PUMaC 2022 · 代数（A 组） · 第 6 题

PUMaC 2022 — Algebra (Division A) — Problem 6

Let x, y, z be positive real numbers satisfying 4 x − 2 xy + y = 64, y − 3 yz + 3 z = 36, and √ 2 2 4 x + 3 z = 49. If the maximum possible value of 2 xy + yz − 4 zx can be expressed as n for some positive integer n , find n . √ 1 3

解析

Let x, y, z be positive real numbers satisfying 4 x − 2 xy + y = 64, y − 3 yz + 3 z = 36, and √ 2 2 4 x + 3 z = 49. If the maximum possible value of 2 xy + yz − 4 zx can be expressed as n for some positive integer n , find n . Proposed by Sunay Joshi 2 Answer: 2205 √ 2 Consider the substitution a = 2 x , b = y , c = z 3. The system of equations becomes a + √ 2 2 2 2 2 2 2 2 b − ab = 8 , b + c − bc 3 = 6 , and c + a = 7 . The desired quantity becomes ab + √ 1 2 4 1 3 1 1 1 √ √ √ bc − ca = ( ab + bc − ca ). By the Law of Cosines, the values a, b, c can be 2 2 2 2 2 3 3 3 interpreted geometrically as follows. Consider a quadrilateral ABCD with AB = a , AC = b , ◦ ◦ AD = c , ∠ BAC = 60 , and ∠ CAD = 30 . Then the given equalities imply that BC = 8, CD = 6, and BD = 7. By the sine area formula, the desired quantity can be seen to equal 4 √ ([ BAC ] + [ DAC ] − [ BAD ]). 3 We now distinguish two configurations: (1) if A, C lie on the same side of line BD , and (2) if A, C lie on opposite sides of line BD . In either case, the absolute value of the desired quantity 4 √ is [ BCD ], and configuration (2) attains the positive (hence maximum) value. Since the 3 √ 21 15 sides of △ BCD are 6, 7, 8, Heron’s formula implies that [ BCD ] = . Hence our quantity 4 √ √ √ 4 21 15 √ is · = 21 5 = 2205, and our answer is 2205. 4 3 √ 1 3