HMMT 十一月 2023 · GEN 赛 · 第 7 题
HMMT November 2023 — GEN Round — Problem 7
题目详情
- Compute all ordered triples ( x, y, z ) of real numbers satisfying the following system of equations: xy + z = 40 xz + y = 51 x + y + z = 19 . √
解析
- Compute all ordered triples ( x, y, z ) of real numbers satisfying the following system of equations: xy + z = 40 xz + y = 51 x + y + z = 19 . Proposed by: Ethan Liu, Pitchayut Saengrungkongka Answer: (12 , 3 , 4) , (6 , 5 . 4 , 7 . 6) Solution 1: By adding the first two equations, we can get xy + z + xz + y = ( x + 1)( y + z ) = 91 . From the third equation we have ( x + 1) + ( y + z ) = 19 + 1 = 20 , 2 so x + 1 and y + z are the two roots of t − 20 t + 91 = 0 by Vieta’s theorem. As the quadratic equation can be decomposed into ( t − 7)( t − 13) = 0 , we know that either x = 6 , y + z = 13 or x = 12 , y + z = 7 . • If x = 12 , by the first equation we have 12 y + z = 40 , and substituting y + z = 7 we have 11 y = 33 , y = 3 and z = 4 . • If x = 6 , by the first equation we have 6 y + z = 40 , and substituting y + z = 13 we have 5 y = 27 , y = 5 . 4 and z = 7 . 6 . Hence, the two solutions are (12 , 3 , 4) and (6 , 5 . 4 , 7 . 6) . Solution 2: Viewing x as a constant, the equations become three linear equations in two variables y and z . This system can only have a solution if x 1 40 1 x 51 det = 0 . 1 1 19 − x Expanding out the determinant, we have 2 x (19 − x ) + 51 + 40 − 51 x − 40 x − (19 − x ) = 0 3 2 = ⇒ x − 19 x + 90 x − 72 = 0 2 = ⇒ ( x − 1)( x − 18 x + 72) = 0 = ⇒ ( x − 1)( x − 6)( x − 12) = 0 so x = 1 , 6 , or 12 . If x = 1 , the system has no solutions, and if x = 6 or 12 , we can find y and z as in the first solution. √