HMMT 二月 2010 · CALC 赛 · 第 7 题
HMMT February 2010 — CALC Round — Problem 7
题目详情
- [ 6 ] Let a , a , and a be nonzero complex numbers with non-negative real and imaginary parts. Find 1 2 3 the minimum possible value of | a + a + a | 1 2 3 √ . 3 | a a a | 1 2 3 ∞ ∞ ∑ ∑ 1
解析
- [ 6 ] Let a , a , and a be nonzero complex numbers with non-negative real and imaginary parts. Find 1 2 3 the minimum possible value of | a + a + a | 1 2 3 √ . 3 | a a a | 1 2 3 √ √ 3 iθ π Answer: 3 2 Write a in its polar form re where 0 ≤ θ ≤ . Suppose a , a and r are fixed so 1 2 3 2 iφ that the denominator is constant. Write a + a as se . Since a and a have non-negative real and 2 3 2 3 π imaginary parts, the angle φ lies between 0 and . Consider the function 2 2 iθ iφ 2 2 2 f ( θ ) = | a + a + a | = | re + se | = r + 2 rs cos( θ − φ ) + s . 1 2 3 π π ′′ ′′ Its second derivative is f ( θ ) = − 2 rs (cos( θ − φ ))). Since − ≤ ( θ − φ ) ≤ , we know that f ( θ ) < 0 2 2 π and f is concave. Therefore, to minimize f , the angle θ must be either 0 or . Similarly, each of a , a 1 2 2 and a must be either purely real or purely imaginary to minimize f and the original fraction. 3 By the AM-GM inequality, if a , a and a are all real or all imaginary, then the minimum value of 1 2 3 the fraction is 3. Now suppose only two of the a ’s, say, a and a are real. Since the fraction is i 1 2 homogenous, we may fix a + a - let the sum be 2. The term a a in the denominator acheives its 1 2 1 2 maximum only when a and a are equal, i.e. when a = a = 1. Then, if a = ki for some real 1 2 1 2 3 number k , then the expression equals √ 2 k + 4 √ . 3 k √ √ 3 Squaring and taking the derivative, we find that the minimum value of the fraction is 3 2, attained √ when k = 2. With similar reasoning, the case where only one of the a ’s is real yields the same i minimum value. ∞ ∞ ∑ ∑ 1