Difference between revisions of "2023 AMC 12B Problems/Problem 9"
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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 | + | 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> |
− | Next, we | + | 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 | + | So now we have <math>2</math> squares. |
− | Finally, we add one more absolute value and | + | 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> | |
− | ~Technodoggo ~Minor formatting change: e_is_2.71828 | + | ~Technodoggo ~Minor formatting change: e_is_2.71828 ~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)} | + | <math>\boxed{\text{(B) 8}}.</math> |
~ Brian__Liu | ~ Brian__Liu | ||
Line 41: | Line 41: | ||
==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)} | + | 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 | ||
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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)} | + | 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== |
Latest revision as of 21:27, 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.