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

$\textbf{(A) } 21 \qquad\textbf{(B)}\ 22 \qquad\textbf{(C)}\ 23 \qquad\textbf{(D)}\ 24 \qquad\textbf{(E)}\ 25$

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

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 $\frac{16 \pi }{2} = 8\pi \approx \boxed{\textbf{(E) }25}$ . ~samrocksnature + ddot1

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2002 AMC 12B Problems/Problem 25

Contents

Problem

Solution 1

Solution 2

See also