HMMT 二月 2012 · 代数 · 第 4 题
HMMT February 2012 — Algebra — Problem 4
题目详情
- During the weekends, Eli delivers milk in the complex plane. On Saturday, he begins at z and delivers 3 5 7 2013 milk to houses located at z , z , z , . . . , z , in that order; on Sunday, he begins at 1 and delivers 2 4 6 2012 milk to houses located at z , z , z , . . . , z , in that order. Eli always walks directly (in a straight line) between two houses. If the distance he must travel from his starting point to the last house is √ 2 2012 on both days, find the real part of z .
解析
- During the weekends, Eli delivers milk in the complex plane. On Saturday, he begins at z and delivers 3 5 7 2013 milk to houses located at z , z , z , . . . , z , in that order; on Sunday, he begins at 1 and delivers 2 4 6 2012 milk to houses located at z , z , z , . . . , z , in that order. Eli always walks directly (in a straight line) between two houses. If the distance he must travel from his starting point to the last house is √ 2 2012 on both days, find the real part of z . 1005 Answer: Note that the distance between two points in the complex plane, m and n , is | m − n | . 1006 We have that 1006 1006 ∑ ∑ ∣ ∣ ∣ ∣ √ 2 k +1 2 k − 1 2 k 2 k − 2 ∣ ∣ ∣ ∣ z − z = z − z = 2012 . k =1 k =1 Algebra Test However, noting that 1006 1006 ∑ ∑ ∣ ∣ ∣ ∣ 2 k 2 k − 2 2 k +1 2 k − 1 ∣ ∣ ∣ ∣ | z | · z − z = z − z , k =1 k =1 √ we must have | z | = 1. Then, since Eli travels a distance of 2012 on each day, we have 1006 1006 1006 ∑ ∑ ∑ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ 2 k − 2 2 k 2 k − 2 2 2 k − 2 2 ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ z − z = z − 1 · z = z − 1 · | z | k =1 k =1 k =1 √ ∣ ∣ 2 ∣ ∣ = 1006 z − 1 = 2012 , √ ∣ ∣ 2012 2 2 ∣ ∣ so z − 1 = . Since | z | = 1, we can write z = cos ( θ ) + i sin ( θ ) and then z = cos (2 θ ) + i sin (2 θ ). 1006 Hence, √ √ √ ∣ ∣ 2012 2 2 2 ∣ ∣ z − 1 = (cos (2 θ ) − 1) + sin (2 θ ) = 2 − 2 cos (2 θ ) = , 1006 2 2 1005 so 2 − 2 cos (2 θ ) = . The real part of z , cos (2 θ ), is thus . 1006 1006