HMMT 十一月 2011 · 冲刺赛 · 第 29 题

英文解析

29. [ 14 ] Let ABC be a triangle with AB = 4, BC = 8, and CA = 5. Let M be the midpoint of BC , andlet D be the point on the circumcircle of ABC so that segment AD intersects the interior of ABC ,
    and ∠ BAD = ∠ CAM . Let AD intersect side BC at X . Compute the ratio AX/AD .
    Answer: Let E be the intersection of AM with the circumcircle of ABC . We note that, by 9
    equal angles ADC ∼ ABM , so that 41
    AB 20
    AD = AC ( ) = .
    AM AM
    Using the law of cosines on ABC , we get that
    2 2 2
    4 + 8 − 5 55
    cos B = = .
    2(4)(8) 64
    Then, using the law of cosines on ABM, we get that
    √
    √
    3 20 2
    2 2
    AM = 4 + 4 − 2(4)(4) cos B = √ ⇒ AD = .
    Applying Power of a Point on M ,23
    √ √
    16 2 41 2
    ( AM )( M E ) = ( BM )( M C ) ⇒ M E = ⇒ AE = .
    3 6
    Then, we note that AXB ∼ ACE , so that
    √
    AC 60 2 AX 9
    AX = AB ( ) = ⇒ =
    AE 41 AD 41
    Guts Round
    B X M CA
    D E