# Difference between revisions of "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

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

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

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 $\frac{1}{2} (\pi \cdot 4^2) = 8\pi \approx 25 \Rightarrow \mathrm{(E)}$.