HMMT 十一月 2013 · 冲刺赛 · 第 25 题

英文解析

25. [ 13 ] Let a, b be positive reals with a > b > a . Place two squares of side lengths a, b next to eachother, such that the larger square has lower left corner at (0 , 0) and the smaller square has lower left 2
    corner at ( a, 0). Draw the line passing through (0 , a ) and ( a + b, 0). The region in the two squaresalying above the line has area 2013. If ( a, b ) is the unique pair maximizing a + b , compute .
    5 ab
    Answer: Let t = ∈ (1 , 2); we will rewrite the sum a + b as a function of t . The area
    3 b
    √
    2 2
    a − ab +2 b 4026
    2 2
    condition easily translates to = 2013, or b ( t − t + 2) = 4026 ⇐⇒ b = . Thus
    2 t − t +22
    √
    4026
    a + b is a function f ( t ) = (1 + t ) of t , and our answer is simply the value of t maximizing f ,
    t − t +22
    (1+ t )22
    or equivalently g ( t ) = = , over the interval (1 , 2). (In general, such maximizers/maximumsf
    4026 t − t +22
    need not exist, but we shall prove there’s a unique maximum here.)
    (1+ t )2
    We claim that λ = is the maximum of . Indeed,16
    7 t − t +22
    ( λ − 1) t − ( λ + 2) t + (2 λ − 1)2
    λ − g ( t ) =
    t − t + 22
    2 2
    1 9 t − 30 t + 25 1 (3 t − 5) = = ≥ 0
    1 7
    7 t − t + 2 722
    ( t − ) +
    2 4
    for all reals t (not just t ∈ (1 , 2)), with equality at t = ∈ (1 , 2).5
    Comment. To motivate the choice of λ , suppose λ were the maximum of f , attained at t ∈ (1 , 2);3
    2 20
    then h ( t ) = λ ( t − t + 2) − ( t + 1) is quadratic and nonnegative on (1 , 2), but zero at t = t . If g is anontrivial quadratic (nonzero leading coefficient), then t must be a double root, so g has determinant 0
26. Of course, g could also be constant or linear over (1 , 2), but we can easily rule out both of these 0
    possibilities.
    Alternatively, we can simply take a derivative of f to find critical points.
    2 2 2