# 2023 AMC 12B Problems/Problem 9

The following problem is from both the 2023 AMC 10B #13 and 2023 AMC 12B #9, so both problems redirect to this page.

## Problem

What is the area of the region in the coordinate plane defined by

$| | x | - 1 | + | | y | - 1 | \le 1$?

$\text{(A)}\ 2 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 12$

## Solution 1

First consider, $|x-1|+|y-1| \le 1.$ We can see that it is a square with a radius of $1$ (diagonal $\sqrt{2}$). The area of the square is $\sqrt{2}^2 = 2.$

Next, we insert an absolute value sign into the equation and get $|x-1|+||y|-1| \le 1.$ This will double the square reflecting over x-axis.

So now we have $2$ squares.

Finally, we add one more absolute value and obtain $||x|-1|+||y|-1| \le 1.$ This will double the squares as we reflect the $2$ squares we already have over the y-axis.

Concluding, we have $4$ congruent squares. The total area is $4\cdot2 =$ $\boxed{\text{(B)} 8}$

~Technodoggo ~Minor formatting change: e_is_2.71828 ~Grammar and clarity: NSAoPS

## Solution 2 (Graphing)

We first consider the lattice points that satisfy $||x|-1| = 0$ and $||y|-1| = 1$. The lattice points satisfying these equations are $(1,0), (1,2), (1,-2), (-1,0), (-1,2),$ and $(-1,-2).$ By symmetry, we also have points $(0,1), (2,1), (-2,1), (0,-1), (2,-1),$ and $(-2,-1)$ when $||x|-1| = 1$ and $||y|-1| = 0$. Graphing and connecting these points, we form 5 squares. However, we can see that any point within the square in the middle does not satisfy the given inequality (take $(0,0)$, for instance). As noted in the above solution, each square has a diagonal $2$ for an area of $\frac{2^2}{2} = 2$, so the total area is $4\cdot2 =$ $\boxed{\text{(B)} 8}.$

~ Brian__Liu

## Note

This problem is very similar to a past AIME problem (1997 P13)

~ CherryBerry

## Solution 3 (Logic)

The value of $|x|$ and $|y|$ can be a maximum of 1 when the other is 0. Therefore the value of $x$ and $y$ range from -2 to 2. This forms a diamond shape which has area $4 \times \frac{2^2}{2}$ which is $\boxed{\text{(B)} 8}.$

~ darrenn.cp ~ DarkPheonix

## Solution 4

We start by considering the graph of $|x|+|y|\leq 1$. To get from this graph to $||x|-1|+||y|-1| \leq 1$ we have to translate it by $\pm 1$ on the $x$ axis and $\pm 1$ on the $y$ axis.

Graphing $|x|+|y|\leq 1$ we get a square with side length of $\sqrt{2}$, so the area of one of these squares is just $2$.

We have to multiply by $4$ since there are $4$ combinations of shifting the $x$ and $y$ axis.

So we have $2\times 4$ which is $\boxed{\text{(B)} 8}$.

~ESAOPS

## Video Solution

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

## See Also

 2023 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 12 Followed byProblem 14 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
 2023 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 8 Followed byProblem 10 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.