2016 AMC 10B Problem 21

Step 1: use symmetry. Changing $x$ to $-x$ does not change $x^2$ or $|x|$, and the same goes for $y$. So the shape looks the same in all four quadrants. Find the area in the first quadrant and multiply by $4$.

Step 2: the first-quadrant curve. When $x$ and $y$ are not negative, $|x| = x$ and $|y| = y$, so the equation is
$$x^2 - x + y^2 - y = 0.$$
Complete the square in $x$ and in $y$ (add $\frac14$ twice to both sides):
$$\left(x - \tfrac12\right)^2 + \left(y - \tfrac12\right)^2 = \tfrac12.$$
This is a circle with center $\left(\frac12, \frac12\right)$ and radius $\sqrt{\frac12} = \frac{\sqrt2}{2}$. It passes through $(0,0)$, $(1,0)$ and $(0,1)$.

Step 3: split the first-quadrant piece.

1. The segment from $(1,0)$ to $(0,1)$ has length $\sqrt2$, which is twice the radius. So it is a diameter, and it cuts the circle into two halves.
2. The half away from the origin lies completely in the first quadrant. That is a half-circle.
3. The other half, on the origin's side, bulges below the $x$-axis and to the left of the $y$-axis, which is outside the first quadrant. The part of it that stays inside the quadrant is exactly the triangle with corners $(0,0)$, $(1,0)$, $(0,1)$.
4. First-quadrant area:
$$\underbrace{\tfrac12 \cdot 1 \cdot 1}_{\text{triangle}} + \underbrace{\tfrac12 \cdot \pi \left(\tfrac{\sqrt2}{2}\right)^2}_{\text{half-circle}} = \tfrac12 + \tfrac{\pi}{4}.$$

Step 4: multiply by 4. Total area $= 4\left(\frac12 + \frac{\pi}{4}\right) = 2 + \pi$.

The answer is $\boxed{\textbf{(B)}\ \pi + 2}$.

Absolute values usually mean "split the plane into four quadrants," and symmetry lets you do just one of them. Completing the square turns the equation in that quadrant into a circle. Overall the shape is a tilted square with corners $(\pm1, 0)$ and $(0, \pm1)$, area $2$, with a half-circle on each of its four sides; four half-circles make two whole circles of area $\frac{\pi}{2}$ each.