HMMT 十一月 2021 · 团队赛 · 第 7 题
HMMT November 2021 — Team Round — Problem 7
题目详情
- [45] Let f ( x ) = x + 3 x − 1 have roots a, b, c . Given that 1 1 1
3 3 3 3 3 3 a + b b + c c + a m can be written as , where m, n are positive integers and gcd( m, n ) = 1, find 100 m + n . n
解析
- [45] Let f ( x ) = x + 3 x − 1 have roots a, b, c . Given that 1 1 1
3 3 3 3 3 3 a + b b + c c + a m can be written as , where m, n are positive integers and gcd( m, n ) = 1, find 100 m + n . n Proposed by: Milan Haiman Answer: 3989 3 Solution: We know that a = − 3 a + 1 and similarly for b, c , so 1 1 1 1 = = = . 3 3 a + b 2 − 3 a − 3 b 2 + 3 c 3(2 / 3 + c ) Now, 13 89 3 2 f ( x − 2 / 3) = x − 2 x + x − 3 27 has roots a + 2 / 3, b + 2 / 3, and c + 2 / 3. Thus the answer is, by Vieta’s formulas, 1 ( a + 2 / 3)( b + 2 / 3) + ( a + 2 / 3)( c + 2 / 3) + ( b + 2 / 3)( c + 2 / 3) 13 / 3 39 = = . 3 ( a + 2 / 3)( b + 2 / 3)( c + 2 / 3) 3 · 89 / 27 89