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 can use casework on the sign of and .
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
A somewhat faster variant of solution 1 is to use a bit of symmetry in order to show that the remaining three cases are identical to Case 1 in the above solution, up to rotations by about the origin. This allows us to quickly sketch the region after solving Case 1.
Upon simplifying Case 1, we obtain which is a circle of radius 3 centered at . We remark that only the points on the semicircle where work here, since Case 1 assumes and . Next, we observe that an ordered pair is a solution to the given equation if and only if any of its rotations about the origin is a solution. This follows as the value of is invariant to rotations, since simply represents the square of the distance to the origin (which is unchanged upon rotation), and is the sum of the distances to the lines and , multiplied by (also unchanged upon rotation).
By the above observation, we can quickly sketch the remainder of the region, and the area is as above.
~scrabbler94
Solution 3 (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:
File:Screenshot 2023-06-14 194749
Video Solution by OmegaLearn (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 |
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