Difference between revisions of "2023 AMC 10B Problems/Problem 13"

Revision as of 17:34, 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, $|x-1|+|y-1| <= 1.$ 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 $|x-1|+||y|-1| <= 1.$ This will double the square reflecting over x-axis.

So now we got 2 squares.

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

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

Revision as of 17:33, 15 November 2023 (view source) E is 2.71828 (talk \| contribs) m (→‎Problem 13) ← Older edit		Revision as of 17:34, 15 November 2023 (view source) E is 2.71828 (talk \| contribs) (→‎Solution) Newer edit →
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	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\| <= 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 = 8</math>.	+	In the end, we got 4 squares. The total area is <math>4\cdot2 = </math>\ text{(B)}\ 8$.