South Point RMO Screening Paper

Let $f(x,y) = |x+y| + |x-y|$ . We note that $y(-x,y) = |-x+y| + |-x,-y| = |x+y| + |x-y| = f(x,y)$ , therefore, $f(x,y)$ is an even function. We need only consider $f(x,y)$ for $x \ge 0$ and $x \le 0$ will have the same result.

Consider $x > y > 0$ , then $f(x) = x+y + x - y = 2x$ and for $f(x,y) = 11$ $\implies x = 5.5, y < 5.5$ .

Now for $y \ge x > 0$ , then $f(x) = x + y + y - x = 2y$ and for $f(x,y) = 11$ $\implies y = 5.5, x < 5.5$ .

Therefore, the graph of $|x+y| + |x-y| = 11$ is a square with sides $(y = \pm 5.5, -5.5 \le x \le 5.5)$ and $(x = \pm 5.5, -5.5 \le y \le 5.5)$ . Therefore the area enclosed by the graph is $11 \times 11 = \boxed{121}$ .

@Chew-Seong Cheong please give a solution.