HMMT 二月 2015 · 冲刺赛 · 第 26 题
HMMT February 2015 — Guts Round — Problem 26
题目详情
- [ 17 ] Let a = 17 and b = i 19, where i = − 1. Find the maximum possible value of the ratio | a − z | / | b − z | over all complex numbers z of magnitude 1 (i.e. over the unit circle | z | = 1).
解析
- [ 17 ] Let a = 17 and b = i 19, where i = − 1. Find the maximum possible value of the ratio | a − z | / | b − z | over all complex numbers z of magnitude 1 (i.e. over the unit circle | z | = 1). 4 Answer: Let | a − z | / | b − z | = k. We wish to determine the minimum and maximum value of k . 3 Squaring and expansion give: 2 2 2 | a − z | = | b − z | · k 2 2 2 | a | − 2 a · z + 1 = ( | b | − 2 b · z + 1) k 2 2 2 2 | a | + 1 − ( | b | + 1) k = 2( a − bk ) · z, where · is a dot product of complex numbers, i.e., the dot product of vectors corresponding to the complex numbers in the complex plane. Now, since z has modulus 1 but can assume any direction, the only constraint on the value of k is 2 2 2 2 || a | + 1 − ( | b | + 1) k | ≤ 2 | a − bk | . Squaring again and completing the square, the inequality reduces to: 2 2 2 2 4 2 2 2 ( | a | − 1) + ( | b | − 1) k + 2(4 a · b − ( | a | + 1)( | b | + 1)) k ≤ 0 2 2 2 2 2 2 (( | a | − 1) − ( | b | − 1) k ) − 4 | a − b | k ≤ 0 2 2 2 | ( | a | − 1) − ( | b | − 1) k | ≤ 2 | a − b | k. At this stage all the relevant expressions are constant real numbers. Denote, for simplicity, A = 2 2 2 | a | − 1 , B = | b | − 1, and C = | a − b | . Then, we are looking for k such that | A − Bk | ≤ 2 Ck . If B = 0, A A then k ≥ | | , so the minimum value is | | and the maximum value is + ∞ . Otherwise, consider 2 C 2 C 2 2 2 2 2 C + AB = ( | a | − 2 a · b + | b | ) + ( | a | − 1)( | b | − 1) 2 = | ab | − 2 a · b + 1 2 = | ab | − 2 ℜ ( ab ) + 1 2 = | ab − 1 | √ 2 So let D = | ab − 1 | = C + AB . We may assume B > 0 (the another case is analogous: just substitute A, B with − A, − B ). Then, k is determined by the following inequalities: 2 Bk + 2 Ck − A ≥ 0 2 Bk − 2 Ck − A ≤ 0 − C − D − C + D C − D C + D The first inequality gives k ≤ or k ≥ , and the second gives ≤ k ≤ . Combining, B B B B C − D C + D this gives | | ≤ k ≤ | | , as claimed. B B 2 2 To summarize the general answer, let A = | a | − 1 , B = | b | − 1 , C = | a − b | , D = | ab − 1 | . Then, if A C − D C + D | b | = 1, min is | | and max is + ∞ ; otherwise, min is | | and max is | | . 2 C B B √ √ ∣ √ √ ∣ √ ∣ ∣ In the special case a = 17 and b = 19 i , we have A = 16, B = 18, C = 17 − 19 i = 36 = 6, √ C + D 6+18 4 and D = 17 · 19 + 1 = 18. Thus the answer is = = . B 18 3