Difference between revisions of "2016 AMC 12B Problems/Problem 18"
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Notice the circle intersect the axes at points <math>(0, 1)</math> and <math>(1, 0)</math>. Find the area of this circle in the first quadrant. The area is made of a semi-circle with radius of <math>\frac{\sqrt{2}}{2}</math> and a triangle: | Notice the circle intersect the axes at points <math>(0, 1)</math> and <math>(1, 0)</math>. Find the area of this circle in the first quadrant. The area is made of a semi-circle with radius of <math>\frac{\sqrt{2}}{2}</math> and a triangle: | ||
<cmath>A = \frac{\pi}{4} +\frac{1}{2}</cmath> | <cmath>A = \frac{\pi}{4} +\frac{1}{2}</cmath> | ||
| − | <asy>draw((0,-1.5)--(0,1.5),EndArrow);draw((-1.5,0)--(1.5,0),EndArrow);draw((0,1)--(1,0)--(0,-1)--(-1,0)--cycle,dotted); | + | <asy>draw((0,-1.5)--(0,1.5),EndArrow);draw((-1.5,0)--(1.5,0),EndArrow);draw((0,1)--(1,0)--(0,-1)--(-1,0)--cycle,dotted);for(int i=0;i<4;++i){draw(rotate(i*90,(0,0))*arc((1/2,1/2),sqrt(2)/2,0,360));dot(rotate(i*90,(0,0))*(1/2,1/2));}</asy> |
| − | for(int i=0;i<4;++i){draw(rotate(i*90,(0,0))*arc((1/2,1/2),sqrt(2)/2, | ||
Because of symmetry, the area is the same in all four quadrants. | Because of symmetry, the area is the same in all four quadrants. | ||
The answer is <math>\boxed{\textbf{(B)}\ \pi + 2}</math> | The answer is <math>\boxed{\textbf{(B)}\ \pi + 2}</math> | ||
Revision as of 20:52, 21 February 2016
Problem
What is the area of the region enclosed by the graph of the equation 
Solution
Consider the case when 
, 
.
![\[x^2+y^2=x+y\]](http://latex.artofproblemsolving.com/5/7/9/579c6d245f6bbc2bf41fcc62182d7d032c5e7ee0.png)
![\[(x - \frac{1}{2})^2+(y - \frac{1}{2})^2=\frac{1}{2}\]](http://latex.artofproblemsolving.com/2/e/5/2e585b5d19dc88abd1cd8d083f79cf210658ac93.png)
Notice the circle intersect the axes at points 
and 
. Find the area of this circle in the first quadrant. The area is made of a semi-circle with radius of 
and a triangle:
![\[A = \frac{\pi}{4} +\frac{1}{2}\]](http://latex.artofproblemsolving.com/a/4/5/a45b86b9a5f727c22135e8c436797d6b574c59df.png)
![[asy]draw((0,-1.5)--(0,1.5),EndArrow);draw((-1.5,0)--(1.5,0),EndArrow);draw((0,1)--(1,0)--(0,-1)--(-1,0)--cycle,dotted);for(int i=0;i<4;++i){draw(rotate(i*90,(0,0))*arc((1/2,1/2),sqrt(2)/2,0,360));dot(rotate(i*90,(0,0))*(1/2,1/2));}[/asy]](http://latex.artofproblemsolving.com/f/9/3/f9317abe87ca3a306ea5acb12917e9b32f903fb0.png)
Because of symmetry, the area is the same in all four quadrants.
The answer is 
See Also
| 2016 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 17 |
Followed by Problem 19 |
| 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. 
