2016 AMC 12B Problems/Problem 18

Revision as of 21:25, 21 February 2016 by Ychen (talk | contribs) (Solution)

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: $\frac{\pi}{4} +\frac{1}{2}$. Because of symmetry, the area is the same in all four quadrants. The answer is $\boxed{\textbf{(B)}\ \pi + 2}$

See Also

2016 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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All AMC 12 Problems and Solutions

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