Difference between revisions of "2002 AMC 12B Problems/Problem 25"
m (→Solution: center?) |
(fixed some relations symbols; product is negative not positive) |
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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 15:56, 7 June 2009
Problem
Let , and let denote the set of points in the coordinate plane such that The area of is closest to
Solution
The first condition gives us that
which is a circle centered at with radius . The second condition gives us that
Thus either
or
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 |