2016 AMC 12B Problems/Problem 18: Difference between revisions
m →Solution: I replaced $x >= 0$, $ y >= 0$ with $x \geq 0$, $ y \geq 0$. |
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==Solution== | ==Solution== | ||
Consider the case when <math>x | Consider the case when <math>x \geq 0</math>, <math> y \geq 0</math>. | ||
<cmath>x^2+y^2=x+y</cmath> | <cmath>x^2+y^2=x+y</cmath> | ||
<cmath>(x - \frac{1}{2})^2+(y - \frac{1}{2})^2=\frac{1}{2}</cmath> | <cmath>(x - \frac{1}{2})^2+(y - \frac{1}{2})^2=\frac{1}{2}</cmath> | ||
Revision as of 00:16, 2 February 2018
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);draw(arc((1/2,1/2),sqrt(2)/2,135, 315),dotted);for(int i=0;i<4;++i){draw(rotate(i*90,(0,0))*arc((1/2,1/2),sqrt(2)/2,-45,135)); dot(rotate(i*90,(0,0))*(1/2,1/2));}[/asy]](http://latex.artofproblemsolving.com/7/7/1/77121b786ebf303837357aa5aab877da041db676.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 | |
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