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

(→‎Solution)
(→‎Solution 1)
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== Solution 1 ==
 
== Solution 1 ==
 +
This is what the diagram looks like:
 
<asy>
 
<asy>
 
size(10cm);  
 
size(10cm);  
Line 13: Line 14:
 
filldraw((-3,3)--(3,3)--(3,-3)--(-3,-3)--cycle, grey);
 
filldraw((-3,3)--(3,3)--(3,-3)--(-3,-3)--cycle, grey);
 
</asy>
 
</asy>
 +
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>
 +
~ Bryguy
  
  

Revision as of 17:10, 11 February 2021

Problem 19

The area of the region bounded by the graph of\[x^2+y^2 = 3|x-y| + 3|x+y|\]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

This is what the diagram looks like: [asy] size(10cm); filldraw(arc((-3,0),3,180,180) -- cycle, gray); filldraw(arc((0,3),3,0,180) -- cycle, gray); filldraw(arc((3,0),3,180,180) -- cycle, gray); filldraw(arc((0,-3),3,180,180) -- cycle, gray); filldraw((-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 $\boxed{\textbf{(E) }54}$ ~ 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)

https://youtu.be/EHHpB6GIGPc

~ pi_is_3.14

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