Difference between revisions of "2023 AMC 12B Problems/Problem 9"

Revision as of 19:36, 15 November 2023

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

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

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

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

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

2023 AMC 10B (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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

@@ Line 1: / Line 1: @@
+{{duplicate|[[2023 AMC 10B Problems/Problem 13|2023 AMC 10B #13]] and [[2023 AMC 12B Problems/Problem 9|2023 AMC 12B #9]]}}
+==Problem==
+What is the area of the region in the coordinate plane defined by
+<math>| | x | - 1 | + | | y | - 1 | \le 1</math>?
+<math>\text{(A)}\ 2 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 12</math>
+==Solution==
+First consider, <math>|x-1|+|y-1| \le 1.</math>
+We can see that it's a square with radius 1 (diagonal 2). The area of the square is <math>\sqrt{2}^2 = 2.</math>
+Next, we add one more absolute value and get <math>|x-1|+||y|-1| \le 1.</math> This will double the square reflecting over x-axis.
+So now we got 2 squares.
+Finally, we add one more absolute value and get <math>||x|-1|+||y|-1| \le 1.</math> This will double the squares reflecting over y-axis.
+In the end, we got 4 squares.  The total area is <math>4\cdot2 = </math> <math>\boxed{\text{(B)} 8}</math>
+~Technodoggo ~Minor formatting change: e_is_2.71828
 ==See Also==
+{{AMC10 box|year=2023|ab=B|num-b=12|num-a=14}}
 {{AMC12 box|year=2023|ab=B|num-b=8|num-a=10}}
 {{MAA Notice}}