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

Revision as of 21:27, 18 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.

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 1

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

Solution 2 (Graphing)

We first consider the lattice points that satisfy $||x|-1| = 0$ and $||y|-1| = 1$ . The lattice points satisfying these equations are $(1,0), (1,2), (1,-2), (-1,0), (-1,2),$ and $(-1,-2).$ By symmetry, we also have points $(0,1), (2,1), (-2,1), (0,-1), (2,-1),$ and $(-2,-1)$ when $||x|-1| = 1$ and $||y|-1| = 0$ . Graphing and connecting these points, we form 5 squares. However, we can see that any point within the square in the middle does not satisfy the given inequality (take $(0,0)$ , for instance). As noted in the above solution, each square has a diagonal $2$ for an area of $\frac{2^2}{2} = 2$ , so the total area is $4\cdot2 =$ $\boxed{\text{(B)} 8}.$

~ Brian__Liu

Note

This problem is very similar to a past AIME problem (1997 P13)

https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_13

~ CherryBerry

Solution 3 (Logic)

The value of $|x|$ and $|y|$ can be a maximum of 1 when the other is 0. Therefore the value of $x$ and $y$ range from -2 to 2. This forms a diamond shape which has area $4 \times \frac{2^2}{2}$ which is $\boxed{\text{(B)} 8}.$

~ darrenn.cp ~ DarkPheonix

Solution 4

We start by considering the graph of $|x|+|y|\leq 1$ . To get from this graph to $||x|-1|+||y|-1| \leq 1$ we have to translate it by $\pm 1$ on the x axis and $\pm 1$ on the y axis.

Graphing $|x|+|y|\leq 1$ we get a square with side length of $sqrt{2}$ , so the area of one of these squares is just $2$ .

We have to multiply by $4$ since there are $4$ combinations of shifting the x and y axis.

So we have $2\times 4$ which is $\boxed{\text{(B)} 8}$ .

~ESAOPS

Video Solution 1 by OmegaLearn

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

Video Solution 2 by MegaMath

https://www.youtube.com/watch?v=300yLhj4BI0&t=1s

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 45: / Line 45: @@
 ~ darrenn.cp
 ~ DarkPheonix
+==Solution 4==
+We start by considering the graph of <math>|x|+|y|\leq 1</math>. To get from this graph to <math>||x|-1|+||y|-1| \leq 1</math> we have to translate it by <math>\pm 1</math> on the x axis and <math>\pm 1</math> on the y axis.
+Graphing <math>|x|+|y|\leq 1</math> we get a square with side length of <math>sqrt{2}</math>, so the area of one of these squares is just <math>2</math>.
+We have to multiply by <math>4</math> since there are <math>4</math> combinations of shifting the x and y axis.
+So we have <math>2\times 4</math> which is <math>\boxed{\text{(B)} 8}</math>.
+~ESAOPS
 ==Video Solution 1 by OmegaLearn==

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