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

(→‎Solution 1)
(→‎Solution 1)
Line 8: Line 8:
 
<asy>
 
<asy>
 
size(10cm);  
 
size(10cm);  
−
filldraw(arc((-3,0),3,180,180) -- cycle, gray);
+
Label f;
−
filldraw(arc((0,3),3,0,180) -- cycle, gray);
+
f.p=fontsize(7);
−
filldraw(arc((3,0),3,180,180) -- cycle, gray);
+
xaxis(-8,8,Ticks(f, 1.0));
−
filldraw(arc((0,-3),3,180,180) -- cycle, gray);
+
yaxis(-8,8,Ticks(f, 1.0));
−
filldraw((-3,3)--(3,3)--(3,-3)--(-3,-3)--cycle, grey);
+
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>
 
</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>.
 
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>.

Revision as of 21:06, 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); 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 $\boxed{\textbf{(E) }54}$ ~ 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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