2016 AMC 10B Problem 21

Below is the professionally curated solution for Problem 21 of the 2016 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2016 AMC 10B solutions, or check the answer key.

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Concepts:absolute valuecirclearea decomposition

Difficulty rating: 1860

21.

What is the area of the region enclosed by the graph of the equation x2+y2=x+y?x^2+y^2=|x|+|y|?

 π+2 \ \pi+\sqrt{2}

 π+2 \ \pi+2

 π+22 \ \pi+2\sqrt{2}

 2π+2 \ 2\pi+\sqrt{2}

 2π+22 \ 2\pi+2\sqrt{2}

Solution:

The equation is symmetric in all four quadrants. In the first quadrant it becomes x2+y2=x+y,x^2+y^2=x+y, or (x12)2+(y12)2=12.(x-\tfrac12)^2+(y-\tfrac12)^2=\tfrac12.

In the first quadrant, the enclosed region is the triangle under x+y=1x+y=1, with area 12\frac12, plus a semicircle of radius 1/2\sqrt{1/2}, with area π4\frac\pi4.

Multiplying by 44, the total area is 4(12+π4)=2+π.4\left(\frac12+\frac\pi4\right)=2+\pi.

Thus, the correct answer is B.

Problem 21 in Other Years