2002 AMC 12B Problem 25

Below is the professionally curated solution for Problem 25 of the 2002 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 12B solutions, or check the answer key.

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Concepts:completing the square circle area

Difficulty rating: 2260

25.

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

$21$

$22$

$23$

$24$

$25$

Solution:

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

Also $f(x)-f(y)=(x-y)(x+y+6),$ 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.

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