HMMT 二月 2011 · ALGGEO 赛 · 第 16 题

英文解析

16. Let ABCD be a quadrilateral inscribed in the unit circle such that ∠ BAD is 30 degrees. Let m denotethe minimum value of CP + P Q + CQ , where P and Q may be any points lying along rays AB and
    AD , respectively. Determine the maximum value of m .
    Answer: 2
    Algebra & Geometry Individual Test
    RHAG
    PFQE
    B D
    For a fixed quadrilateral ABCD as described, we first show that m , the minimum possible length of c
    CP + P Q + QC , equals the length of AC . Reflect B , C , and P across line AD to points E , F , and
    R , respectively, and then reflect D and F across AE to points G and H , respectively. These two
    °
    reflections combine to give a 60 rotation around A , so triangle ACH is equilateral. It also follows
    °
    that RH is a 60 rotation of P C around A , so, in particular, these segments have the same length.
    Because QR = QP by reflection,
    CP + P Q + QC = CQ + QR + RH.
    The latter is the length of a broken path CQRH from C to H , and by the “shortest path is a straightline” principle, this total length is at least as long as CH = CA . (More directly, this follows from the triangle inequality: ( CQ + QR )+ RH ≥ CR + RH ≥ CH ). Therefore, the lower bound m ≥ AC indeedholds. To see that this is actually an equality, note that choosing Q as the intersection of segment CHwith ray AD , and choosing P so that its reflection R is the intersection of CH with ray AE , alignspath CQRH with segment CH , thus obtaining the desired minimum m = AC .
    We may conclude that the largest possible value of m is the largest possible length of AC , namely 2:
    the length of a diameter of the circle.
    2 π 2 π