Difference between revisions of "2016 AMC 12B Problems/Problem 18"
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| − | Consider the case when <math>x > 0, y > 0</math>. | + | Consider the case when <math>x > 0</math>, <math> y > 0</math>. |
<math>x^2+y^2=x+y</math>. | <math>x^2+y^2=x+y</math>. | ||
<math>(x - \frac{1}{2})^2+(y - \frac{1}{2})^2=\frac{1}{2}</math>. | <math>(x - \frac{1}{2})^2+(y - \frac{1}{2})^2=\frac{1}{2}</math>. | ||
| − | 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 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: |
<math>\frac{\pi}{4} +\frac{1}{2}</math>. | <math>\frac{\pi}{4} +\frac{1}{2}</math>. | ||
Because of symmetry, the area is the same in all four quadrants. | Because of symmetry, the area is the same in all four quadrants. | ||
Revision as of 20:25, 21 February 2016
Problem
What is the area of the region enclosed by the graph of the equation 
Solution
Consider the case when 
, 
.
.
.
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:
.
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. 
