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

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{{duplicate|[[2023 AMC 10B Problems/Problem 13|2023 AMC 10B #13]] and [[2023 AMC 12B Problems/Problem 9|2023 AMC 12B #9]]}}
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==Problem==
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What is the area of the region in the coordinate plane defined by
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<math>| | x | - 1 | + | | y | - 1 | \le 1</math>?
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<math>\text{(A)}\ 2 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 12</math>
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==Solution 1==
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First consider, <math>|x-1|+|y-1| \le 1.</math>
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We can see that it is a square with a side length of <math>1</math> (diagonal <math>\sqrt{2}</math>). The area of the square is <math>\sqrt{2}^2 = 2.</math>
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Next, we insert an absolute value sign into the equation and get <math>|x-1|+||y|-1| \le 1.</math> This will double the square reflecting over x-axis.
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So now we have <math>2</math> squares.
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Finally, we add one more absolute value and obtain <math>||x|-1|+||y|-1| \le 1.</math> This will double the squares as we reflect the <math>2</math> squares we already have over the y-axis.
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Concluding, we have <math>4</math> congruent squares. Thus, the total area is <math>4\cdot2 = </math> <math>\boxed{\text{(B) 8}}</math>
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~Technodoggo ~Minor formatting change: e_is_2.71828, mathkiddus ~Grammar and clarity: NSAoPS
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==Solution 2 (Graphing)==
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We first consider the lattice points that satisfy <math>||x|-1| = 0</math> and <math>||y|-1| = 1</math>. The lattice points satisfying these equations
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are <math>(1,0), (1,2), (1,-2), (-1,0), (-1,2),</math> and <math>(-1,-2).</math> By symmetry, we also have points <math>(0,1), (2,1), (-2,1), (0,-1),
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(2,-1),</math> and  <math>(-2,-1)</math>  when  <math>||x|-1| = 1</math> and <math>||y|-1| = 0</math>.  Graphing and connecting these points, we form 5 squares. However,
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we can see that any point within the square in the middle does not satisfy the given inequality (take <math>(0,0)</math>, for instance). As
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noted in the above solution, each square has a diagonal <math>2</math> for an area of <math>\frac{2^2}{2} = 2</math>, so the total area is <math>4\cdot2 =</math>
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<math>\boxed{\text{(B) 8}}.</math> 
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~ Brian__Liu
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==Note==
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This problem is very similar to a past AIME problem (1997 P13)
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https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_13
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~ CherryBerry
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==Solution 3 (Logic)==
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The value of <math>|x|</math> and <math>|y|</math> can be a maximum of 1 when the other is 0. Therefore the value of <math>x</math> and <math>y</math> range from -2 to 2. This forms a diamond shape which has area <math>4 \times \frac{2^2}{2}</math> which is <math>\boxed{\text{(B) 8}}.</math> 
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~ darrenn.cp
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~ DarkPheonix
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==Solution 4==
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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 <math>x</math> axis and <math>\pm 1</math> on the <math>y</math> axis.
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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>.
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We have to multiply by <math>4</math> since there are <math>4</math> combinations of shifting the <math>x</math> and <math>y</math> axis.
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So we have <math>2\times 4</math> which is <math>\boxed{\text{(B) 8}}</math>.
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~ESAOPS
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==Solution 5 (Desperate)==
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There are <math>2</math> sets of <math>2</math> absolute value bars. Each of those <math>2</math> absolute value bars can take on <math>2</math> values, so we have <math>2 \cdot 2 \cdot 2 = 8</math> cases. We guess that the answer is divisible by <math>8</math>. The only answer choice that is divisible by <math>8</math> is <math>\boxed{\text{(B)}~8}</math>.
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~ cxsmi
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==Video Solution 1 by OmegaLearn==
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https://youtu.be/300Ek9E-RrA
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==Video Solution 2 by MegaMath==
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https://www.youtube.com/watch?v=300yLhj4BI0&t=1s
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==Video Solution==
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https://youtu.be/Tic8qo-iQq4
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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==See Also==
 
==See Also==
{{AMC12 box|year=2023|ab=B|num-b=4|num-a=6}}
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{{AMC10 box|year=2023|ab=B|num-b=12|num-a=14}}
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{{AMC12 box|year=2023|ab=B|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 11:14, 3 November 2024

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 is a square with a side length of $1$ (diagonal $\sqrt{2}$). The area of the square is $\sqrt{2}^2 = 2.$

Next, we insert an absolute value sign into the equation and get $|x-1|+||y|-1| \le 1.$ This will double the square reflecting over x-axis.

So now we have $2$ squares.

Finally, we add one more absolute value and obtain $||x|-1|+||y|-1| \le 1.$ This will double the squares as we reflect the $2$ squares we already have over the y-axis.

Concluding, we have $4$ congruent squares. Thus, the total area is $4\cdot2 =$ $\boxed{\text{(B) 8}}$

~Technodoggo ~Minor formatting change: e_is_2.71828, mathkiddus ~Grammar and clarity: NSAoPS

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

Solution 5 (Desperate)

There are $2$ sets of $2$ absolute value bars. Each of those $2$ absolute value bars can take on $2$ values, so we have $2 \cdot 2 \cdot 2 = 8$ cases. We guess that the answer is divisible by $8$. The only answer choice that is divisible by $8$ is $\boxed{\text{(B)}~8}$.

~ cxsmi

Video Solution 1 by OmegaLearn

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

Video Solution 2 by MegaMath

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

Video Solution

https://youtu.be/Tic8qo-iQq4


~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

See Also

2023 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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