HMMT 二月 2010 · GEN2 赛 · 第 7 题
HMMT February 2010 — GEN2 Round — Problem 7
题目详情
- [ 5 ] Suppose that x and y are complex numbers such that x + y = 1 and that x + y = 20. Find the 2 2 sum of all possible values of x + y .
解析
- [ 5 ] Suppose that x and y are complex numbers such that x + y = 1 and that x + y = 20. Find the 2 2 sum of all possible values of x + y . 2 2 2 2 Answer: − 90 We have x + y + 2 xy = 1. Define a = 2 xy and b = x + y for convenience. Then √ 2 b − 1 ± 1 2 2 2 a + b = 1 and b − a = x + y − 2 xy = ( x − y ) = 2 b − 1 so that x, y = . Then 2 √ √ ( ) ( ) 20 20 2 b − 1 + 1 2 b − 1 − 1 20 20 x + y = + 2 2 √ √ 1 20 20 = [( 2 b − 1 + 1) + ( 2 b − 1 − 1) ] 20 2 [ ( ) ( ) ] √ √ √ 2 20 20 20 18 16 = ( 2 b − 1) + ( 2 b − 1) + ( 2 b − 1) + . . . 20 2 2 4 [ ( ) ( ) ] 2 20 20 10 9 8 = (2 b − 1) + (2 b − 1) + (2 b − 1) + . . . 20 2 2 4 = 20 We want to find the sum of distinct roots of the above polynomial in b ; we first prove that the original 20 20 20 20 polynomial is square-free. The conditions x + y = 1 and x + y = 20 imply that x +(1 − x ) − 20 = 0; 20 20 ′ let p ( x ) = x + (1 − x ) − 20. p is square-free if and only if GCD ( p, p ) = c for some constant c : ′ 20 20 19 19 GCD ( p, p ) = GCD ( x + (1 − x ) − 20 , 20( x − (1 − x ) )) 20 19 19 19 19 = GCD ( x − x (1 − x ) + (1 − x ) − 20 , 20( x − (1 − x ) )) 19 19 19 = GCD ((1 − x ) − 20 , x − (1 − x ) ) 19 19 = GCD ((1 − x ) − 20 , x − 20) √ 19 19 2 πik 19 k The roots of x − 20 are 20 exp( ) for some k = 0 , 1 , . . . , 18; the roots of (1 − x ) − 20 are 19 √ 19 2 πik 19 19 k 1 − 20 exp( ) for some k = 0 , 1 , . . . , 18. If x − 20 and (1 − x ) − 20 share a common root, 19 √ √ 19 19 2 πim 2 πin m n then there exist integers m, n such that 20 exp( ) = 1 − 20 exp( ); since the imaginary 19 19 √ 19 2 πim 1 1 m m parts of both sides must be the same, we have m = n and 20 exp( ) = = ⇒ 20 = , a 19 19 2 2 contradiction. Thus we have proved that the polynomial in x has no double roots. Since for each b 2 2 2 there exists a unique pair ( x, y ) (up to permutations) that satisfies x + y = b and ( x + y ) = 1, the polynomial in b has no double roots. n n Let the coefficient of b in the above equation be [ b ]. By Vieta’s Formulas, the sum of all possible ( ) ( ( ) ( ) ) 9 [ b ] 10 20 2 2 2 2 10 10 9 9 9 values of b = x + y is equal to − . [ b ] = 2 and [ b ] = − 2 + 2 ; thus 10 20 20 [ b ] 2 2 1 2 10 9 20 9 9 2 − 2 ( ) ( ) [ b ] 1 2 − = − = − 90. 10 10 [ b ] 2