Difference between revisions of "2023 AMC 12B Problems/Problem 9"
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==Video Solution 1 by OmegaLearn== | ==Video Solution 1 by OmegaLearn== | ||
https://youtu.be/300Ek9E-RrA | https://youtu.be/300Ek9E-RrA | ||
+ | |||
+ | ==Video Solution 2 by MegaMath== | ||
+ | |||
+ | https://www.youtube.com/watch?v=300yLhj4BI0&t=1s | ||
==See Also== | ==See Also== |
Revision as of 22:01, 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
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.