HMMT 二月 2020 · ALGNT 赛 · 第 9 题

HMMT February 2020 — ALGNT Round — Problem 9

Let P ( x ) = x + x + 2, which has 2020 distinct roots. Let Q ( x ) be the monic polynomial of degree ( ) 2020 whose roots are the pairwise products of the roots of P ( x ). Let α satisfy P ( α ) = 4. Compute 2 2 2 the sum of all possible values of Q ( α ) .

解析

Let P ( x ) = x + x + 2, which has 2020 distinct roots. Let Q ( x ) be the monic polynomial of degree ( ) 2020 whose roots are the pairwise products of the roots of P ( x ). Let α satisfy P ( α ) = 4. Compute 2 2 2 the sum of all possible values of Q ( α ) . Proposed by: Milan Haiman 2019 Answer: 2020 · 2 ∏ 2 Solution: Let P ( x ) have degree n = 2020 with roots r , . . . , r . Let R ( x ) = ( x − r ). Then 1 n i i ( ) ∏ ∏ ∏ x n 2 r P = ( x − r r ) = Q ( x ) R ( x ) . i j i r i i i j 2 n Using R ( x ) = ( − 1) P ( x ) P ( − x ) and Vieta, we obtain ( ) 2 ∏ x 2 2 n P ( x ) P ( − x ) Q ( x ) = P (0) P . r i i Plugging in x = α , we use the facts that P ( α ) = 4, P ( − α ) = 4 − 2 α , and also ( ) 2 4040 2 2 2 2 α α α ( α − 2) α 2( − α − r ) i P = + + 2 = − + + 2 = . 2020 r r r r + 2 r r ( r + 2) i i i i i i i This will give us 2 n 2 ∏ 2( − α − r ) 2 P ( − α ) i 2 2 n n P ( α ) P ( − α ) Q ( α ) = 2 = 2 · . r ( r + 2) P (0) P ( − 2) i i i Therefore, n 4 P ( − α ) 2 2 Q ( α ) = P (0) P ( − 2) P ( α ) n 4 (4 − 2 α ) = 2020 2 · 2 · 4 4041 2 (2 − α ) = 2023 2 2018 = 2 (2 − α ) . We can check that P ( x ) − 4 has no double roots (e.g. by checking that it shares no roots with its 2019 derivative), which means that all possible α are distinct. Therefore, adding over all α gives 2020 · 2 , because the sum of the roots of P ( x ) − 4 is 0.