HMMT 二月 2014 · 代数 · 第 9 题
HMMT February 2014 — Algebra — Problem 9
题目详情
- 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 , p 2 compute ( ab + bc + ca ) . (Here, i = 1.)
解析
- 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 , √ 2 compute ( ab + bc + ca ) . (Here, i = − 1.) − 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 every valid 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 then the 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 1 ◦ 2 2 ( ( ab + bc + ca ) sin 120 ) = [ ABC ] 2 √ √ ∏ √ √ √ √ ( x + y + z ) ( x + y − z ) cyc = 4 2 √ √ 1 √ √ 2 2 = [( x + y ) − z ][( x − y ) − z ] 16 1 2 2 = [( x − y ) + z − 2 z ( x + y )] 16 1 2 2 2 2 2 = ( ) [( x + y + z ) − 2( x + y + z )] , 4 as desired. 2 2 Solution 2: Let s = a + b + c . We have x − y = b + bc − ca − a = ( b − a ) s and cyclic, so ∑ ∑ 2 x + as = y + bs = z + cs (they all equal a + bc ). Now add all equations to get ∑ ∑ ∑ 1 2 2 2 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 )] = ∑ ∑ ∑ ∑ 1 2 2 2 2 2 2 ( x + y + z ) − 2 ( z − y ) = 6 yz − 3 x . But 2 s − ( x + y + z ) = s − ( b − c ) = ( a + b + 2 ∑ 1 2 2 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 = , s ∑ u − x 2 etc. Then s = , or s = 3 u − s ( x + y + z ). But we get another equation in s, u by just plugging s 2 2 in directly to a + ab + b = z (and after everything is in terms of s , we can finish without too much trouble).