HMMT 二月 2007 · 冲刺赛 · 第 28 题

英文解析

28. [ 15 ] Compute the circumradius of cyclic hexagon ABCDEF , which has side lengths AB = BC =
    2 , CD = DE = 9, and EF = F A = 12.
    ′ ′8
    Answer: 8 . Construct point E on the circumcircle of ABCDEF such that DE = EF = 12 and
    √ √
    ′ ′ ′
    ′ ′ 2 2 2
    E F = DE = 9; then BE is a diameter. Let BE = d . Then CE = BE − BC = d − 4 and
    √ √
    ′
    ′ 2 ′ 2 2
    BD = BE − DE = d − 144. Applying Ptolemy’s theorem to BCDE now yields
    √
    2 2
    9 · d + 2 · 12 = ( d − 4)( d − 144)
    4 2 2
    Squaring and rearranging, we find 0 = d − 229 d − 432 d = d ( d − 16)( d + 16 d + 27). Since d is apositive real number, d = 16, and the circumradius is 8.
    5 4 2