HMMT 二月 2011 · 冲刺赛 · 第 29 题

英文解析

29. [ 16 ] Let ABC be a triangle such that AB = AC = 182 and BC = 140. Let X lie on AC such that
    CX = 130. Let the line through X perpendicular to BX at X meet AB at X . Define X , X , . . . ,1
    1 1 1 1 2 2 3
    as follows: for n odd and n ≥ 1, let X be the intersection of AB with the perpendicular to X Xn +1 n − 1 nthrough X ; for n even and n ≥ 2, let X be the intersection of AC with the perpendicular ton n +1
    X X through X . Find BX + X X + X X + . . . .
    n − 1 n n 1 1 2 2 3
    1106
    Answer:
    Let M and N denote the perpendiculars from X and A to BC , respectively. Since triangle ABC5
    is isosceles, we have M is the midpoint of BC . Moreover, since AM is parallel to X N , we have 1
    N C M C X N 70 51 = ⇔ = = , so N C = 50. Moreover, since X N ⊥ BC , we find X C = 120 by 1
    1 1
    X C AC 130 182 13
    the Pythagorean Theorem. Also, BN = BC − N C = 140 − 50 = 90, so by the Pythagorean Theorem,1
    X B = 150 .
    We want to compute X X = X B tan( ∠ ABX ). We have 1
    2 1 1 1
    ( ) ( )
    12 4
    1 + tan( ∠ ABC ) tan( ∠ X BC )−
    5 31
    ( ) ( )
    tan( ∠ ABX ) = tan( ∠ ABC − ∠ X BC ) = =
    1 1
    12 4
    tan( ∠ ABC ) − tan( ∠ X BC )
    1 +
    5 31
    1616 = = .15
    6363
    16 6515
    Hence X X = 150 · , and by the Pythagorean Theorem again, X B = 150 · .
    2 1 2
    63 63
    Next, notice that is constant for every nonnegative integer n (where we let B = X ). Indeed,n aX
    AX0
    n +2
    since X X is parallel to X X for each n , the dilation taking X to X for some n alson n +1 n +2 n +3 n n +2
    takes X to X for all k .
    k k +2
    Guts Round
    X X
    AX n +2 n +3
    Since 4 AX X ∼ 4 AX X with ratio for each even n , we can compute that =nn +2 n +3 n n +1
    AX X Xn +2 n n +1
    150 ·65
    n +2AX = 1 − for every nonnegative integer n . Notice we use all three sides of the above similar 63
    AX 182
    triangles.n
    We now split our desired sum into two geometric series, one with the even terms and one with the oddterms, to obtain
    150 ·16
    63150
    BX + X X + . . . = ( BX + X X + . . . ) + ( X X + X X + . . . ) = +
    1 1 2 1 2 3 1 2 3 4
    65 65
    150 · 150 ·
    63 63
    182 182
    · 15079
    1106 = = .63
    150 ·65
    63 5
    182