Difference between revisions of "2023 AMC 12B Problems/Problem 9"
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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)} | + | <math>\boxed{\text{(B) 8}}.</math> |
~ Brian__Liu | ~ Brian__Liu |
Revision as of 20:26, 27 April 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 radius 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. The total area is
~Technodoggo ~Minor formatting change: e_is_2.71828 ~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 3 by paixiao
https://www.youtube.com/watch?v=EvA2Nlb7gi4&t=175s
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.