Area in coordinate Plane: Inequalities

Question

The area of the region in the coordinate plane satisfying the inequality

$$n \le |x + y| + | x- y| \le n + 3$$

is $2019$ for some integer $n$. What is the value of $n$?

Answer 1

If $-y<x< y$ and $y>0$ then $|x+y| + |x-y| = 2y$

There are 3 other cases, but there is a symmetry to the region.

Your region looks like:

$\frac {n}{2}\le |y| \le \frac {n+3}{2}$ and $|x| \le |y|$ or $\frac {n}{2}\le |x| \le \frac {n+3}{2}$ and $|y|\le |x|$

And the area is $(n+3)^2 - n^2 = 6n+ 9$

Answer 2

Map the given $(x,y)$-region $B$ to the $(u,v)$-plane via $$T:\quad (x,y)\mapsto\left\{\eqalign{u&=x-y\cr v&=x+y\cr}\right.\quad.$$ The image domain $B'$ is given by $$B':=\bigl\{(u,v)\bigm| n\leq|u|+|v|\leq n+3\bigr\}$$ and has ${\rm area}(B')=4\cdot\bigl(3n+{9\over2}\bigr)$. On the other hand we know that ${\rm area}(B')=|{\rm det}(T)|\>{\rm area}(B)=4038$. It follows that $n=335$.