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

(→‎Video Solution 2 by MegaMath)
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==Solution 1==
 
==Solution 1==
 
First consider, <math>|x-1|+|y-1| \le 1.</math>
 
First consider, <math>|x-1|+|y-1| \le 1.</math>
−
We can see that it's a square with radius 1 (diagonal <math>\sqrt{2}</math>). The area of the square is <math>\sqrt{2}^2 = 2.</math>
+
We can see that it is a square with a side length of <math>\sqrt{2}</math> (diagonal <math>2</math>). 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| \le 1.</math> This will double the square reflecting over x-axis.
+
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.
  
−
So now we got 2 squares.
+
So now we have <math>2</math> squares.
  
−
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.
+
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.
  
−
In the end, we got 4 squares. The total area is <math>4\cdot2 = </math> <math>\boxed{\text{(B)} 8}</math>
+
Concluding, we have <math>4</math> congruent squares. Thus, the total area is <math>4\cdot2 = </math> <math>\boxed{\text{(B) 8}}</math>
  
−
~Technodoggo ~Minor formatting change: e_is_2.71828
+
~Technodoggo ~Minor formatting change: e_is_2.71828, mathkiddus ~Grammar and clarity: NSAoPS
  
 
==Solution 2 (Graphing)==
 
==Solution 2 (Graphing)==
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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
 
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
 
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>
 
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>
−
<math>\boxed{\text{(B)} 8}.</math>   
+
<math>\boxed{\text{(B) 8}}.</math>   
  
 
~ Brian__Liu
 
~ Brian__Liu
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==Solution 3 (Logic)==
 
==Solution 3 (Logic)==
−
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>   
+
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>   
  
 
~ darrenn.cp
 
~ darrenn.cp
Line 53: Line 53:
 
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.
 
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.
  
−
So we have <math>2\times 4</math> which is <math>\boxed{\text{(B)} 8}</math>.
+
So we have <math>2\times 4</math> which is <math>\boxed{\text{(B) 8}}</math>.
  
 
~ESAOPS
 
~ESAOPS
 +
 +
==Solution 5 (Desperate)==
 +
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>.
 +
 +
~ cxsmi
  
 
==Video Solution 1 by OmegaLearn==
 
==Video Solution 1 by OmegaLearn==
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https://www.youtube.com/watch?v=300yLhj4BI0&t=1s
 
https://www.youtube.com/watch?v=300yLhj4BI0&t=1s
  
−
==Video Solution 3 by paixiao==
+
==Video Solution==
 +
 
 +
https://youtu.be/Tic8qo-iQq4
 +
 
  
−
https://www.youtube.com/watch?v=EvA2Nlb7gi4&t=175s
+
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
  
 
==See Also==
 
==See Also==

Latest revision as of 11:12, 9 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 $\sqrt{2}$ (diagonal $2$). 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 (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. AMC logo.png

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