PUMaC 2021 · 代数（A 组） · 第 4 题

PUMaC 2021 — Algebra (Division A) — Problem 4

The roots of a monic cubic polynomial p are positive real numbers forming a geometric se- quence. Suppose that the sum of the roots is equal to 10. Under these conditions, the largest m possible value of | p ( − 1) | can be written as , where m, n are relatively prime integers. Find n m + n .

解析

The roots of a monic cubic polynomial p are positive real numbers forming a geometric se- quence. Suppose that the sum of the roots is equal to 10. Under these conditions, the largest m possible value of | p ( − 1) | can be written as , where m, n are relatively prime integers. Find n m + n . Proposed by: Frank Lu Answer: 2224 Because the cubic has roots that form a geometric sequence, we may write the cubic in the d form p ( x ) = ( x − d )( x − dr )( x − ) , where d, r are positive numbers. We are given that r 1 d d + dr + = 10 , and we want to find the maximum of | p ( − 1) | . In other words, we want to find r d 1 2 3 the maximum value of 1 + d + dr + + d (1 + r + ) + d . r r d 3 3 But given that d + dr + = 10 , notice that we can write this as d +10 d +10+1 = d +10 d +11 . r 3 2 Notice in fact that we can write d + 10 d = ( d + 10) d, both of which are increasing with d. Therefore, it suffices for us to find the largest possible value of d, as this will then give us the largest possible value of | p ( − 1) | . d 1 However, given that d + dr + = 10 , we see that d is maximized when 1 + r + is minimized. r r 1 10 But r + ≥ 2; thus, we have that d = maximizes p ( − 1) . Substituting this value in, we get r 3 1000 100 1000+900+297 2197 our maximal value for | p ( − 1) | as + + 11 = = . Our sum for m + n 27 3 27 27 is thus 2224 . 2021 P 2 πj cπ