HMMT 二月 2016 · 冲刺赛 · 第 27 题

英文解析

27. [ 14 ] Find the smallest possible area of an ellipse passing through (2 , 0), (0 , 3), (0 , 7), and (6 , 0).
    Proposed by: Calvin Deng
    √
    56 π 3
    Answer:
    Let Γ be an ellipse passing through A = (2 , 0) , B = (0 , 3) , C = (0 , 7) , D = (6 , 0), and let P = (0 , 0) be 9
    Area of Γthe intersection of AD and BC . is unchanged under an affine transformation, so we just
    Area of ABCD
    P A 1 P B 3
    have to minimize this quantity over situations where Γ is a circle and = and = . In fact,
    P D 3 BC 7
    √ √
    we may assume that P A = 7 , P B = 3 , P C = 7 , P D = 3 7. If ∠ P = θ , then we can compute lengthsto get
    √
    Area of Γ 32 − 20 7 cos θ + 21 cos θ2
    r = = π √
    Area of ABCD3
    9 7 · sin θ
    Let x = cos θ . Then if we treat r as a function of x ,
    √
    ′
    r 3 x 42 x − 20 7
    0 = = + √
    r 1 − x 22
    32 − 20 x 7 + 21 x
    √ √ √
    which means that 21 x − 40 x 7 + 138 x − 20 7 = 0. Letting y = x 7 gives 3
    3 2 2
    0 = 3 y − 40 y + 138 y − 140 = ( y − 2)(3 y − 34 y + 70)
    √
    The other quadratic has roots that are greater than 7, which means that the minimum ratio isy
    2 Area of Γ
    √ √
    attained when cos θ = x = = . Plugging that back in gives that the optimum is
    Area of ABCD
    7 7
    √ √
    28 π 3 56 π 3
    , so putting this back into the original configuration gives Area of Γ ≥ . If you want to
    81 9
    8 7
    check on Geogebra, this minimum occurs when the center of Γ is ( , ).
    3 3