2002 AMC 12B Problem 25

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

Complete the square:

f(x)+f(y)

=(x+3)^2+(y+3)^2-16

The first condition is a disk of radius

4

; the second,

(x-y)(x+y+6)\le0,

keeps half of it

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.

Problem 25 in other years