HMMT 二月 2014 · 冲刺赛 · 第 12 题

HMMT February 2014 — Guts Round — Problem 12

[ 6 ] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT 2014, 22 FEBRUARY 2014 — GUTS ROUND Organization Team Team ID#

解析

[ 6 ] Find a nonzero monic polynomial P ( x ) with integer coefficients and minimal degree such that √ √ 3 3 P (1 − 2 + 4) = 0. (A polynomial is called monic if its leading coefficient is 1.) √ √ √ √ √ 3 3 3 3 3 3 3 2 √ Answer: x − 3 x + 9 x − 9 Note that (1 − 2 + 4)(1 + 2) = 3, so 1 − 2 + 4 = . 3 1+ 2 √ 3 3 3 3 2 Now, if f ( x ) = x − 2, we have f ( 2) = 0, so if we let g ( x ) = f ( x − 1) = ( x − 1) − 2 = x − 3 x +3 x − 3, √ √ √ 3 3 3 3 27 27 9 3 √ then g (1+ 2) = f ( 2) = 0. Finally, we let h ( x ) = g ( ) = − + − 3 so h ( ) = g (1+ 2) = 0. 3 2 3 x x x x 1+ 2 3 x 3 2 To make this a monic polynomial, we multiply h ( x ) by − to get x − 3 x + 9 x − 9. 3