# 2002 AMC 12B Problems/Problem 25

## Problem

Let $f(x) = x^2 + 6x + 1$, and let $R$ denote the set of points $(x,y)$ in the coordinate plane such that $$f(x) + f(y) \le 0 \qquad \text{and} \qquad f(x)-f(y) \le 0$$ The area of $R$ is closest to $\mathrm{(A)}\ 21 \qquad\mathrm{(B)}\ 22 \qquad\mathrm{(C)}\ 23 \qquad\mathrm{(D)}\ 24 \qquad\mathrm{(E)}\ 25$

## Solution 1

The first condition gives us that $$x^2 + 6x + 1 + y^2 + 6y + 1 \le 0 \Longrightarrow (x+3)^2 + (y+3)^2 \le 16$$

which is a circle centered at $(-3,-3)$ with radius $4$. The second condition gives us that $$x^2 + 6x + 1 - y^2 - 6y - 1 \le 0 \Longrightarrow (x^2 - y^2) + 6(x-y) \le 0 \Longrightarrow (x-y)(x+y+6) \le 0$$

Thus either $$x - y \ge 0,\quad x+y+6 \le 0$$

or $$x - y \le 0,\quad x+y+6 \ge 0$$

Each of those lines passes through $(-3,-3)$ and has slope $\pm 1$, as shown above. Therefore, the area of $R$ is half of the area of the circle, which is $\frac{1}{2} (\pi \cdot 4^2) = 8\pi \approx 25 \Rightarrow \mathrm{(E)}$.

## Solution 2

Similar to Solution 1, we proceed to get the area of the circle satisfying $f(x)+f(y) \le 0$, or $16 \pi$.

Since $f(x)-f(y) \le 0 \implies f(x) \le f(y)$, we have that by symmetry, if $(x,y)$ is in $R$, then $(y,x)$ is not, and vice versa. Therefore, the shaded part of the circle above the line $y=x$ has the same area as the unshaded part below $y=x$, and the unshaded part above $y=x$ has the same area as the shaded part below $y=x$. This means that exactly half the circle is shaded, allowing us to divide by two to get $\frac{16 \pi }{2} = \boxed{8 \pi}$. ~samrocksnature + ddot1

## See also

 2002 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 24 Followed byLast problem 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 12 Problems and Solutions

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