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

Revision as of 21:57, 9 January 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 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: $A = \frac{\pi}{4} +\frac{1}{2}$ $[asy]draw((0,-1.5)--(0,1.5),EndArrow);draw((-1.5,0)--(1.5,0),EndArrow);draw((0,1)--(1,0)--(0,-1)--(-1,0)--cycle,dotted);draw(arc((1/2,1/2),sqrt(2)/2,135, 315),dotted);for(int i=0;i<4;++i){draw(rotate(i*90,(0,0))*arc((1/2,1/2),sqrt(2)/2,-45,135)); dot(rotate(i*90,(0,0))*(1/2,1/2));}[/asy]$ Because of symmetry, the area is the same in all four quadrants. The answer is $\boxed{\textbf{(B)}\ \pi + 2}$

2016 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 14: / Line 14: @@
 Because of symmetry, the area is the same in all four quadrants.
 The answer is <math>\boxed{\textbf{(B)}\ \pi + 2}</math>
-===Note===
 ==See Also==
 {{AMC12 box|year=2016|ab=B|num-b=17|num-a=19}}
 {{MAA Notice}}

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

Revision as of 21:57, 9 January 2024

Problem

Solution

See Also