2021 AMC 10A Problem 19

The equation is unchanged by $x \to -x$, by $y \to -y$, and by swapping $x$ and $y$, so it is enough to understand the sector where $x \ge |y|$ (between the lines $y = x$ and $y = -x$ on the right) and copy the picture four times.

In that sector $x - y \ge 0$ and $x + y \ge 0$, so the equation becomes
$$x^2 + y^2 = 3(x - y) + 3(x + y) = 6x \quad\Longleftrightarrow\quad (x-3)^2 + y^2 = 9,$$
a circle of radius $3$ centered at $(3, 0)$. It passes through the origin and meets the sector's boundary lines at $(3, 3)$ and $(3, -3)$; the part of the circle inside the sector is the right semicircle from $(3,3)$ through $(6,0)$ to $(3,-3)$.

The other three sectors contribute, by symmetry, the semicircles bulging left, up, and down from the points $(\pm 3, \pm 3)$. The closed curve therefore consists of four semicircular arcs of radius $3$ attached to the sides of the square with vertices $(\pm3, \pm3)$.

Area $=$ square $+$ four half-disks:
$$6^2 + 4 \cdot \tfrac12 \pi \cdot 3^2 = 36 + 18\pi.$$
So $m + n = 36 + 18 = 54$.

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

Absolute values split the plane into sectors where the equation is a plain conic; symmetry means only one sector needs work. Recognizing $x^2 + y^2 = 6x$ as a circle through the origin, and checking exactly which arc lies in the sector, turns an intimidating graph into a square with semicircular "caps," a shape whose area is immediate.

• Taking the whole disk of radius 3 in each quadrant (area 4 * 9pi) or forgetting the central square, giving 18pi alone.
• Splitting by the signs of $x$ and $y$ (quadrants) instead of the signs of $x - y$ and $x + y$; the sector lines are $y = \pm x$.