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

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==Solution==
 
==Solution==
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Consider the case when <math>x > 0, y > 0.  </math>
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Consider the case when <math>x > 0, y > 0</math>.
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<math>x^2+y^2=x+y.  </math>
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<math>x^2+y^2=x+y</math>.
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<math>(x - 0.5)^2+(y - 0.5)^2=0.5.  </math>
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<math>(x - \frac{1}{2})^2+(y - \frac{1}{2})^2=\frac{1}{2}</math>.
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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> :
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Notice the circle intersect the axe at point <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:
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<math>0.5 + 0.25\pi</math>
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<math>\frac{çpi}{4} +\frac{1}{2}</math>
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Because of symmetry, that area is the same in all four quadrants.
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Because of symmetry, the area is the same in all four quadrants.
 
The answer is <math>\boxed{\textbf{(B)}\ 2 + \pi}</math>
 
The answer is <math>\boxed{\textbf{(B)}\ 2 + \pi}</math>
  

Revision as of 20:21, 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 axe at point $(0, 1)$ and $(1, 0)$. Find the area of this circle in the first quadrant. The area is made of a semi-circle and a triangle: $\frac{çpi}{4} +\frac{1}{2}$ (Error compiling LaTeX. Unknown error_msg) Because of symmetry, the area is the same in all four quadrants. The answer is $\boxed{\textbf{(B)}\ 2 + \pi}$

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

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