HMMT 二月 2015 · 冲刺赛 · 第 32 题

英文解析

32. [ 20 ] A wealthy king has his blacksmith fashion him a large cup, whose inside is a cone of height 9
    inches and base diameter 6 inches (that is, the opening at the top of the cup is 6 inches in diameter).
    At one of his many feasts, he orders the mug to be filled to the brim with cranberry juice.
    For each positive integer n , the king stirs his drink vigorously and takes a sip such that the heightof fluid left in his cup after the sip goes down by inches. Shortly afterwards, while the king is 1
    n 2
    distracted, the court jester adds pure Soylent to the cup until it’s once again full. The king takes sipsprecisely every minute, and his first sip is exactly one minute after the feast begins.
    As time progresses, the amount of juice consumed by the king (in cubic inches) approaches a numberr . Find r .
    √
    216 π − 2187 33
    Answer: First, we find the total amount of juice consumed. We can simply subtract
    8 π2
    the amount of juice remaining at infinity from the initial amount of juice in the cup, which of courseis simply the volume of the cup; we’ll denote this value by V .
    Since volume in the cup varies as the cube of height, the amount of juice remaining in the cup after mminutes is
    ( )
    ( ) ( )3
    m m 3
    ∏ ∏1
    9 −
    n 21
    V · = V · 1 − .
    9 9 n 2
    n =1 n =1
    Guts
    We can now factor the term inside the product to find
    ( )
    ( )3
    3 m
    ∏
    (3 n + 1)(3 n − 1) (3 m + 1)!
    V = V .
    2 3 m 3
    9 n 3 ( m !)
    n =1
    If remains to evaluate the limit of this expression as m goes to infinity.
    However, by Stirling’s approximation, we have
    √
    3 n +1 3 n +1
    ( ) · 2 π (3 n + 1)
    (3 m + 1)!
    lim = lim √e
    3 m 3
    3 n
    3 n 3
    m →∞ m →∞
    3 ( m !)
    ( ) (2 πn )
    √e
    ( )
    3 n
    (3 n + 1) 3 3 n + 1 = limm →∞
    2 πne 3 n
    √
    3 3 = .
    2 π
    Therefore the total amount of juice the king consumes is
    ( ) ( )
    √ 3 √ √
    ( )
    2 3 3
    3 3 3 · π · 9 8 π − 81 3 216 π − 2187 3
    V − V = = .
    3 2
    2 π 3 8 π 8 π
    (3 m +1)!
    Remark. We present another way to calculate the limit at m → ∞ of f ( m ) = . We have
    3 m 3
    3 ( m !)
    2 4
    ( m + )( m + )
    (3 m + 4)!
    3 3
    f ( m + 1) = = f ( m ) ,
    3 m +3 3 2
    3 ( m + 1)! ( m + 1)
    whence we can write
    2 4
    c Γ( m + )Γ( m + )
    3 3
    f ( m ) =
    Γ( m + 1)2
    for some constant c . We can find c by equating the expressions at m = 0; we have
    2 4
    c Γ( )Γ( )
    3 3
    1 = f (0) = ,
    Γ(1)2
    2 4
    so that c = Γ(1) / Γ( )Γ( ).2
    3 3
    Of course, Γ(1) = 0! = 1. We can evaluate the other product as follows:
    ( ) ( ) ( ) ( )
    2 4 1 2 1 1 π 2 π
    Γ Γ = Γ Γ = · = √ .
    3 3 3 3 3 3 sin π/ 3
    3 3
    Here the first step follows from Γ( n + 1) = n Γ( n ), while the second follows from Euler’s reflection
    √
    formula. Thus c = 3 3 / 2 π . We can now compute
    √
    2 4 2 4
    c Γ( m + )Γ( m + ) Γ( m + )Γ( m + )
    3 3
    3 3 3 3
    lim f ( m ) = lim = lim .
    2 2
    m →∞ m →∞ m →∞
    Γ( m + 1) 2 π Γ( m + 1)
    √
    Since lim Γ( n + α ) / [Γ( n ) n ] = 1, this final limit is 1 and f ( m ) → 3 3 / 2 π as m → ∞ .αn →∞
    ( )
    1000