HMMT 十一月 2025 · 团队赛 · 第 9 题

HMMT November 2025 — Team Round — Problem 9

[55] Let a , b , and c be positive real numbers such that √ √ √ a + b + c = 7 , √ √ √ a + 1 + b + 1 + c + 1 = 8 , √ √ √ √ √ √ ( a + 1 + a )( b + 1 + b )( c + 1 + c ) = 60 . Compute a + b + c .

解析

[55] Let a , b , and c be positive real numbers such that √ √ √ a + b + c = 7 , √ √ √ a + 1 + b + 1 + c + 1 = 8 , √ √ √ √ √ √ ( a + 1 + a )( b + 1 + b )( c + 1 + c ) = 60 . Compute a + b + c . Proposed by: Pitchayut Saengrungkongka 199 Answer: 8 Solution: Define √ √ √ √ √ √ x = a + 1 + a, y = b + 1 + b, z = c + 1 + c √ √ 1 1 1 We are given xyz = 60. We have = a + 1 − a , and likewise for and , so x y z √ √ √ √ √ √ x + y + z = ( a + 1 + b + 1 + c + 1) + ( a + b + c ) = 8 + 7 = 15 √ √ √ √ √ √ 1 1 1

- = ( a + 1 + b + 1 + c + 1) − ( a + b + c ) = 8 − 7 = 1 . x y z Thus, 1 1 1 xy + yz + xz = xyz + + = 60 · 1 = 60 . x y z We have 2 1 1 1 1 2 a = x − = x + − 2 , 2 4 x 4 x and likewise for b and c . Thus, 1 1 1 1 2 2 2 a + b + c = x + y + z + + + − 6 . 2 2 2 4 x y z © 2025 HMMT We have 2 2 2 2 2 x + y + z = ( x + y + z ) − 2( xy + yz + xz ) = 15 − 2 · 60 = 105 , 2 1 1 1 1 1 1 x + y + z 15 1 2
- = + + − 2 · = 1 − 2 · = . 2 2 2 x y z x y z xyz 60 2 Hence, the answer is 1 1 199 a + b + c = 105 + − 6 = . 4 2 8