HMMT 二月 2006 · 冲刺赛 · 第 41 题

英文解析

41. Let Γ denote the circumcircle of triangle ABC . Point D is on AB such that CD bisects
    ∠ ACB . Points P and Q are on Γ such that P Q passes through D and is perpendicularto CD . Compute P Q , given that BC = 20 , CA = 80 , AB = 65.
    √
    Answer: 4 745
    Solution: Suppose that P lies between A and B and Q lies between A and C , and letline P Q intersect lines AC and BC at E and F respectively. As usual, we write a, b, cfor the lengths of BC, CA, AB . By the angle bisector theorem, AD/DB = AC/CB so
    ( ) ( )
    bc ac ac bcthat AD = and BD = . Now by Stewart’s theorem, c · CD + c =2
    a + b a + b a + b a + b
    2 2
    2 2
    ab (( a + b ) − c )
    a bc ab c + from which CD = . Now observe that triangles CDE and 2
    a + b a + b ( a + b )2
    CA ED F B
    CDF are congruent, so ED = DF . By Menelaus’ theorem, = 1 so that
    AE DF BCb ( b − a )
    CA AE 2 ab = . Since CF = CE while b > a , it follows that AE = so that EC = .
    BC F B a + b a + b
    √
    √
    2 2
    ab ( c − ( a − b ) )
    2 2
    Finally, DE = CE − CD = . Plugging in a = 20 , b = 80 , c = 65,
    a + bwe see that AE = 48 , EC = 32 , DE = 10 as well as AD = 52 , BD = 13. Now let
    P D = x, QE = y . By power of a point about D and E , we have x ( y + 10) = 676 andy ( x + 10) = 1536. Subtracting one from the other, we see that y = x + 86. Therefore,
    √ √
    x + 96 x − 676 = 0, from which x = − 48 + 2 745. Finally, P Q = x + y + 10 = 4 745.2