# 2021 AMC 10A Problems/Problem 19

## 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 can use casework on the sign of $|x-y|$ and $|x+y|$.

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 $\boxed{\textbf{(E) }54}$

~Bryguy

## Solution 2

A somewhat faster variant of solution 1 is to use a bit of symmetry in order to show that the remaining three cases are identical to Case 1 in the above solution, up to rotations by $90^{\circ}$ about the origin. This allows us to quickly sketch the region after solving Case 1.

Upon simplifying Case 1, we obtain $(x-3)^2 + y^2 = 3^2$ which is a circle of radius 3 centered at $(3,0)$. We remark that only the points on the semicircle where $x \ge 3$ work here, since Case 1 assumes $x-y \ge 0$ and $x+y \ge 0$. Next, we observe that an ordered pair is a solution to the given equation if and only if any of its $90^{\circ}$ rotations about the origin is a solution. This follows as the value of $x^2+y^2-3(|x-y|+|x+y|)$ is invariant to $90^{\circ}$ rotations, since $x^2+y^2$ simply represents the square of the distance to the origin (which is unchanged upon rotation), and $|x-y|+|x+y|$ is the sum of the distances to the lines $y=x$ and $y=-x$, multiplied by $\sqrt{2}$ (also unchanged upon $90^{\circ}$ rotation).

By the above observation, we can quickly sketch the remainder of the region, and the area is $\boxed{\textbf{(E) }54}$ as above.

~scrabbler94

## Solution 3 (Guessing)

Assume $y$ = $0$. We get that $x$ = $6$. That means that this figure must contain the points $(0,6), (6,0), (0, -6), (-6, 0)$. Now, assume that $x$ = $y$. We get that $x$ = $3 \sqrt 3$. We get the points $(3,3), (3,-3), (-3, 3), (-3, -3)$.

Since this contains $x^2 + y^2$, assume that there are circles. Therefore, we can guess that there is a center square with area $6 \cdot 6$ = $36$ and $4$ semicircles with radius $3$. We get $4$ semicircles with area $4.5 \pi$, and therefore the answer is $36+18$ = $\boxed {(E)54}$

~Arcticturn

## Remark

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

~ pi_is_3.14

~IceMatrix

## See Also

 2021 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 18 Followed byProblem 20 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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