HMMT 二月 2014 · 团队赛 · 第 5 题

HMMT February 2014 — Team Round — Problem 5

[ 25 ] Prove that there exists a nonzero complex number c and a real number d such that 1 1 c = d 2 2 1 + z + z 1 + z + z 2 for all z with | z | = 1 and 1 + z + z 6 = 0. (Here, | z | denotes the absolute value of the complex number p 2 2 z , so that | a + bi | = a + b for real numbers a, b .)

解析

[ 25 ] Prove that there exists a nonzero complex number c and a real number d such that ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ 1 1 ∣ ∣ ∣ ∣ ∣ ∣ − − c = d ∣ ∣ ∣ ∣ ∣ ∣ 2 2 1 + z + z 1 + z + z 2 for all z with | z | = 1 and 1 + z + z 6 = 0. (Here, | z | denotes the absolute value of the complex number √ 2 2 z , so that | a + bi | = a + b for real numbers a, b .) ∣ ∣ ∣ ∣ 4 1 it it Answer: Let f ( z ) = . Parametrize z = e = cos t + i sin t and let g ( t ) = f ( e ), ∣ ∣ 2 3 1+ z + z 1 cos t − i sin t 0 ≤ t < 2 π . Writing out in terms of t and simplifying, we find that g ( t ) = . Letting 2 1+ z + z 1+2 cos t x ( t ) = ℜ ( g ( t )) and y ( t ) = ℑ ( g ( t )) (the real and imaginary parts of g ( t ), respectively), what we wish to prove is equivalent to showing that { ( x ( t ) , y ( t )) | 0 ≤ t < 2 π } is a hyperbola with one focus at (0 , 0). However ( ) 2 2 2 9 x ( t ) − − 3 y ( t ) = 1 3 holds for all t , so from this equation we find that the locus of points ( x ( t ) , y ( t )) is a hyperbola, with 2 2 4 4 center ( , 0) and focal length , so the foci are at (0 , 0) and ( , 0). Hence c = . 3 3 3 3