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HMMT 十一月 2014 · GEN 赛 · 第 8 题

HMMT November 2014 — GEN Round — Problem 8

专题
Discrete Math / 离散数学
难度
L3
来源
HMMT

题目详情

  1. Let a , b , c , x be reals with ( a + b )( b + c )( c + a ) 6 = 0 that satisfy 2 2 2 2 2 2 a a b b c c = + 20 , = + 14 , and = + x. a + b a + c b + c b + a c + a c + b Compute x .
解析
  1. Let a , b , c , x be reals with ( a + b )( b + c )( c + a ) 6 = 0 that satisfy 2 2 2 2 2 2 a a b b c c = + 20 , = + 14 , and = + x. a + b a + c b + c b + a c + a c + b Compute x . General Test Answer: − 34 Note that 2 2 2 2 2 2 2 2 2 2 2 2 a b c a b c a − b b − c c − a
    • − − − = + + a + b b + c c + a c + a a + b b + c a + b b + c c + a = ( a − b ) + ( b − c ) + ( c − a ) = 0 . Thus, when we sum up all the given equations, we get that 20 + 14 + x = 0. Therefore, x = − 34.