Difference between revisions of "2016 AMC 12B Problems/Problem 18"

Revision as of 21:27, 21 February 2016

Problem

What is the area of the region enclosed by the graph of the equation $x^2+y^2=|x|+|y|?$

$\textbf{(A)}\ \pi+\sqrt{2} \qquad\textbf{(B)}\ \pi+2 \qquad\textbf{(C)}\ \pi+2\sqrt{2} \qquad\textbf{(D)}\ 2\pi+\sqrt{2} \qquad\textbf{(E)}\ 2\pi+2\sqrt{2}$

Solution

Consider the case when $x > 0$ , $y > 0$ . $x^2+y^2=x+y$ $(x - \frac{1}{2})^2+(y - \frac{1}{2})^2=\frac{1}{2}$ Notice the circle intersect the axes at points $(0, 1)$ and $(1, 0)$ . Find the area of this circle in the first quadrant. The area is made of a semi-circle with radius of $\frac{\sqrt{2}}{2}$ and a triangle: $A = \frac{\pi}{4} +\frac{1}{2}$ Because of symmetry, the area is the same in all four quadrants. The answer is $\boxed{\textbf{(B)}\ \pi + 2}$

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.

@@ Line 10: / Line 10: @@
 <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:
-<cmath>\frac{\pi}{4} +\frac{1}{2}</cmath>
+<cmath>A = \frac{\pi}{4} +\frac{1}{2}</cmath>
 Because of symmetry, the area is the same in all four quadrants.
 The answer is <math>\boxed{\textbf{(B)}\ \pi + 2}</math>

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Difference between revisions of "2016 AMC 12B Problems/Problem 18"

Revision as of 21:27, 21 February 2016

Problem

Solution

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