HMMT 二月 2014 · 冲刺赛 · 第 25 题

英文解析

25. [ 17 ] Let ABC be an equilateral triangle of side length 6 inscribed in a circle ω . Let A , A be the
    1 2
    points (distinct from A ) where the lines through A passing through the two trisection points of BCmeet ω . Define B , B , C , C similarly. Given that A , A , B , B , C , C appear on ω in that order,
    1 2 1 2 1 2 1 2 1 2
    find the area of hexagon A A B B C C .
    1 2 1 2 1 2
    √
    846 3 ′ ′ ′
    Answer: Let A be the point on BC such that 2 BA = A C . By law of cosines on triangle
    √49
    ′ ′ ′ 2 ∗ 4 4
    √ √
    AA B , we find that AA = 2 7. By power of a point, A A = = . Using side length ratios,
    2 7 71
    √
    √4
    2 7+
    7 18AA
    √1
    A A = 2 = 2 = .
    1 2 ′
    AA 7
    2 7
    Now our hexagon can be broken down into equilateral triangle A B C and three copies of triangle
    1 1 1
    A C C . Since our hexagon has rotational symmetry, ∠ C = 120, and using law of cosines on this
    1 1 2 2
    18 30
    triangle with side lengths and 6, a little algebra yields A C = (this is a 3-5-7 triangle with an
    1 2
    7 7
    angle 120).
    √ √ √
    6 3 3 846 32
    1 18 30
    The area of the hexagon is therefore + 3 ∗ =
    4 2 7 7 2 49