Difference between revisions of "2002 AMC 12B Problems/Problem 25"
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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 == | + | |
+ | == Solution 1== | ||
The first condition gives us that | The first condition gives us that | ||
<cmath>x^2 + 6x + 1 + y^2 + 6y + 1 \le 0 \Longrightarrow (x+3)^2 + (y+3)^2 \le 16</cmath> | <cmath>x^2 + 6x + 1 + y^2 + 6y + 1 \le 0 \Longrightarrow (x+3)^2 + (y+3)^2 \le 16</cmath> | ||
Line 27: | Line 28: | ||
[[Image:2002_12B_AMC-25.png|center]] | [[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 25 \ | + | 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>. |
+ | |||
+ | == 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>. | ||
+ | |||
+ | 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 | ||
== See also == | == See also == | ||
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 14:48, 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 |
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