HMMT 二月 2015 · 冲刺赛 · 第 9 题

HMMT February 2015 — Guts Round — Problem 9

[ 6 ] Let a, b, c be integers. Define f ( x ) = ax + bx + c . Suppose there exist pairwise distinct integers u, v, w such that f ( u ) = 0, f ( v ) = 0, and f ( w ) = 2. Find the maximum possible value of the 2 discriminant b − 4 ac of f . 2 2015 2014

解析

[ 6 ] Let a, b, c be integers. Define f ( x ) = ax + bx + c . Suppose there exist pairwise distinct integers u, v, w such that f ( u ) = 0, f ( v ) = 0, and f ( w ) = 2. Find the maximum possible value of the 2 discriminant b − 4 ac of f . Answer: 16 By the factor theorem, f ( x ) = a ( x − u )( x − v ), so the constraints essentially boil down to 2 = f ( w ) = a ( w − u )( w − v ). (It’s not so important that u 6 = v ; we merely specified it for a shorter problem statement.) 2 2 2 2 2 2 2 We want to maximize the discriminant b − 4 ac = a [( u + v ) − 4 uv ] = a ( u − v ) = a [( w − v ) − ( w − u )] . Clearly a | 2. If a > 0, then ( w − u )( w − v ) = 2 /a > 0 means the difference | u − v | is less than 2 /a , whereas if a < 0, since at least one of | w − u | and | w − v | equals 1, the difference | u − v | of factors is greater than 2 / | a | . So the optimal choice occurs either for a = − 1 and | u − v | = 3, or a = − 2 and | u − v | = 2. The latter 2 2 wins, giving a discriminant of ( − 2) · 2 = 16. 1 proven most easily with complex numbers or the product-to-sum identity on sin( a − b ) sin( a + b ) (followed by the double angle formula for cosine) Guts 2 2015 2014