HMMT 二月 2022 · 冲刺赛 · 第 36 题
HMMT February 2022 — Guts Round — Problem 36
题目详情
- [20] For a cubic polynomial P ( x ) with complex roots z , z , z , let 1 2 3 max( | z − z | , | z − z | , | z − z | ) 1 2 1 3 2 3 M ( P ) = . min( | z − z | , | z − z | , | z − z | ) 1 2 1 3 2 3 3 2 Over all polynomials P ( x ) = x + ax + bx + c , where a, b, c are nonnegative integers at most 100 and P ( x ) has no repeated roots, the twentieth largest possible value of M ( P ) is m . Estimate A = ⌊ m ⌋ . An 1 / 2 estimate of E earns max(0 , ⌊ 20 − 20 | 3 ln( A/E ) | ⌋ ) points.
解析
- [20] For a cubic polynomial P ( x ) with complex roots z , z , z , let 1 2 3 max( | z − z | , | z − z | , | z − z | ) 1 2 1 3 2 3 M ( P ) = . min( | z − z | , | z − z | , | z − z | ) 1 2 1 3 2 3 3 2 Over all polynomials P ( x ) = x + ax + bx + c , where a, b, c are nonnegative integers at most 100 and P ( x ) has no repeated roots, the twentieth largest possible value of M ( P ) is m . Estimate A = ⌊ m ⌋ . 1 / 2 An estimate of E earns max(0 , ⌊ 20 − 20 | 3 ln( A/E ) | ⌋ ) points. Proposed by: Daniel Zhu Answer: 8097 ′ 2 Solution: Consider fixing a and b . Then, we know that P ( x ) = 3 x + 2 ax + b , which has a root at 2 approximately r ≈ − b/ 2 a , which is rather small compared to 100. Then P ( r ) ≈ − b / 4 a . Assuming that this is greater than about − 100, then the value of c that produces the roots that are closest together is the closest integer to − P ( r ) (the chance when this creates a double root is pretty rare. Let − P ( r ) = c + s , so that we can now assume that s is uniformly distributed in ( − 1 / 2 , 1 / 2). One can p show that the difference between these roots is about 2 | s | /a . Since these roots are rather small, by p 1 3 Vieta’s formulas the other root is near − a , so M ( P ) is about a / | s | . 2 It’s clear from this discussion that a needs to be reasonably large for M ( P ) to be large. Thus the condition P ( r ) > − 100 is satisfied close to all the time—we will henceforth ignore it. Set some L and let’s consider the expected number of P so that M ( P ) > L . Then, for a given a, b , 3 2 4 2 we need | s | < a / (2 L ) . Summing over all a and b , we find the probability is 2 · 100 · 100 / 4 · 1 / (2 L ) . Setting this equal to 20 gives us 10 10 2 L = = ⇒ L ≈ 7900 . 160 This is good enough for 14 points.