HMMT 二月 2008 · 冲刺赛 · 第 24 题

英文解析

24. [ 10 ] Suppose that ABC is an isosceles triangle with AB = AC . Let P be the point on side AC so that
    AP = 2 CP . Given that BP = 1, determine the maximum possible area of ABC .
    Answer: Let Q be the point on AB so that AQ = 2 BQ , and let X be the intersection of BP9
    and CQ . The key observation that, as we will show, BX and CX are fixed lengths, and the ratio of 10
    areas [ ABC ] / [ BCX ] is constant. So, to maximize [ ABC ], it is equivalent to maximize [ BCX ].
    Using Menelaus’ theorem on ABP , we have
    BX · P C · AQ = 1 .
    XP · CA · QB
    Since P C/CA = 1 / 3 and AQ/QB = 2, we get BX/XP = 3 / 2. It follows that BX = 3 / 5. Bysymmetry, CX = 3 / 5.
    Also, we have
    [ ABC ] = 3[ BP C ] = 3 · [ BXC ] = 5[ BXC ] .5
    °63
    Note that [ BXC ] is maximized when ∠ BXC = 90 (one can check that this configuration is indeed
    ( )
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    possible). Thus, the maximum value of [ BXC ] is BX · CX = = . It follows that the
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    maximum value of [ ABC ] is .9
    11 HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUNDth