HMMT 二月 2024 · ALGNT 赛 · 第 5 题

HMMT February 2024 — ALGNT Round — Problem 5

Compute the unique ordered pair ( x, y ) of real numbers satisfying the system of equations x 1 y 1 √ − = 7 and √ + = 4 . 2 2 2 2 x y x + y x + y n !

解析

Compute the unique ordered pair ( x, y ) of real numbers satisfying the system of equations x 1 y 1 √ − = 7 and √ + = 4 . 2 2 2 2 x y x + y x + y Proposed by: Pitchayut Saengrungkongka ( ) 13 13 Answer: − , 96 40 Solution 1: Consider vectors √ ( ) ( ) 2 2 x/ x + y − 1 /x √ and . 2 2 1 /y y/ x + y ( ) √ √ 7 2 2 They are orthogonal and add up to , which have length 7 + 4 = 65. The first vector has 4 √ length 1, so by Pythagorean’s theorem, the second vector has length 65 − 1 = 8, so we have √ 1 1 2 2

= 64 = ⇒ x + y = ± 8 xy. 2 2 x y However, the first equation indicates that x < 0, while the second equation indicates that y > 0, so √ 2 2 xy < 0. Thus, x + y = − 8 xy . Plugging this into both of the starting equations give 1 1 1 1 − − = 7 and − + = 4 . 8 y x 8 x y ( ) 13 13 Solving this gives ( x, y ) = − , , which works. 96 40 Solution 2: Let x = r cos θ and y = r sin θ . Then our equations read 1 cos θ − = 7 r cos θ 1 sin θ + = 4 . r sin θ Multiplying the first equation by cos θ and the second by sin θ , and then adding the two gives 7 cos θ + 4 sin θ = 1. This means 2 2 2 4 sin θ = 1 − 7 cos θ = ⇒ 16 sin θ = 1 − 14 cos θ + 49 cos θ = ⇒ 65 cos θ − 14 cos θ − 15 = 0 . 3 5 3 This factors as (13 cos θ + 5)(5 cos θ − 3) = 0, so cos θ is either or − . This means either cos θ = 5 13 5 4 5 12 and sin θ = − , or cos θ = − and sin θ = . 5 13 13 The first case, plugging back in, makes r a negative number, a contradiction, so we take the second ( ) 1 13 1 13 13 13 case. Then x = = − and y = = . The answer is ( x, y ) = − , . cos θ − 7 96 4 − sin θ 40 96 40 n !