Difference between revisions of "2021 AMC 10A Problems/Problem 19"

Revision as of 09:36, 29 October 2021

Problem

The area of the region bounded by the graph of $x^2+y^2 = 3|x-y| + 3|x+y|$ is $m+n\pi$ , where $m$ and $n$ are integers. What is $m + n$ ?

$\textbf{(A)} ~18\qquad\textbf{(B)} ~27\qquad\textbf{(C)} ~36\qquad\textbf{(D)} ~45\qquad\textbf{(E)} ~54$

Solution 1

In order to attack this problem, we need to consider casework:

Case 1: $|x-y|=x-y, |x+y|=x+y$

Substituting and simplifying, we have $x^2-6x+y^2=0$ , i.e. $(x-3)^2+y^2=3^2$ , which gives us a circle of radius $3$ centered at $(3,0)$ .

Case 2: $|x-y|=y-x, |x+y|=x+y$

Substituting and simplifying again, we have $x^2+y^2-6y=0$ , i.e. $x^2+(y-3)^2=3^2$ . This gives us a circle of radius $3$ centered at $(0,3)$ .

Case 3: $|x-y|=x-y, |x+y|=-x-y$

Doing the same process as before, we have $x^2+y^2+6y=0$ , i.e. $x^2+(y+3)^2=3^2$ . This gives us a circle of radius $3$ centered at $(0,-3)$ .

Case 4: $|x-y|=y-x, |x+y|=-x-y$

One last time: we have $x^2+y^2+6x=0$ , i.e. $(x+3)^2+y^2=3^2$ . This gives us a circle of radius $3$ centered at $(-3,0)$ .

After combining all the cases and drawing them on the Cartesian Plane, this is what the diagram looks like:

$[asy] size(10cm); Label f; f.p=fontsize(7); xaxis(-8,8,Ticks(f, 1.0)); yaxis(-8,8,Ticks(f, 1.0)); draw(arc((-3,0),3,90,270) -- cycle, gray); draw(arc((0,3),3,0,180) -- cycle, gray); draw(arc((3,0),3,-90,90) -- cycle, gray); draw(arc((0,-3),3,-180,0) -- cycle, gray); draw((-3,3)--(3,3)--(3,-3)--(-3,-3)--cycle, grey); [/asy]$ Now, the area of the shaded region is just a square with side length $6$ with four semicircles of radius $3$ . The area is $6\cdot6+4\cdot \frac{9\pi}{2} = 36+18\pi$ . The answer is $36+18$ which is $\boxed{\textbf{(E) }54}$

Solution by Bryguy

Remark

This problem asks for the area of the union of these four circles:

Solution 2

Video Solution (Using Absolute Value Properties to Graph)

https://youtu.be/EHHpB6GIGPc

~ pi_is_3.14

Video Solution by The Power Of Logic (Graphing)

https://youtu.be/-pa72wBA85Y

Video Solution by TheBeautyofMath

https://youtu.be/U6obY_kio0g

~IceMatrix

@@ Line 46: / Line 46: @@
 [[File:Image 2021-02-11 111327.png|center|600px]]
+==Solution 2==
 == Video Solution (Using Absolute Value Properties to Graph) ==

2021 AMC 10A (Problems • Answer Key • Resources)
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