Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 4)
(→‎Solution 1)
 
(3 intermediate revisions by 3 users not shown)
Line 11: Line 11:
 
==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 is a square with a radius of <math>1</math> (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>1</math> (diagonal <math>\sqrt{2}</math>). The area of the square is <math>\sqrt{2}^2 = 2.</math>
  
 
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.
 
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.
Line 19: Line 19:
 
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.
 
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.
  
−
Concluding, we have <math>4</math> congruent 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 ~Grammar and clarity: NSAoPS
+
~Technodoggo ~Minor formatting change: e_is_2.71828, mathkiddus ~Grammar and clarity: NSAoPS
  
 
==Solution 2 (Graphing)==
 
==Solution 2 (Graphing)==
Line 68: Line 68:
  
 
https://www.youtube.com/watch?v=300yLhj4BI0&t=1s
 
https://www.youtube.com/watch?v=300yLhj4BI0&t=1s
−
 
−
==Video Solution 3 by paixiao==
 
−
 
−
https://www.youtube.com/watch?v=EvA2Nlb7gi4&t=175s
 
  
 
==Video Solution==
 
==Video Solution==

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 (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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12B_Problems/Problem_9&oldid=231360"