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

(Solution)
(Solution)
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==Solution==
 
==Solution==
Consider the case when <math>x > 0, y > 0</math>
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Consider the case when <math>x > 0, y > 0</math>
<math>x^2+y^2=x+y</math>
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<math>x^2+y^2=x+y</math>
<math>(x - 0.5)^2+(y - 0.5)^2=0.5</math>
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<math>(x - 0.5)^2+(y - 0.5)^2=0.5</math>
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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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> :
 
<math>0.5 + 0.25\pi</math>
 
<math>0.5 + 0.25\pi</math>
 
Because of symmetry, that area is the same in all four quadrants.
 
Because of symmetry, that area is the same in all four quadrants.
The answer is <math>2 + \pi</math> B
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The answer is <math>\boxed{\textbf{(B)}\ 2 + \pi}</math>
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2016|ab=B|num-b=17|num-a=19}}
 
{{AMC12 box|year=2016|ab=B|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:13, 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 $x > 0, y > 0.$ $x^2+y^2=x+y.$ $(x - 0.5)^2+(y - 0.5)^2=0.5.$ Find the area of this circle in the first quadrant. Notice the circle intersect the axe at point $(0, 1) and (1, 0).$ : $0.5 + 0.25\pi$ Because of symmetry, that area is the same in all four quadrants. The answer is $\boxed{\textbf{(B)}\ 2 + \pi}$

See Also

2016 AMC 12B (ProblemsAnswer KeyResources)
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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