2023 AMC 10B Problems/Problem 13: Difference between revisions

Revision as of 17:42, 15 November 2023

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 $4\cdot2 =$ $\boxed{\text{(B)} 8}$

~Minor formatting change: e_is_2.71828

@@ Line 8: / Line 8: @@
 == Solution ==
-First consider, <math>|x-1|+|y-1| <= 1.</math>
+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| <= 1.</math> This will double the square reflecting over x-axis.
+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| <= 1.</math> This will double the squares reflecting over y-axis.
+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>
+~Minor formatting change: e_is_2.71828