2021 AMC 10A Problem 19

Step 1: use the symmetry. Changing $x$ to $-x$ just swaps the two absolute values $|x - y|$ and $|x + y|$, so the equation stays the same. The same happens if you change $y$ to $-y$ or swap $x$ and $y$. So the picture looks the same after flipping it over either axis or over the line $y = x$. It is enough to understand one piece and copy it.

Step 2: remove the absolute values in one piece.

1. The lines $y = x$ and $y = -x$ cut the plane into four wedges: right, top, left and bottom. Work in the right wedge, where $x$ is at least as big as both $y$ and $-y$.
2. There $x - y \ge 0$ and $x + y \ge 0$, so the absolute value signs can simply be dropped:
$$x^2 + y^2 = 3(x - y) + 3(x + y) = 6x.$$
3. Move $6x$ to the left and complete the square by adding $9$ to both sides:
$$x^2 - 6x + 9 + y^2 = 9, \quad\text{so}\quad (x - 3)^2 + y^2 = 9.$$
This is a circle with center $(3, 0)$ and radius $3$.
4. Where does it cross the edges of the wedge? On the line $y = x$: $x^2 + x^2 = 6x$ gives $x = 3$, the point $(3, 3)$. On $y = -x$ it gives $(3, -3)$.
5. The part of the circle inside the right wedge is its right half, the arc from $(3, 3)$ out through $(6, 0)$ to $(3, -3)$.

Step 3: copy and find the area.

6. The other three wedges give the same half-circle facing up, left and down. The four arcs join at $(3, 3)$, $(-3, 3)$, $(-3, -3)$ and $(3, -3)$, which are the corners of a $6 \times 6$ square.
7. So the region is that square with a half-disk of radius $3$ attached to each side:
$$6 \times 6 + 4 \times \tfrac12 \times \pi \times 3^2 = 36 + 18\pi.$$
8. So $m = 36$, $n = 18$ and $m + n = 54$.

The answer is $\boxed{\textbf{(E)}\ 54}$.

Absolute values split the plane into pieces where the equation is a plain one, and symmetry means only one piece needs real work. The key care point is step 5: only part of the circle lies in the wedge, and checking exactly which part shows that the shape is a square with four round caps, not four whole circles.