HMMT 十一月 2022 · 冲刺赛 · 第 15 题

HMMT November 2022 — Guts Round — Problem 15

[9] Vijay chooses three distinct integers a, b, c from the set { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 } . If k is the min- imum value taken on by the polynomial a ( x − b )( x − c ) over all real numbers x , and l is the minimum value taken on by the polynomial a ( x − b )( x + c ) over all real numbers x , compute the maximum possible value of k − l. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2022, November 12, 2022 — GUTS ROUND Organization Team Team ID#

解析

[9] Vijay chooses three distinct integers a, b, c from the set { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 } . If k is the minimum value taken on by the polynomial a ( x − b )( x − c ) over all real numbers x , and l is the minimum value taken on by the polynomial a ( x − b )( x + c ) over all real numbers x , compute the maximum possible value of k − l. Proposed by: Preston Bushnell Answer: 990 Solution: Quadratics are minimized at the average of their roots, so 2 b + c b + c b − c c − b a ( b − c ) k = a − b − c = a = − , and 2 2 2 2 4 2 b − c b − c − b − c b + c a ( b + c ) l = a − b + c = a = − . 2 2 2 2 4 Therefore, a 2 2 k − l = − ( b − c ) − ( b + c ) = abc 4 Thus, k − l = abc is maximized when a , b , and c are 9, 10, and 11 are some order, so the answer is 9 · 10 · 11 = 990.