The area of the region bounded by the graph of\[x^2+y^2 = 3|x-y| + 3|x+y|\]is $m+n\pi$, where $m$ and $n$ are integers. What is $m + n$?
$\textbf{(A)} ~18\qquad\textbf{(B)} ~27\qquad\textbf{(C)} ~36\qquad\textbf{(D)} ~45\qquad\textbf{(E)} ~54$
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:
[asy] size(10cm);  Label f;  f.p=fontsize(7); xaxis(-8,8,Ticks(f, 1.0));  yaxis(-8,8,Ticks(f, 1.0));  draw(arc((-3,0),3,90,270) -- cycle, gray); draw(arc((0,3),3,0,180) -- cycle, gray); draw(arc((3,0),3,-90,90) -- cycle, gray); draw(arc((0,-3),3,-180,0) -- cycle, gray); draw((-3,3)--(3,3)--(3,-3)--(-3,-3)--cycle, grey); [/asy]
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 $\boxed{\textbf{(E) }54}$
~Bryguy