Difference between revisions of "2016 AMC 12B Problems/Problem 18"
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==Solution== | ==Solution== | ||
| − | Consider the case when <math>x > 0, y > 0</math> | + | Consider the case when <math>x > 0, y > 0. </math> |
| − | <math>x^2+y^2=x+y</math> | + | <math>x^2+y^2=x+y. </math> |
| − | <math>(x - 0.5)^2+(y - 0.5)^2=0.5</math> | + | <math>(x - 0.5)^2+(y - 0.5)^2=0.5. </math> |
| − | Find the area of this circle in the first quadrant. Notice the circle intersect the axe at point <math>(0, 1) and (1, 0)</math> : | + | Find the area of this circle in the first quadrant. Notice the circle intersect the axe at point <math>(0, 1) and (1, 0). </math> : |
<math>0.5 + 0.25\pi</math> | <math>0.5 + 0.25\pi</math> | ||
Because of symmetry, that area is the same in all four quadrants. | Because of symmetry, that area is the same in all four quadrants. | ||
| − | The answer is <math>2 + \pi</math> | + | The answer is <math>\boxed{\textbf{(B)}\ 2 + \pi}</math> |
==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}} | ||
Revision as of 20:13, 21 February 2016
Problem
What is the area of the region enclosed by the graph of the equation 
Solution
Consider the case when 
Find the area of this circle in the first quadrant. Notice the circle intersect the axe at point 
:
Because of symmetry, that 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. 
