2021 AMC 10A Problems/Problem 19
Contents
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
Solution by Bryguy
Remark
This problem asks for the area of the union of these four circles:
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
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 | |
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All AMC 10 Problems and Solutions |
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