PUMaC 2023 · 代数（A 组） · 第 8 题
PUMaC 2023 — Algebra (Division A) — Problem 8
题目详情
- Given a positive integer m , define the polynomial 2 2 2 2 m 3 m − 2 2 m 4 3 2 P ( z ) = z − z + z − z + 1 . m 2 2 2 m + 1 m + 1 m + 1 Let S be the set of roots of the polynomial P ( z ) · P ( z ) · P ( z ) · P ( z ). Let w be the point in 5 7 8 18 P P a 2 the complex plane which minimizes | z − w | . The value of | z − w | equals for relatively b z ∈ S z ∈ S prime positive integers a and b . Compute a + b . Name: Team: Write answers in table below: Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 1
解析
- Given a positive integer m , define the polynomial 2 2 2 2 m 3 m − 2 2 m 4 3 2 P ( z ) = z − z + z − z + 1 . m 2 2 2 m + 1 m + 1 m + 1 Let S be the set of roots of the polynomial P ( z ) · P ( z ) · P ( z ) · P ( z ). Let w be the point in 5 7 8 18 P P a 2 the complex plane which minimizes | z − w | . The value of | z − w | equals for relatively b z ∈ S z ∈ S prime positive integers a and b . Compute a + b . Proposed by Owen Yang and Atharva Pathak Answer: 171 4 1 We claim that w = . To show this, we prove that the roots of P come in pairs ( z , z ), m 1 2 2 1 1 ( z , z ) on the unit circle such that z , z , are collinear and such that z , z , are collinear. 3 4 1 2 3 4 2 2 By the triangle inequality any minimizer w of the sum of distances must lie on the lines z z 1 2 1 and z z , so that w = . 3 4 2 1 We now prove these claims. Note that the coefficients of P are symmetric, so that P can m 2 m z 1 2 be regarded as a polynomial in z + . Applying this trick and rescaling by z , we obtain the z factorization m m − i m m + i 2 2 P ( z ) = ( z − z + )( z − z + ) (10) m m + i m + i m − i m − i The roots z , z of the first factor are given as 1 2 p √ 2 2 2 m ± m − 4( m + 1) m ± i 3 m + 4 z = = (11) 2( m + i ) 2( m + i ) so that √ p 2 1 − i ± i 3 m + 4 i 2 z − = = ( − 1 ± 3 m + 4) (12) 2 2( m + i ) 2( m + i ) √ 2 − 1+ 3 m +4 1 √ with ratio ∈ R , implying the collinearity of z , z , . Further, note that the 1 2 2 2 − 1 − 3 m +4 modulus of z , z are given as 1 2 √ 2 2 2 2 | m ± i 3 m + 4 | m + (3 m + 4) 2 | z | = = = 1 (13) 2 2 | 2( m + i ) | 4( m + 1) implying that z , z lie on the unit circle. Since the second quadratic factor is obtained by 1 2 conjugating the first, we obtain the same results for the remaining roots z , z . The above 3 4 1 claims follow, so that w = . 2 P 1 2 It remains to compute | z − | , where z runs over the roots of P , P , P , P . Let z be 5 7 8 18 2 1 2 1 1 5 1 1 a root of P ( z ). Then | z − | = ( z − )(¯ z − ) = − ( z + ), since | z | = 1. By Vieta, m 2 2 2 4 2 z P P 2 1 2 m z = = , where the sum runs over all four roots of P , and where we used the fact 2 m z m +1 2 2 m 2 that the coefficients of P are symmetric. Therefore P contributes 5 − = 3 + to 2 2 m m m +1 m +1 the desired sum. Summing over m ∈ { 5 , 7 , 8 , 18 } , we find 1 1 1 1 1 1 1 1 3 · 4 + 2( + + + ) = 12 + 2( + + + ) (14) 2 2 2 2 5 + 1 7 + 1 8 + 1 18 + 1 26 50 65 325 2 = 12 + (15) 13 158 = (16) 13 so that a + b = 158 + 13 = 171, our answer. 5