HMMT 二月 2014 · 团队赛 · 第 5 题

英文解析

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 + zfor all z with | z | = 1 and 1 + z + z 6 = 0. (Here, | z | denotes the absolute value of the complex number 2
   √
   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 ),
   ∣ ∣
   3 1+ z + z 2
   1 cos t − i sin t
   0 ≤ t < 2 π . Writing out in terms of t and simplifying, we find that g ( t ) = . Letting
   1+ z + z 1+2 cos t 2
   x ( t ) = ℜ ( g ( t )) and y ( t ) = ℑ ( g ( t )) (the real and imaginary parts of g ( t ), respectively), what we wish toprove is equivalent to showing that { ( x ( t ) , y ( t )) | 0 ≤ t < 2 π } is a hyperbola with one focus at (0 , 0).
   However
   ( )
   9 x ( t ) − − 3 y ( t ) = 1222
   holds for all t , so from this equation we find that the locus of points ( x ( t ) , y ( t )) is a hyperbola, with 3
   2 2 4 4
   center ( , 0) and focal length , so the foci are at (0 , 0) and ( , 0). Hence c = .
   3 3 3 3