HMMT 十一月 2014 · 冲刺赛 · 第 31 题
HMMT November 2014 — Guts Round — Problem 31
题目详情
- [ 17 ] Flat Albert and his buddy Mike are watching the game on Sunday afternoon. Albert is drinking lemonade from a two-dimensional cup which is an isosceles triangle whose height and base measure 9cm and 6cm; the opening of the cup corresponds to the base, which points upwards. Every minute after the game begins, the following takes place: if n minutes have elapsed, Albert stirs his drink vigorously 1 and takes a sip of height cm. Shortly afterwards, while Albert is busy watching the game, Mike adds 2 n cranberry juice to the cup until it’s once again full in an attempt to create Mike’s cranberry lemonade. Albert takes sips precisely every minute, and his first sip is exactly one minute after the game begins. After an infinite amount of time, let A denote the amount of cranberry juice that has been poured (in 27 square centimeters). Find the integer nearest A . 2 π 2 n
解析
- [ 17 ] Flat Albert and his buddy Mike are watching the game on Sunday afternoon. Albert is drinking lemonade from a two-dimensional cup which is an isosceles triangle whose height and base measure 9cm and 6cm; the opening of the cup corresponds to the base, which points upwards. Every minute after the game begins, the following takes place: if n minutes have elapsed, Albert stirs his drink vigorously 1 and takes a sip of height cm. Shortly afterwards, while Albert is busy watching the game, Mike adds 2 n cranberry juice to the cup until it’s once again full in an attempt to create Mike’s cranberry lemonade. Albert takes sips precisely every minute, and his first sip is exactly one minute after the game begins. After an infinite amount of time, let A denote the amount of cranberry juice that has been poured (in 27 square centimeters). Find the integer nearest A . 2 π 1 Answer: 26 Let A = (6)(9) = 27 denote the area of Albert’s cup; since area varies as the square 0 2 of length, at time n Mike adds ( ) ( ) 2 1 A 1 − 1 − 2 9 n whence in all, he adds ( ) ∞ ∑ 2 1 2 A ζ (2) A ζ (4) 1 0 0 A − = − = 6 ζ (2) − ζ (4) 0 2 4 9 n 81 n 9 81 3 n =1 2 4 4 π π π 2 where ζ is the Riemann zeta function. Since ζ (2) = and ζ (4) = , we find that A = π − , so 6 90 270 2 27 A π = 27 − , which gives an answer 26. 2 π 10 Note that while the value of ζ (2) is needed to reasonable precision, we only need the fact that 0 . 5 < 9 ζ (4) < 1 . 5 in order to obtain a sufficiently accurate approximation. This is not hard to obtain 2 π because the terms of the expansion ζ (4) decrease rapidly. 2 n