HMMT 二月 2020 · 冲刺赛 · 第 16 题

HMMT February 2020 — Guts Round — Problem 16

[8] Determine all triplets of real numbers ( x, y, z ) satisfying the system of equations 2 2 x y + y z = 1040 2 2 x z + z y = 260 ( x − y )( y − z )( z − x ) = − 540 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT February 2020, February 15, 2020 — GUTS ROUND Organization Team Team ID#

解析

[8] Determine all triplets of real numbers ( x, y, z ) satisfying the system of equations 2 2 x y + y z = 1040 2 2 x z + z y = 260 ( x − y )( y − z )( z − x ) = − 540 . Proposed by: Krit Boonsiriseth Answer: (16 , 4 , 1), (1 , 16 , 4) Solution: Call the three equations (1) , (2) , (3). (1) / (2) gives y = 4 z . (3) + (1) − (2) gives 2 2 2 ( y − z ) x = 15 z x = 240 , 2 so z x = 16. Therefore 81 2 2 2 2 z ( x + 2 z ) = x z + z y + 4 z x = 5 49 2 2 2 2 z ( x − 2 z ) = x z + z y − 4 z x = 5 ∣ ∣ ∣ ∣ x +2 z 9 z so = . Thus either x = 16 z or x = . ∣ ∣ x − 2 z 7 4 3 3 If x = 16 z , then (1) becomes 1024 z + 16 z = 1040 , so ( x, y, z ) = (16 , 4 , 1). z 1 3 3 If x = , then (1) becomes z + 16 z = 1040, so ( x, y, z ) = (1 , 16 , 4). 4 4