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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</math>, <math> y > 0</math>.  
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Consider the case when <math>x \geq 0</math>, <math> y \geq 0</math>.  
 
<cmath>x^2+y^2=x+y</cmath>  
 
<cmath>x^2+y^2=x+y</cmath>  
 
<cmath>(x - \frac{1}{2})^2+(y - \frac{1}{2})^2=\frac{1}{2}</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 intersects 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>A = \frac{\pi}{4} +\frac{1}{2}</cmath>  
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<cmath>A = \frac{\pi}{4} +\frac{1}{2}</cmath>
−
<asy>draw((0,-1.5)--(0,1.5),EndArrow);draw((-1.5,0)--(1.5,0),EndArrow);draw((0,1)--(1,0)--(0,-1)--(-1,0)--cycle,dotted);for(int i=0;i<4;++i){draw(rotate(i*90,(0,0))*arc((1/2,1/2),sqrt(2)/2,-45,135)); draw(rotate(i*90,(1/2,1/2))*arc((1/2,1/2),sqrt(2)/2,135, 315),dotted);dot(rotate(i*90,(0,0))*(1/2,1/2));}</asy>
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−
Because of symmetry, the area is the same in all four quadrants.
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<asy>draw((0,-1.5)--(0,1.5),EndArrow);draw((-1.5,0)--(1.5,0),EndArrow);draw((0,1)--(1,0)--(0,-1)--(-1,0)--cycle,dotted);draw(arc((1/2,1/2),sqrt(2)/2,135, 315),dotted);for(int i=0;i<4;++i){draw(rotate(i*90,(0,0))*arc((1/2,1/2),sqrt(2)/2,-45,135)); dot(rotate(i*90,(0,0))*(1/2,1/2));}</asy>
−
The answer is <math>\boxed{\textbf{(B)}\ \pi + 2}</math>
+
 
 +
Because of symmetry, the area is the same in all four quadrants. The answer is <math>\boxed{\textbf{(B)}\ \pi + 2}</math>
 +
 
 +
==Video Solution by CanadaMath (Problem 11-20)==
 +
Fast Forward to 33:32 for problem 18
 +
https://www.youtube.com/watch?v=4osvFatUv1o
 +
 
 +
~THEMATHCANADIAN
 +
 
  
 
==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}}

Latest revision as of 00:30, 10 November 2024

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 $x \geq 0$, $y \geq 0$. \[x^2+y^2=x+y\] \[(x - \frac{1}{2})^2+(y - \frac{1}{2})^2=\frac{1}{2}\] Notice the circle intersects 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: \[A = \frac{\pi}{4} +\frac{1}{2}\]

[asy]draw((0,-1.5)--(0,1.5),EndArrow);draw((-1.5,0)--(1.5,0),EndArrow);draw((0,1)--(1,0)--(0,-1)--(-1,0)--cycle,dotted);draw(arc((1/2,1/2),sqrt(2)/2,135, 315),dotted);for(int i=0;i<4;++i){draw(rotate(i*90,(0,0))*arc((1/2,1/2),sqrt(2)/2,-45,135)); dot(rotate(i*90,(0,0))*(1/2,1/2));}[/asy]

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

Video Solution by CanadaMath (Problem 11-20)

Fast Forward to 33:32 for problem 18 https://www.youtube.com/watch?v=4osvFatUv1o

~THEMATHCANADIAN


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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