Difference between revisions of "2002 AMC 12B Problems/Problem 25"
Dairyqueenxd (talk | contribs) (→Problem) |
|||
Line 4: | Line 4: | ||
The area of <math>R</math> is closest to | The area of <math>R</math> is closest to | ||
− | <math>\ | + | <math>\textbf{(A) } 21 |
− | \qquad\ | + | \qquad\textbf{(B)}\ 22 |
− | \qquad\ | + | \qquad\textbf{(C)}\ 23 |
− | \qquad\ | + | \qquad\textbf{(D)}\ 24 |
− | \qquad\ | + | \qquad\textbf{(E)}\ 25</math> |
+ | |||
== Solution 1== | == Solution 1== | ||
The first condition gives us that | The first condition gives us that |
Revision as of 13:45, 7 March 2022
Contents
Problem
Let , and let denote the set of points in the coordinate plane such that The area of is closest to
Solution 1
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 .
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
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.