE and AF are equal tangents, and CP : P F = EP : P F = c : b by similarity. Thenby Menelaus’ theorem,
    ′ ′
    BQ CP F A BQ c b
    1 = = ,
    ′ ′
    QC P F AB QC b c
    ( )
    2 1 2 2 2
    so that BQ = QC and AQ is actually a median. So, AQ = 2 b + 2 c − a = 148 . Now by Powerof a Point, AB · AX = AQ = AC · AY, so AX = 148 / 15 and AY = 148 / 13. Moreover, BXCY is 24
    cyclic as 4 ABC ∼ 4 AY X. Thus, XZ · Y Z = BZ · CZ , and it suffices to compute BZ/CZ . Menelausonce more gives
    BZ CY AX
    1 = ,
    ZC Y A XBwhence, BZ/CZ = ( AY /AX )( BX/CY ) = (15 / 13)((77 · 13) / (21 · 15)) = 11 / 3. We write CZ = 3 d and 13
    BZ = 11 d . Because AX < AB and AY < AC , Z does not lie on segment BC . Given the configuration
    1617
    information, BC = 8 d = 14, so d = 7 / 4, and finally b BZ · CZ c = b 33 d c = b c = 101 .2
    1416