Difference between revisions of "2023 AMC 12B Problems/Problem 9"
(→Solution 4) |
|||
| Line 47: | Line 47: | ||
==Solution 4== | ==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 x axis and <math>\pm 1</math> on the y axis. | + | 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>. | + | 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 x and y 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>. | ||
Revision as of 21:28, 18 November 2023
- 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's a square with radius 1 (diagonal 
). The area of the square is 
Next, we add one more absolute value and get 
This will double the square reflecting over x-axis.
So now we got 2 squares.
Finally, we add one more absolute value and get 
This will double the squares reflecting over y-axis.
In the end, we got 4 squares. The total area is 
~Technodoggo ~Minor formatting change: e_is_2.71828
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
Video Solution 1 by OmegaLearn
Video Solution 2 by MegaMath
https://www.youtube.com/watch?v=300yLhj4BI0&t=1s
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. 
