Difference between revisions of "2016 AMC 12B Problems/Problem 18"
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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> | ||
| − | Notice the circle | + | Notice the circle intersects 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);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> | <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> | ||
| − | Because of symmetry, the area is the same in all four quadrants. | + | |
| − | The answer is <math>\boxed{\textbf{(B)}\ \pi + 2}</math> | + | Because of symmetry, the area is the same in all four quadrants. The answer is <math>\boxed{\textbf{(B)}\ \pi + 2}</math> |
| + | |||
| + | ==Video Solution by CanadaMath (Problem 11-20)== | ||
| + | Fast Forward to 33:32 for problem 18 | ||
| + | https://www.youtube.com/watch?v=4osvFatUv1o | ||
| + | |||
| + | ~THEMATHCANADIAN | ||
| + | |||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2016|ab=B|num-b=17|num-a=19}} | {{AMC12 box|year=2016|ab=B|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 00:30, 10 November 2024
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 intersects 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 
Video Solution by CanadaMath (Problem 11-20)
Fast Forward to 33:32 for problem 18 https://www.youtube.com/watch?v=4osvFatUv1o
~THEMATHCANADIAN
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. 
