Difference between revisions of "2021 AMC 10A Problems/Problem 19"
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== Video Solution by OmegaLearn (Using Absolute Value Properties to Graph) == | == Video Solution by OmegaLearn (Using Absolute Value Properties to Graph) == | ||
Revision as of 21:48, 14 June 2023
Contents
Problem
The area of the region bounded by the graph of![\[x^2+y^2 = 3|x-y| + 3|x+y|\]](http://latex.artofproblemsolving.com/8/4/b/84b498bfa9802780599eea78d4507cbe55897db6.png)
is 
, 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:
![[asy] size(10cm); Label f; f.p=fontsize(7); xaxis(-8,8,Ticks(f, 1.0)); yaxis(-8,8,Ticks(f, 1.0)); draw(arc((-3,0),3,90,270) -- cycle, gray); draw(arc((0,3),3,0,180) -- cycle, gray); draw(arc((3,0),3,-90,90) -- cycle, gray); draw(arc((0,-3),3,-180,0) -- cycle, gray); draw((-3,3)--(3,3)--(3,-3)--(-3,-3)--cycle, grey); [/asy]](http://latex.artofproblemsolving.com/6/3/5/6352083edc8b0551fe5673e3d75e6debba348a74.png)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
