HMMT 二月 2006 · 冲刺赛 · 第 26 题
HMMT February 2006 — Guts Round — Problem 26
题目详情
英文原题
- [9] Let a ≥ b ≥ c be real numbers such that
2 2 2
a bc + ab c + abc + 8 = a + b + c
2 2 2 2 2 2
a b + a c + b c + b a + c a + c b + 3 abc = − 4
2 2 2 2 2 2
a b c + ab c + a bc = 2 + ab + bc + ca
If a + b + c > 0, then compute the integer nearest to a .5
解析
英文解析
- Let a ≥ b ≥ c be real numbers such that
2 2 2
a bc + ab c + abc + 8 = a + b + c
2 2 2 2 2 2
a b + a c + b c + b a + c a + c b + 3 abc = − 4
2 2 2 2 2 2
a b c + ab c + a bc = 2 + ab + bc + ca
If a + b + c > 0, then compute the integer nearest to a .5
Answer: 1279
Solution: We factor the first and third givens, obtaining the system
2 2 2
a bc + ab c + abc − a − b − c = ( abc − 1)( a + b + c ) = − 8
2 2 2 2 2 2
a b + a c + b c + b a + c a + c b + 3 abc = ( ab + bc + ca )( a + b + c ) = − 4
2 2 2 2 2 2
a b c + ab c + a bc − ab − bc − ca = ( abc − 1)( ab + bc + ca ) = 2
Writing X = a + b + c, Y = ab + bc + ca, Z = abc − 1, we have XZ = − 8 , XY = − 4 , Y Z =8 2