AMC10 2021 A

In order to attack this problem, we can use casework on the sign of $|x-y|$ and $|x+y|$. Case 1: $|x-y|=x-y, |x+y|=x+y$ Substituting and simplifying, we have $x^2-6x+y^2=0$, i.e. $(x-3)^2+y^2=3^2$, which gives us a circle of radius $3$ centered at $(3,0)$. Case 2: $|x-y|=y-x, |x+y|=x+y$ Substituting and simplifying again, we have $x^2+y^2-6y=0$, i.e. $x^2+(y-3)^2=3^2$. This gives us a circle of radius $3$ centered at $(0,3)$. Case 3: $|x-y|=x-y, |x+y|=-x-y$ Doing the same process as before, we have $x^2+y^2+6y=0$, i.e. $x^2+(y+3)^2=3^2$. This gives us a circle of radius $3$ centered at $(0,-3)$. Case 4: $|x-y|=y-x, |x+y|=-x-y$ One last time: we have $x^2+y^2+6x=0$, i.e. $(x+3)^2+y^2=3^2$. This gives us a circle of radius $3$ centered at $(-3,0)$. After combining all the cases and drawing them on the Cartesian Plane, this is what the diagram looks like: Now, the area of the shaded region is just a square with side length $6$ with four semicircles of radius $3$. The area is $6\cdot6+4\cdot \frac{9\pi}{2} = 36+18\pi$. The answer is $36+18$ which is $\boxed{\textbf{(E) }54}$