PUMaC 2022 · 团队赛 · 第 7 题

PUMaC 2022 — Team Round — Problem 7

Pick x, y, z to be real numbers satisfying ( − x + y + z ) − = 4( y − z ) , ( x − y + z ) − = 4( z − x ) , 3 4 p 2 1 2 and ( x + y − z ) − = 4( x − y ) . If the value of xy + yz + zx can be written as for relatively 5 q prime positive integers p, q , find p + q . 1

解析

Pick x, y, z to be real numbers satisfying ( − x + y + z ) − = 4( y − z ) , ( x − y + z ) − = 4( z − x ) , 3 4 p 2 1 2 and ( x + y − z ) − = 4( x − y ) . If the value of xy + yz + zx can be written as for relatively 5 q prime positive integers p, q , find p + q . Proposed by Sunay Joshi Answer: 1727 1 1 1 For convenience, let A = , B = , and C = . Isolating the constant on the right-hand side of 3 4 5 2 2 the first equation, we find ( − x + y + z ) − 4( y − z ) = A . By difference of squares, this becomes ( − x + 3 y − z )( − x − y + 3 z ) = A . Consider the substitution M = 3 x − y − z , N = − x + 3 y − z , P = − x − y + 3 z . Then our system reduces to N P = A , M P = B , M N = C . Multiplying the √ three together and taking the square root, we find M N P = s ABC , where s ∈ {± 1 } . Hence √ √ √ 1 1 1 M = s ABC , N = s ABC , P = s ABC . By our definition of M, N, P , we also have A B C √ √ 2 M + N + P s ABC 2 1 1 s ABC M + N + P = x + y + z , hence x = = ( + + ) = 15 and similarly 4 4 A B C 4 √ √ s ABC s ABC 2 y = 16 and z = 17 . Since s = 1, it follows that the desired quantity equals 4 4 2 2 s ABC 3 · 16 − 1 767 xy + yz + zx = (15 · 16 + 16 · 17 + 17 · 15) = = 16 16 · 60 960 Hence our answer is 767 + 960 = 1727.