Question

Let $f(x)=x^2+6x+1,$ and let $R$ denote the set of points $(x,y)$ in the coordinate plane such that $f(x)+f(y)\le0$ and $f(x)-f(y)\le0.$ The area of $R$ is closest to

$21$

$22$

$23$

$24$

$25$

Answer: E

Concepts:completing the square circle area

Difficulty rating: 2260

Solution:

Completing the square, $\begin{gathered} f(x)+f(y) \\ {}=(x+3)^2+(y+3)^2-16, \end{gathered}$ so the first condition is the disk of radius $4$ centered at $(-3,-3).$

Also $\begin{gathered} f(x)-f(y) \\ {}=(x-y)(x+y+6), \end{gathered}$ so the second condition $(x-y)(x+y+6)\le0$ describes two half-planes bounded by the perpendicular lines through $(-3,-3)$ of slopes $1$ and $-1.$ These cut the disk into two equal halves.

Thus $R$ has area $\tfrac12\pi\cdot4^2=8\pi\approx25.13,$ which is closest to $25.$

Thus, the correct answer is E.

2002 AMC 12B Problem 25

Answer: E

Difficulty rating: 2260

Solution:

Problem 25 in Other Years