Difference between revisions of "2021 AMC 10A Problems/Problem 19"
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The area is <math>6\cdot6+4\cdot \frac{9\pi}{2} = 36+18\pi</math>. The answer is <math>36+18</math> which is <math>\boxed{\textbf{(E) }54}</math> | The area is <math>6\cdot6+4\cdot \frac{9\pi}{2} = 36+18\pi</math>. The answer is <math>36+18</math> which is <math>\boxed{\textbf{(E) }54}</math> | ||
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==Solution 2 (Guessing)== | ==Solution 2 (Guessing)== | ||
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~Arcticturn | ~Arcticturn | ||
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+ | == Remark == | ||
+ | This problem asks for the area of <b>the union of these four circles</b>: | ||
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+ | [[File:Image 2021-02-11 111327.png|center|600px]] | ||
== Video Solution (Using Absolute Value Properties to Graph) == | == Video Solution (Using Absolute Value Properties to Graph) == |
Revision as of 08:23, 23 August 2022
Contents
[hide]Problem
The area of the region bounded by the graph ofis , where and are integers. What is ?
Solution 1
In order to attack this problem, we need to consider casework:
Case 1:
Substituting and simplifying, we have , i.e. , which gives us a circle of radius centered at .
Case 2:
Substituting and simplifying again, we have , i.e. . This gives us a circle of radius centered at .
Case 3:
Doing the same process as before, we have , i.e. . This gives us a circle of radius centered at .
Case 4:
One last time: we have , i.e. . This gives us a circle of radius centered at .
After combining all the cases and drawing them on the Cartesian Plane, this is what the diagram looks like:
Now, the area of the shaded region is just a square with side length with four semicircles of radius . The area is . The answer is which is
~Bryguy
Solution 2 (Guessing)
Assume = . We get that = . That means that this figure must contain the points . Now, assume that = . We get that = . We get the points .
Since this contains , assume that there are circles. Therefore, we can guess that there is a center square with area = and semicircles with radius . We get semicircles with area , and therefore the answer is =
~Arcticturn
Remark
This problem asks for the area of the union of these four circles:
Video Solution (Using Absolute Value Properties to Graph)
~ pi_is_3.14
Video Solution by The Power Of Logic (Graphing)
Video Solution by TheBeautyofMath
~IceMatrix
See Also
2021 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.