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]]}} | ||
+ | |||
+ | ==Problem== | ||
+ | |||
+ | What is the area of the region in the coordinate plane defined by | ||
+ | |||
+ | <math>| | x | - 1 | + | | y | - 1 | \le 1</math>? | ||
+ | |||
+ | <math>\text{(A)}\ 2 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 12</math> | ||
+ | |||
+ | ==Solution 1== | ||
+ | First consider, <math>|x-1|+|y-1| \le 1.</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. | ||
+ | |||
+ | So now we have <math>2</math> squares. | ||
+ | |||
+ | 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. Thus, the total area is <math>4\cdot2 = </math> <math>\boxed{\text{(B) 8}}</math> | ||
+ | |||
+ | ~Technodoggo ~Minor formatting change: e_is_2.71828, mathkiddus ~Grammar and clarity: NSAoPS | ||
+ | |||
+ | ==Solution 2 (Graphing)== | ||
+ | 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 | ||
+ | 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), | ||
+ | (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, | ||
+ | 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> | ||
+ | <math>\boxed{\text{(B) 8}}.</math> | ||
+ | |||
+ | ~ 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 <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 | ||
+ | ~ 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 <math>x</math> axis and <math>\pm 1</math> on the <math>y</math> 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 <math>x</math> and <math>y</math> axis. | ||
+ | |||
+ | So we have <math>2\times 4</math> which is <math>\boxed{\text{(B) 8}}</math>. | ||
+ | |||
+ | ~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== | ||
+ | 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== | ==See Also== | ||
− | {{AMC12 box|year=2023|ab=B|num-b= | + | {{AMC10 box|year=2023|ab=B|num-b=12|num-a=14}} |
+ | {{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.
Contents
Problem
What is the area of the region in the coordinate plane defined by
?
Solution 1
First consider, We can see that it is a square with a side length of (diagonal ). The area of the square is
Next, we insert an absolute value sign into the equation and get This will double the square reflecting over x-axis.
So now we have squares.
Finally, we add one more absolute value and obtain This will double the squares as we reflect the squares we already have over the y-axis.
Concluding, we have congruent squares. Thus, the total area is
~Technodoggo ~Minor formatting change: e_is_2.71828, mathkiddus ~Grammar and clarity: NSAoPS
Solution 2 (Graphing)
We first consider the lattice points that satisfy and . The lattice points satisfying these equations are and By symmetry, we also have points and when and . 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 , for instance). As noted in the above solution, each square has a diagonal for an area of , so the total area is
~ 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 and can be a maximum of 1 when the other is 0. Therefore the value of and range from -2 to 2. This forms a diamond shape which has area which is
~ darrenn.cp ~ DarkPheonix
Solution 4
We start by considering the graph of . To get from this graph to we have to translate it by on the axis and on the axis.
Graphing we get a square with side length of , so the area of one of these squares is just .
We have to multiply by since there are combinations of shifting the and axis.
So we have which is .
~ESAOPS
Solution 5 (Desperate)
There are sets of absolute value bars. Each of those absolute value bars can take on values, so we have cases. We guess that the answer is divisible by . The only answer choice that is divisible by is .
~ cxsmi
Video Solution 1 by OmegaLearn
Video Solution 2 by MegaMath
https://www.youtube.com/watch?v=300yLhj4BI0&t=1s
Video Solution
~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.