Difference between revisions of "2002 AMC 12B Problems/Problem 25"
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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 <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>. | ||
| + | ~SHEN KISLAY KAI | ||
== Solution 2== | == 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 <math>f(x)+f(y) \le 0</math>, or <math>16 \pi</math>. | ||
| − | 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 <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 |
== See also == | == See also == | ||
Latest revision as of 02:49, 26 September 2024
Contents
Problem
Let 
, and let 
denote the set of points 
in the coordinate plane such that
![\[f(x) + f(y) \le 0 \qquad \text{and} \qquad f(x)-f(y) \le 0\]](http://latex.artofproblemsolving.com/a/c/c/acc8ac26f16b1cda3bb82ba42bc6622a0feca8c5.png)
The area of is closest to
Solution 1
The first condition gives us that
![\[x^2 + 6x + 1 + y^2 + 6y + 1 \le 0 \Longrightarrow (x+3)^2 + (y+3)^2 \le 16\]](http://latex.artofproblemsolving.com/b/b/8/bb8ffbea5beac4d271409e70b039ab3fd7fc571a.png)
which is a circle centered at 
with radius 
. The second condition gives us that
![\[x^2 + 6x + 1 - y^2 - 6y - 1 \le 0 \Longrightarrow (x^2 - y^2) + 6(x-y) \le 0 \Longrightarrow (x-y)(x+y+6) \le 0\]](http://latex.artofproblemsolving.com/d/c/f/dcfbc59e8ec63aca25e470670be7d5deb5925eca.png)
Thus either
![\[x - y \ge 0,\quad x+y+6 \le 0\]](http://latex.artofproblemsolving.com/e/6/5/e652e0c8f0c71d919631f0f09e73783e50ad2cae.png)
or
![\[x - y \le 0,\quad x+y+6 \ge 0\]](http://latex.artofproblemsolving.com/e/e/2/ee27e9853791cc1fc519b576ce48f7f479902128.png)
Each of those lines passes through and has slope 
, as shown above. Therefore, the area of is half of the area of the circle, which is 
.
~SHEN KISLAY KAI
Solution 2
Similar to Solution 1, we proceed to get the area of the circle satisfying 
, or 
.
Since 
, we have that by symmetry, if is in , then 
is not, and vice versa. Therefore, the shaded part of the circle above the line 
has the same area as the unshaded part below , and the unshaded part above has the same area as the shaded part below . This means that exactly half the circle is shaded, allowing us to divide by two to get 
. ~samrocksnature + ddot1 +Shen kislay kai
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 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
