AMC12 2002 B

AMC12 2002 B · Q25

AMC12 2002 B · Q25. It mainly tests Graphs (coordinate plane), Geometry misc.

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

设 $f(x) = x^2 + 6x + 1$，并令 $R$ 表示坐标平面中满足 \[f(x) + f(y) \le 0 \qquad \text{且} \qquad f(x)-f(y) \le 0\] 的点 $(x,y)$ 的集合。$R$ 的面积最接近

(A) 21 21

(B) 22 22

(D) 24 24

(E) 25 25

Answer

Correct choice: (E)

正确答案：（E）

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 \[x - y \ge 0,\quad x+y+6 \le 0\] 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 $\frac{1}{2} (\pi \cdot 4^2) = 8\pi \approx \boxed{\textbf{(E) }25}$.

第一个条件给出 \[x^2 + 6x + 1 + y^2 + 6y + 1 \le 0 \Longrightarrow (x+3)^2 + (y+3)^2 \le 16\] 这是一个以 $(-3,-3)$ 为圆心、半径为 $4$ 的圆。第二个条件给出 \[x^2 + 6x + 1 - y^2 - 6y - 1 \le 0 \Longrightarrow (x^2 - y^2) + 6(x-y) \le 0 \Longrightarrow (x-y)(x+y+6) \le 0\] 因此要么 \[x - y \ge 0,\quad x+y+6 \le 0\] 要么 \[x - y \le 0,\quad x+y+6 \ge 0\] 这两条直线都经过 $(-3,-3)$，斜率为 $\pm 1$，如上图所示。因此，$R$ 的面积是圆面积的一半，即 $\frac{1}{2} (\pi \cdot 4^2) = 8\pi \approx \boxed{\textbf{(E) }25}$。

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