HMMT 二月 2016 · 代数 · 第 10 题

英文解析

10. Let a, b and c be positive real numbers such that
    2 2
    a + ab + b = 9
    2 2
    b + bc + c = 52
    2 2
    c + ca + a = 49 .
    2 2
    49 b − 33 bc + 9 c
    Compute the value of .
    a 2
    Proposed by: Alexander Katz
    Answer: 52
    Consider a triangle ABC with Fermat point P such that AP = a, BP = b, CP = c . Then
    2 2 2 °
    AB = AP + BP − 2 AP · BP cos(120 )
    by the Law of Cosines, which becomes
    2 2 2
    AB = a + ab + b
    √
    and hence AB = 3. Similarly, BC = 52 and AC = 7.
    Furthermore, we have
    2 2 2
    BC = 52 = AB + BC − 2 AB · BC cos ∠ BAC
    2 2 = 3 + 7 − 2 · 3 · 7 cos ∠ BAC = 58 − 42 cos ∠ BAC
    And so cos ∠ BAC = .1
    ′ ′ ′7
    Invert about A with arbitrary radius r . Let B , P , C be the images of B, P, C respectively. Since
    ′ ′ ° ′ ′ ° ′ ′ ′ °
    ∠ AP B = ∠ AB P = 120 and ∠ AP C = ∠ AC P = 120 , we note that ∠ B P C = 120 − ∠ BAC ,
    and so
    ′ ′ ′ °
    cos ∠ B P C = cos(120 − ∠ BAC )
    ° °
    = cos 120 cos ∠ BAC − sin 120 sin ∠ BAC
    ( )
    √ √
    ( )
    1 1 3 4 3 = − +
    2 7 2 7 =11
    Furthermore, using the well-known result 14
    r BC2
    ′ ′
    B C =
    AB · ACfor an inversion about A , we have
    BP r 2
    ′ ′
    B P =
    AB · APbr 2
    a · 3=
    br 2
    3 a=
    √
    2 2
    cr r 52
    ′ ′ ′ ′ ′ ′ ′
    and similarly P C = , B C = . Applying the Law of Cosines to B P C gives us
    7 a 21
    ′ ′ 2 ′ ′ 2 ′ ′ 2 ′ ′ ′ ′ °
    B C = B P + P C − 2 B P · P C cos(120 − ∠ BAC )
    4 2 4 2 4 4
    52 r b r c r 11 bcr = ⇒ = + −
    2 2 2 2
    21 9 a 49 a 147 a
    2 2
    52 b c 11 bc = ⇒ = + −
    2 2 2 2
    21 9 a 49 a 147 a
    2 2
    52 49 b − 33 bc + 9 c = ⇒ =
    2 2 2
    21 21 a
    2 2
    49 b − 33 bc +9 cand so = 52 .
    a 2
    Motivation: the desired sum looks suspiciously like the result of some Law of Cosines, so we should try
    7 b 3 c 33 bc 11
    building a triangle with sides and . Getting the − term is then a matter of setting cos θ = .
    a a a 14
    Now there are two possible leaps: noticing that cos θ = cos(120 − ∠ BAC ) , or realizing that it’s pretty
    7 b bdifficult to contrive a side of – but it’s much easier to contrive a side of . Either way leads to thea 3 anatural inversion idea, and the rest is a matter of computation.