HMMT 二月 2014 · 代数 · 第 9 题

英文解析

9. Given that a , b , and c are complex numbers satisfying
   2 2
   a + ab + b = 1 + i
   2 2
   b + bc + c = − 2
   2 2
   c + ca + a = 1 ,
   √
   compute ( ab + bc + ca ) . (Here, i = − 1.)2 − 11 − 4 i 11+4 i
   2 2 2 2
   Answer: OR − More generally, suppose a + ab + b = z , b + bc + c = x ,
   3 3
   2 2
   c + ca + a = y for some complex numbers a, b, c, x, y, z .
   We show that
   1 1
   ° 2 2 2 2 2 2
   f ( a, b, c, x, y, z ) = ( ( ab + bc + ca ) sin 120 ) − ( ) [( x + y + z ) − 2( x + y + z )]
   2 4
   holds in general. Plugging in x = − 2, y = 1, z = 1 + i will then yield the desired answer,
   16 1
   2 2 2 2 2
   ( ab + bc + ca ) = [( x + y + z ) − 2( x + y + z )]
   3 16
   2 2
   i − 2(4 + 1 + (1 + i ) ) − 1 − 2(5 + 2 i ) − 11 − 4 i = = = .
   3 3 3
   2 2
   Solution 1: Plug in x = b + bc + c , etc. to get a polynomial g in a, b, c (that agrees with f for everyvalid choice of a, b, c, x, y, z ). It suffices to show that g ( a, b, c ) = 0 for all positive reals a, b, c , as thenthe polynomial g will be identically 0.
   But this is easy: by the law of cosines, we get a geometrical configuration with a point P inside a
   ° 2
   triangle ABC with P A = a , P B = b , P C = c , ∠ P AB = ∠ P BC = ∠ P CA = 120 , x = BC ,
   2 2
   y = CA , z = AB . By Heron’s formula, we have
   ° 2 21
   ( ( ab + bc + ca ) sin 120 ) = [ ABC ]
   √ √ ∏ √ √2
   √ √
   ( x + y + z ) ( x + y − z ) =cyc
   √ √24
   √ √1
   2 2 = [( x + y ) − z ][( x − y ) − z ]
   2 2116 = [( x − y ) + z − 2 z ( x + y )]
   2 2 2 2 2116 = ( ) [( x + y + z ) − 2( x + y + z )] ,
   as desired.4
   2 2
   Solution 2: Let s = a + b + c . We have x − y = b + bc − ca − a = ( b − a ) s and cyclic, so
   ∑ ∑
   x + as = y + bs = z + cs (they all equal a + bc ). Now add all equations to get 2
   ∑ ∑ ∑
   2 2 21
   x + y + z = 2 a + bc = s + ( b − c ) ;
   ∑2
   2 2 4 2 2 2
   multiplying both sides by 4 s yields 4 s ( x + y + z ) = 4 s + 2 ( z − y ) , so [2 s − ( x + y + z )] =
   ∑ ∑ ∑ ∑
   2 2 2 2 2 21
   ( x + y + z ) − 2 ( z − y ) = 6 yz − 3 x . But 2 s − ( x + y + z ) = s − ( b − c ) = ( a + b +
   ∑2
   2 21
   c ) − ( b − c ) = 3( ab + bc + ca ), so
   ∑ ∑2
   2 2 2 2 2 2
   9( ab + bc + ca ) = 6 yz − 3 x = 3[( x + y + z ) − 2( x + y + z )] ,
   which easily rearranges to the desired.
   u − x
   Comment: Solution 2 can be done with less cleverness. Let u = x + as = y + bs = z + cs , so a = ,
   ∑su − xetc. Then s = , or s = 3 u − s ( x + y + z ). But we get another equation in s, u by just plugging 2
   2 2 sin directly to a + ab + b = z (and after everything is in terms of s , we can finish without too muchtrouble).