Difference between revisions of "2021 AMC 10A Problems/Problem 19"
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== Solution 1 == | == Solution 1 == | ||
| − | This is what the diagram looks like: | + | In order to attack this problem, we need to consider casework: |
| + | |||
| + | Case 1: <math>|x-y|=x-y, |x+y|=x+y</math> | ||
| + | |||
| + | Substituting and simplifying, we have <math>x^2-6x+y^2=0</math>, i.e. <math>(x-3)^2+y^2=3^2</math>, which gives us a circle of radius <math>3</math> centered at <math>(3,0)</math>. | ||
| + | |||
| + | Case 2: <math>|x-y|=y-x, |x+y|=x+y</math> | ||
| + | |||
| + | Substituting and simplifying again, we have <math>x^2+y^2-6y=0</math>, i.e. <math>x^2+(y-3)^2=3^2</math>. This gives us a circle of radius <math>3</math> centered at <math>(0,3)</math>. | ||
| + | |||
| + | Case 3: <math>|x-y|=x-y, |x+y|=-x-y</math> | ||
| + | |||
| + | Doing the same process as before, we have <math>x^2+y^2+6y=0</math>, i.e. <math>x^2+(y+3)^2=3^2</math>. This gives us a circle of radius <math>3</math> centered at <math>(0,-3)</math>. | ||
| + | |||
| + | Case 4: <math>|x-y|=y-x, |x+y|=-x-y</math> | ||
| + | |||
| + | One last time: we have <math>x^2+y^2+6x=0</math>, i.e. <math>(x+3)^2+y^2=3^2</math>. This gives us a circle of radius <math>3</math> centered at <math>(-3,0)</math>. | ||
| + | |||
| + | After combining all the cases and drawing them on the Cartesian Plane, this is what the diagram looks like: | ||
| + | |||
<asy> | <asy> | ||
size(10cm); | size(10cm); | ||
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Now, the area of the shaded region is just a square with side length <math>6</math> with four semicircles of radius <math>3</math>. | Now, the area of the shaded region is just a square with side length <math>6</math> with four semicircles of radius <math>3</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> | 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> | ||
| − | + | ||
| + | Solution by Bryguy | ||
Revision as of 21:20, 11 February 2021
Problem 19
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 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:
![[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 
Solution by Bryguy
https://artofproblemsolving.com/wiki/index.php/File:Image_2021-02-11_111327.png (someone please help link file thanks)
Video Solution (Using absolute value properties to graph)
~ pi_is_3.14
