Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2021 AMC 10A Problems/Problem 19

Revision as of 21:20, 11 February 2021 by Bryguy (talk | contribs) (→‎Solution 1)

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

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


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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10A_Problems/Problem_19&oldid=145574"