2023 AMC 10B Problem 13

The absolute values look scary, so peel them off one layer at a time, starting where the signs are simplest.

1. Use symmetry. The inequality only uses $|x|$ and $|y|$. Changing $x$ to $-x$ or $y$ to $-y$ gives the same thing, so the region looks the same in all four quadrants. Find its area in the top-right quadrant ($x \ge 0$, $y \ge 0$) and multiply by $4$.
2. Simplify in that quadrant. There $|x| = x$ and $|y| = y$, so the inequality is
$$|x - 1| + |y - 1| \le 1.$$
This says: (horizontal distance from $x$ to $1$) plus (vertical distance from $y$ to $1$) is at most $1$.
3. Find the shape. Check a few points around $(1, 1)$:
- $(1, 1)$ itself: $0 + 0 = 0$, inside.
- $(0, 1)$, $(2, 1)$, $(1, 0)$, $(1, 2)$: each gives exactly $1$, on the edge.
- $(0, 0)$: $1 + 1 = 2$, outside.

The region is a tilted square (a diamond) with corners $(0,1)$, $(2,1)$, $(1,0)$, $(1,2)$. It fits inside the top-right quadrant.
4. Area of the diamond. Its two diagonals go from $(0,1)$ to $(2,1)$ and from $(1,0)$ to $(1,2)$, each of length $2$. The area of a diamond is half the product of its diagonals:
$$\frac{1}{2} \cdot 2 \cdot 2 = 2.$$
5. All four quadrants. $4 \times 2 = 8$.

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

$|X| + |Y| \le 1$ is always a diamond around the point where $X$ and $Y$ are zero. The inner $|x|$ and $|y|$ just copy that diamond into every quadrant. When a formula only uses $|x|$ and $|y|$, work in one quadrant and multiply.