HMMT 二月 2007 · 冲刺赛 · 第 26 题

英文解析

26. [ 12 ] ABCD is a cyclic quadrilateral in which AB = 4 , BC = 3 , CD = 2, and AD = 5. Diagonals ACand BD intersect at X . A circle ω passes through A and is tangent to BD at X . ω intersects AB and
    AD at Y and Z respectively. Compute Y Z/BD .
    Answer: . Denote the lengths AB, BC, CD , and DA by a, b, c , and d respectively. Because 115
    AX BX a 143
    ABCD is cyclic, 4 ABX ∼ 4 DCX and 4 ADX ∼ 4 BCX . It follows that = = and
    DX CX c
    AX DX d = = . Therefore we may write AX = adk, BX = abk, CX = bck , and DX = cdk for some k .
    BX CX b
    Now, ∠ XDC = ∠ BAX = ∠ Y XB and ∠ DCX = ∠ XBY , so 4 BXY ∼ 4 CDX . Thus, XY =
    BX abk
    2 2
    DX · = cdk · = abdk . Analogously, XY = acdk . Note that XY /XZ = CB/CD . Since
    CD c
    ∠ Y XZ = π − ∠ ZAY = ∠ BCD , we have that 4 XY Z ∼ 4 CBD . Thus, Y Z/BD = XY /CB = adk .2
    Finally, Ptolemy’s theorem applied to ABCD gives
    ( ad + bc ) k · ( ab + cd ) k = ac + bd
    It follows that the answer isad ( ac + bd ) 20 · 23 115 = =
    ( ab + cd )( ad + bc ) 22 · 26 143