HMMT 二月 2026 · ALGNT 赛 · 第 7 题
HMMT February 2026 — ALGNT Round — Problem 7
题目详情
- Positive real numbers x , y , and z satisfy the following equations: xyz = 3 , ( x − y )( y − z )( z − x ) = 4 , ( x + y )( y + z )( z + x ) = 40 . Compute the minimum possible value for x .
解析
- Positive real numbers x , y , and z satisfy the following equations: xyz = 3 , ( x − y )( y − z )( z − x ) = 4 , ( x + y )( y + z )( z + x ) = 40 . Compute the minimum possible value for x . Proposed by: Derek Liu √ − 1 / 3 Answer: (3 + 6) Solution 1: Expanding the third equation gives 2 2 2 2 2 2 ( x y + y z + z x ) + ( xy + yz + zx ) + 2 xyz = 40 . Since xyz = 3 , we have 2 2 2 2 2 2 ( xy + yz + zx ) + ( x y + y z + z x ) = 34 . Expanding the second equation gives 2 2 2 2 2 2 ( x z + y x + z y ) − ( x y + y z + z x ) = 4 . ©2026 HMMT The two equations above yield 2 2 2 x z + y x + z y = 19 , 2 2 2 x y + y z + z x = 15 . Dividing both equations by xyz = 3 , we conclude that y x z 19
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- = , y z x 3 y x z
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- = 5 . z x y y z x Note that xyz = 3 , so x , y , and z are nonzero, and let a = , b = , and c = . Then, a + b + c = 5 x y z and abc = 1 . Furthermore, 1 1 1 19 ab + bc + ca = + + = . c a b 3 Therefore, a , b , and c are the roots of 19 3 2 t − 5 t + t − 1 = 0 . 3 This cubic factors as ( ) 2 1 ( t − 3) t − 2 t + = 0 , 3 √ √ y z with roots 3 , 1 + 2 / 3 , and 1 − 2 / 3 . Since xyz = 3 is fixed, for x to be minimized, both and x x √ √ should be maximized. Their maximal possible values are 3 and 1 / (1 − 2 / 3) = 3 + 6 , respectively, so the minimum possible value for x satisfies ( ) ( ) − 1 / 3 √ √ 3 3 3 + 6 x = 3 = ⇒ x = 3 + 6 . x Solution 2: Note that x , y , and z are nonzero because xyz = 3 , and consider the cubic with roots , y y z , and . Let z x 3 2 y x z f ( t ) = t + ut + vt + w = ( t − )( t − )( t − ) . y z x y x z By Vieta’s formula, − w is the product of the roots of f , which is ( )( )( ) = 1 . Moreover, notice that y z x y x z f (1) = (1 − )(1 − )(1 − ) y z x = ( y − x )( z − y )( x − z ) / ( xyz ) 4 = − , 3 and y x z f ( − 1) = ( − 1 − )( − 1 − )( − 1 − ) y z x = ( − x − y )( − y − z )( − z − x ) / ( xyz ) 40 = − . 3 Expressing these in terms of the coefficients of f yields the system − w = 1 , 4 1 + u + v + w = − , 3 40 − 1 + u − v + w = − . 3 ©2026 HMMT Solving this system of equations gives 19 3 2 f ( t ) = t − t + 5 t − 1 , 3 whence we can finish similar to the first solution.