I'm having trouble following the solution (which is attached below) provided to this problem:
What is the area of the polygon formed by all points $(x,y)$ in the plane satisfying the inequality $||x|-2|+||y|-2| \le 4$?
Can I have an explanation of the solution (you can use a different approach to this problem if needed), especially can someone help me visualise the reflection "in the axes and the origin":
- Why does this work?
How can I picture this quickly and efficiently? (keep in mind that this problem is from a Maths contest)
Note that $$||x|-2|+||y|-2|=||-x|-2|+||y|-2|$$ $$=||x|-2|+||-y|-2|=||-x|-2|+||-y|-2|,$$ so we can solve the equation $$|x-2|+|y-2|\leq4,$$ for the first quadrant $$x\geq0, y\geq0,$$ and symmetically expand it. Then we can consider the next cases.
A. $0\leq x\leq2,\quad 0\leq y\leq2$,
$$4-x-y\leq4,\quad x+y\geq 0.$$ The inner square with corners $(0,0), (2,0), (2,2), (0,2).$
B. $2\leq x,\quad 0\leq y\leq2$,
$$x-2+2-y\leq4,\quad y\geq x-4.$$ The trapeze with corners $(2,0), (4,0), (6,2), (2,2).$
C. $0\leq x\leq2,2\leq y$,
$$2-x+y-2\leq4,\quad y\leq x+4.$$ The trapeze with corners $(0,2), (2,2), (2,6), (0,4).$
D. $2\leq x,2\leq y$,
$$x-2+y-2\leq4,\quad x+y\leq8.$$ The triangle with corners $(2,2), (6,2), (2,6).$
The boundaries of each area identified, the corner points are easily reflected on the other quadrants too easy. The final appearance of the graph gives Wolfram Alpha.