Question

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 $m + n?$

$18$

$27$

$36$

$45$

$54$

Answer: E

Concepts:absolute value circle area decomposition

Difficulty rating: 2150

Solution:

Consider the four sign cases for $x-y$ and $x+y.$ In one case, for example, $|x-y|=x-y$ and $|x+y|=x+y,$ so

$\begin{gathered} x^2+y^2=6x \\ \quad\Longrightarrow\quad (x-3)^2+y^2=9. \end{gathered}$

The other three cases similarly give circles of radius $3$ centered at $(0,3),$ $(-3,0),$ and $(0,-3).$ The relevant arcs form the boundary shown by these four congruent circle pieces.

The region consists of a central square of side length $6,$ together with four semicircles of radius $3.$ The square contributes area $36,$ and the four semicircles have the area of two full radius- $3$ circles, namely $18\pi.$

Therefore the area is $36+18\pi,$ so $m+n=36+18=54.$

Thus, E is the correct answer.

2021 AMC 10A Spring Problem 19

Answer: E

Difficulty rating: 2150

Solution:

Problem 19 in Other Years