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Difference between revisions of "2016 AMC 10B Problems/Problem 21"

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<math>\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}</math>
 
<math>\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}</math>
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==Solution==
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WLOG note that if a point in the first quadrant satisfies the equation, so do its corresponding points in the other three quadrants. Therefore we can assume that <math>x, y \ge 0</math> and multiply by <math>4</math> at the end.
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We can rearrange the equation to get <math>x^2-x+y^2-y=0 \Rightarrow (x-\tfrac12)^2+(y-\tfrac12)^2=(\tfrac{\sqrt2}{2})^2</math>, which describes a circle with center <math>(\tfrac12, \tfrac12)</math> and radius <math>\tfrac{\sqrt2}{2}.</math> It's clear we now want to find the union of four circles with overlap.
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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);
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for(int i=0;i<4;++i){draw(rotate(i*90,(0,0))*arc((1/2,1/2),sqrt(1/2),-45,135));dot(rotate(i*90,(0,0))*(1/2,1/2));}</asy>
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There are several ways to find the area, but note that if you connect <math>(0, 1)</math> to its other three respective points in the other three quadrants, you get a square of area <math>2</math>, along with four half-circles of diameter <math>\sqrt{2}</math>, for a total area of <math>2+2\cdot(\tfrac{\sqrt2}{2})^2\pi = \pi + 2</math> which is <math>\textbf{(B)}</math>.
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==See Also==
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{{AMC10 box|year=2016|ab=B|num-b=20|num-a=22}}
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{{MAA Notice}}

Revision as of 12:53, 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

WLOG note that if a point in the first quadrant satisfies the equation, so do its corresponding points in the other three quadrants. Therefore we can assume that $x, y \ge 0$ and multiply by $4$ at the end.

We can rearrange the equation to get $x^2-x+y^2-y=0 \Rightarrow (x-\tfrac12)^2+(y-\tfrac12)^2=(\tfrac{\sqrt2}{2})^2$, which describes a circle with center $(\tfrac12, \tfrac12)$ and radius $\tfrac{\sqrt2}{2}.$ It's clear we now want to find the union of four circles with overlap.

[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(1/2),-45,135));dot(rotate(i*90,(0,0))*(1/2,1/2));}[/asy] There are several ways to find the area, but note that if you connect $(0, 1)$ to its other three respective points in the other three quadrants, you get a square of area $2$, along with four half-circles of diameter $\sqrt{2}$, for a total area of $2+2\cdot(\tfrac{\sqrt2}{2})^2\pi = \pi + 2$ which is $\textbf{(B)}$.

See Also

2016 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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 10 Problems and Solutions

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