Difference between revisions of "2002 AMC 12B Problems/Problem 25"
m (→Solution: center?) |
(fixed some relations symbols; product is negative not positive) |
||
| Line 19: | Line 19: | ||
Thus either | Thus either | ||
| − | <cmath>x - y \ge 0,\quad x+y+6 \ | + | <cmath>x - y \ge 0,\quad x+y+6 \le 0</cmath> |
or | or | ||
| − | <cmath>x - y \le 0,\quad x+y+6 \ | + | <cmath>x - y \le 0,\quad x+y+6 \ge 0</cmath> |
[[Image:2002_12B_AMC-25.png|center]] | [[Image:2002_12B_AMC-25.png|center]] | ||
Revision as of 14:56, 7 June 2009
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
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 
.
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 | |
