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

Revision as of 16:47, 15 November 2023

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 $4\cdot2 = 8$ .

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-Sticking out your gyat for the rizzler. You're so skibidi. You're so fanum tax. I just wanna be your sigma Ur sus
+== Solution ==
+First consider, <math>|x-1|+|y-1| <= 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.
+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.
+In the end, we got 4 squares.  The total area is <math>4\cdot2 = 8</math>.