2023 AMC 12B Problems/Problem 9

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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

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

Next, we add one more absolute value and get $|x-1|+||y|-1| \le 1.$ This will double the square reflecting over x-axis.

So now we got 2 squares.

Finally, we add one more absolute value and get $||x|-1|+||y|-1| \le 1.$ This will double the squares reflecting over y-axis.

In the end, we got 4 squares. The total area is $4\cdot2 =$ $\boxed{\text{(B)} 8}$

~Technodoggo ~Minor formatting change: e_is_2.71828

Video Solution 1 by OmegaLearn

https://youtu.be/300Ek9E-RrA


See Also

2023 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 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 (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 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

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