HMMT 二月 2014 · 几何 · 第 10 题

英文解析

10. Let ABC be a triangle with AB = 13 , BC = 14, and CA = 15. Let Γ be the circumcircle of ABC , let
    ̂
    O be its circumcenter, and let M be the midpoint of minor arc BC . Circle ω is internally tangent to
    Γ at A , and circle ω , centered at M , is externally tangent to ω at a point T . Ray AT meets segment 1
    2 1
    BC at point S , such that BS − CS = 4 / 15. Find the radius of ω .
    Answer: 1235/108 Let N be the midpoint of BC . Notice that BS − CS = means that 42
    N S = . let lines M N and AS meet at P , and let D be the foot of the altitude from A to BC .215
    2 / 1515
    32 SN 3
    Then BD = 5 and AD = 12, so DN = 2 and DS = . Thus N P = AD = 12 = . Now
    15 SD 32 / 15 4
    √
    √ ( )
    (13)(14)(15)2
    abc 65 65 33 27
    2 2 2
    OB = R = = = , so ON = OB − BN = − 7 = . Thus OP =
    4 A 4(84) 8 8 8 8
    and P M = OM − OP = . By Monge’s theorem, the exsimilicenter of ω and Γ (which is A ), the 19
    ′41
    insimilicenter of ω and ω (which is T ), and the insimilicenter of ω and Γ (call this P ) are collinear.
    1 2 2
    ′
    But notice that this means P = OM ∩ AT = P . From this we getradius of ω M P 38 = = .2
    R OP 27
    65 38 1235
    Thus the radius of ω is · = .
    8 27 1082