HMMT 十一月 2011 · 团队赛 · 第 10 题

英文解析

10. [ 8 ] Let G G G be a triangle with G G = 7, G G = 13, and G G = 15. Let G be a point
    1 2 3 1 2 2 3 3 1 4 − − − →
    outside triangle G G G so that ray G G cuts through the interior of the triangle, G G = G G ,
    1 2 3 1 4 3 4 4 2
    °
    and ∠ G G G = 30 . Let G G and G G meet at G . Determine the length of segment G G .
    3 1 4 3 4 1 2 5 2 5
    Answer:169
    G23
    y 5
    4Gx
    2G
    G G713
    1 3
    We first show that quadrilateral G G G G is cyclic. Note that by the law of cosines,15
    1 2 4 3
    Team Round
    2 2 2
    7 + 15 − 13 1
    cos ∠ G G G = = ,
    2 1 3
    2 · 7 · 15 2
    ° °
    so ∠ G G G = 60 . However, we know that ∠ G G G = 30 , so G G is an angle bisector. Now, let
    2 1 3 3 1 4 1 4
    ̂ ̂
    G G intersect the circumcircle of triangle G G G at X . Then, the minor arcs G X and G X are
    1 4 1 2 3 2 3
    subtended by the equal angles ∠ G G X and ∠ G G X , implying that G X = G X , i.e. X is on the
    2 1 3 1 2 3
    perpendicular bisector of G G , l . Similarly, since G G = G G , G lies on l . However, since l and
    2 3 4 2 4 3 4
    G G are distinct (in particular, G lies on G G but not l ), we in fact have X = G , so G G G G
    1 4 1 1 4 4 1 2 4 3
    is cyclic.
    We now have G G G ∼ G G G since G G G G is cyclic. Now, we have ∠ G G G = ∠ G G G =
    5 2 4 5 3 1 1 2 4 3 4 3 2 4 1 2
    √
    °
    30 , and we may now compute G G = G G = 13 / 3. Let G G = x and G G = y . Now, from
    2 4 4 3 5 2 5 4
    G G G ∼ G G G , we have:
    5 4 2 5 1 3
    √
    x 13 / 3 y
    √ = = .
    15 x + 7
    y + 13 / 3
    Equating the first and second expressions and cross-multiplying, we get
    √ √
    13 3 15 3 xy + = .
    3 13
    Now, equating the first and third expressions and and substituting gives
    ( ) ( )
    √ √ √
    15 3 x 13 3 15 3 x − = x ( x + 7).
    13 3 13
    Upon dividing both sides by x , we obtain a linear equation from which we can solve to get x = 169 / 23.
    Team Round