PUMaC 2022 · 代数（A 组） · 第 5 题
PUMaC 2022 — Algebra (Division A) — Problem 5
题目详情
- Suppose that x, y, z are nonnegative real numbers satisfying the equation p p p √ 1 xyz − (1 − x )(1 − y ) z − (1 − x ) y (1 − z ) − x (1 − y )(1 − z ) = − . 2 √ √ a + b The largest possible value of xy equals , where a , b , and c are positive integers such c 2 2 2 that b is not divisible by the square of any prime. Find a + b + c . 2 2 2 2
解析
- Suppose that x, y, z are nonnegative real numbers satisfying the equation p p p √ 1 xyz − (1 − x )(1 − y ) z − (1 − x ) y (1 − z ) − x (1 − y )(1 − z ) = − . 2 √ √ a + b The largest possible value of xy equals , where a , b , and c are positive integers such c 2 2 2 that b is not divisible by the square of any prime. Find a + b + c . Proposed by Frank Lu Answer: 29 We first observe that x, y, z are required to be real numbers between 0 and 1 . With this in 2 2 2 mind, this suggests the parametrization by x = cos α , y = cos α , and z = cos α , where 1 2 3 π the values of cos α , cos α , cos α lie between 0 and . 1 2 3 2 This means that, substituting in the values, we get the equation cos α cos α cos α − sin α sin α cos α − 1 2 3 1 2 3 sin α cos α sin α − cos α sin α sin α . But we can apply the sum of angles formula to yield 1 2 3 1 2 3 that this is equal to cos( α + α ) cos α − sin( α + α ) sin α = cos( α + α + α ) . It follows 1 2 3 1 2 3 1 2 3 2 π that α + α + α is equal to . 1 2 3 3 √ 1 However, notice that xy = cos α cos α = (cos( α + α ) + cos( α − α )) . From here, notice 1 2 1 2 1 2 2 that given α , we can maximize this value by making α = α . It then suffices to find the α 3 1 2 3 1 such that (cos( α + α ) + 1) is maximized. But to do this, we need to minimize α + α . 1 2 1 2 2 π π We recall, on the other hand, that α ≤ , meaning that we need to have α + α ≥ . 3 1 3 2 6 √ 2+ 3 Using this value gives us our maximum value as . The answer that we seek is then 4 2 2 2 2 + 3 + 4 = 4 + 9 + 16 = 29 . 2 2 2 2