HMMT 二月 2010 · 代数 · 第 5 题

英文解析

5. [ 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
   √ √
   20 201 = [( 2 b − 1 + 1) + ( 2 b − 1 − 1) ]
   [ ( ) ( ) ]220
   √ √ √
   2 20 20
   20 18 16 = ( 2 b − 1) + ( 2 b − 1) + ( 2 b − 1) + . . .
   2 2 420
   [ ( ) ( ) ]
   2 20 20
   10 9 8 = (2 b − 1) + (2 b − 1) + (2 b − 1) + . . .
   2 2 420 = 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)
   √
   2 πik 19
   19 19
   The roots of x − 20 are 20 exp( ) for some k = 0 , 1 , . . . , 18; the roots of (1 − x ) − 20 arek
   √19
   2 πik 19
   19 19
   1 − 20 exp( ) for some k = 0 , 1 , . . . , 18. If x − 20 and (1 − x ) − 20 share a common root,k
   √ √19
   19 19
   2 πim 2 πinm nthen there exist integers m, n such that 20 exp( ) = 1 − 20 exp( ); since the imaginary
   19 19
   √
   2 πim 1 119
   parts of both sides must be the same, we have m = n and 20 exp( ) = = ⇒ 20 = , amm
   19 2 219
   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, thepolynomial 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 20
   9 9
   2 − 29
   ( ) ( )
   [ b ]
   1 2 − = − = − 90.
   10 10
   [ b ] 2
   2010 2009