2002 AMC 12B Problems/Problem 25: Difference between revisions

Latest revision as of 02:00, 9 November 2025

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

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 +Shen kislay kai

Note

The equation $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 0 \hspace{1mm} \text{or} \hspace{1mm} \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 0$ is the equation of a degenerate hyperbola.

@@ Line 1: / Line 1: @@
 == Problem ==
-Let <math>f(x) = x^2 + 6x + 1</math>, and let <math>R</math> denote the [[set]] of [[point]]s <math>(x,y)</math> in the [[coordinate plane]] such that
+Let <imath>f(x) = x^2 + 6x + 1</imath>, and let <imath>R</imath> denote the [[set]] of [[point]]s <imath>(x,y)</imath> in the [[coordinate plane]] such that
 <cmath>f(x) + f(y) \le 0 \qquad \text{and} \qquad f(x)-f(y) \le 0</cmath>
-The area of <math>R</math> is closest to
+The area of <imath>R</imath> is closest to
-<math>\textbf{(A) } 21
+<imath>\textbf{(A) } 21
 \qquad\textbf{(B)}\ 22
 \qquad\textbf{(C)}\ 23
 \qquad\textbf{(D)}\ 24
-\qquad\textbf{(E)}\ 25</math>
+\qquad\textbf{(E)}\ 25</imath>
 == Solution 1==
@@ Line 14: / Line 14: @@
 <cmath>x^2 + 6x + 1 + y^2 + 6y + 1 \le 0 \Longrightarrow (x+3)^2 + (y+3)^2 \le 16</cmath>
-which is a [[circle]] centered at <math>(-3,-3)</math> with [[radius]] <math>4</math>. The second condition gives us that
+which is a [[circle]] centered at <imath>(-3,-3)</imath> with [[radius]] <imath>4</imath>. The second condition gives us that
 <cmath>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</cmath>
@@ Line 28: / Line 28: @@
 [[Image:2002_12B_AMC-25.png|center]]
-Each of those lines passes through <math>(-3,-3)</math> and has slope <math>\pm 1</math>, as shown above. Therefore, the area of <math>R</math> is half of the area of the circle, which is <math>\frac{1}{2} (\pi \cdot 4^2) = 8\pi \approx \boxed{\textbf{(E) }25}</math>.
+Each of those lines passes through <imath>(-3,-3)</imath> and has slope <imath>\pm 1</imath>, as shown above. Therefore, the area of <imath>R</imath> is half of the area of the circle, which is <imath>\frac{1}{2} (\pi \cdot 4^2) = 8\pi \approx \boxed{\textbf{(E) }25}</imath>.
-SHEN KISLAY KAI
+~SHEN KISLAY KAI
 == Solution 2==
-Similar to Solution 1, we proceed to get the area of the circle satisfying <math>f(x)+f(y) \le 0</math>, or <math>16 \pi</math>.
+Similar to Solution 1, we proceed to get the area of the circle satisfying <imath>f(x)+f(y) \le 0</imath>, or <imath>16 \pi</imath>.
-Since <math>f(x)-f(y) \le 0 \implies f(x) \le f(y)</math>, we have that by symmetry, if <math>(x,y)</math> is in <math>R</math>, then <math>(y,x)</math> is not, and vice versa. Therefore, the shaded part of the circle above the line <math>y=x</math> has the same area as the unshaded part below <math>y=x</math>, and the unshaded part above <math>y=x</math> has the same area as the shaded part below <math>y=x</math>. This means that exactly half the circle is shaded, allowing us to divide by two to get <math>\frac{16 \pi }{2} = 8\pi \approx \boxed{\textbf{(E) }25}</math>. ~samrocksnature + ddot1 Shen kislay kai
+Since <imath>f(x)-f(y) \le 0 \implies f(x) \le f(y)</imath>, we have that by symmetry, if <imath>(x,y)</imath> is in <imath>R</imath>, then <imath>(y,x)</imath> is not, and vice versa. Therefore, the shaded part of the circle above the line <imath>y=x</imath> has the same area as the unshaded part below <imath>y=x</imath>, and the unshaded part above <imath>y=x</imath> has the same area as the shaded part below <imath>y=x</imath>. This means that exactly half the circle is shaded, allowing us to divide by two to get <imath>\frac{16 \pi }{2} = 8\pi \approx \boxed{\textbf{(E) }25}</imath>. ~samrocksnature + ddot1 +Shen kislay kai
+==Note==
+The equation <imath> \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 0 \hspace{1mm} \text{or} \hspace{1mm} \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 0 </imath> is the equation of a degenerate hyperbola.
 == See also ==

2002 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last 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
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