Difference between revisions of "2016 AMC 12B Problems/Problem 18"
m (→Solution) |
m (→Solution) |
||
Line 7: | Line 7: | ||
==Solution== | ==Solution== | ||
Consider the case when <math>x > 0</math>, <math> y > 0</math>. | Consider the case when <math>x > 0</math>, <math> y > 0</math>. | ||
− | < | + | <cmath>x^2+y^2=x+y</cmath> |
− | < | + | <cmath>(x - \frac{1}{2})^2+(y - \frac{1}{2})^2=\frac{1}{2}</cmath> |
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>\frac{\pi}{4} +\frac{1}{2}</cmath> |
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 21:26, 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.