2023 AMC 10B Problem 13

The inequality involves $x$ and $y$ only through $|x|$ and $|y|$, so the region is symmetric across both axes. It suffices to find the area in the quadrant $x \ge 0$, $y \ge 0$ and multiply by $4$.

There $|x| = x$ and $|y| = y$, so the condition becomes
$$|x - 1| + |y - 1| \le 1.$$
This is a diamond (a square rotated $45^\circ$) centered at $(1,1)$ with vertices $(0,1)$, $(2,1)$, $(1,0)$, $(1,2)$. Its diagonals both have length $2$, so its area is $\frac{1}{2}\cdot 2 \cdot 2 = 2$. The diamond lies entirely in the closed first quadrant, so nothing is lost or double counted.

Four such diamonds, one per quadrant, give total area $4 \cdot 2 = 8$.

The answer is $\boxed{\textbf{(B)}\ 8}$.

Nested absolute values are handled from the outside in: the outer structure $|X| + |Y| \le 1$ is a diamond in the variables $X = |x| - 1$, $Y = |y| - 1$, and the inner $|x|$, $|y|$ reflect that diamond into all four quadrants. Symmetry reduces the work to one quadrant, where a familiar shape appears.

• Reporting the area of one quadrant's diamond (2) without multiplying by 4 for the four quadrants.
• Treating $|x - 1| + |y - 1| \le 1$ as an axis-aligned $2 \times 2$ square of area $4$ instead of a diamond of area $2$.