PUMaC 2022 · 代数（B 组） · 第 5 题

PUMaC 2022 — Algebra (Division B) — Problem 5

Find the number of real solutions ( x, y ) to the system of equations: ( 2 sin( x − y ) = 0 | x | + | y | = 2 π 2

解析

Find the number of real solutions ( x, y ) to the system of equations: ( 2 sin( x − y ) = 0 | x | + | y | = 2 π Proposed by Ben Zenker Answer: 52 2 2 Note that sin( x − y ) = 0 iff x − y = kπ for some k ∈ Z . Therefore we seek the number of 2 intersections of the parabola y = x − kπ with the square | x | + | y | = 2 π for each k . Since the vertex of the parabola has y -coordinate − πk , it is clear that there are 0 intersections for k ≤ − 3 and 1 intersection for k = − 2. If the vertex of the parabola lies strictly within the square, it is clear that there must be exactly be 2 intersections. This occurs for − 1 ≤ k ≤ 1. When k = 2, the vertex of the parabola is the vertex (0 , − 2 π ) of the square, and one can check that there are 5 intersections, including the vertex. √ For k ≥ 13, there are no intersections, since the x -intercept of the parabola equals x = πk > 2 π . For 3 ≤ k ≤ 12, it is easy to see that there are 4 intersections. Summing, we find a total of 1 + 2 · 3 + 5 + 10 · 4 = 52 intersections, our answer. 2