2016 AMC 10B Problem 21

Question

What is the area of the region enclosed by the graph of the equation

x^2+y^2=|x|+|y|?

\ \pi+\sqrt{2}

\ \pi+2

\ \pi+2\sqrt{2}

\ 2\pi+\sqrt{2}

\ 2\pi+2\sqrt{2}

Work in one quadrant and multiply by

4

The first-quadrant region is a triangle plus a semicircle

The equation is symmetric in all four quadrants. In the first quadrant it becomes

x^2+y^2=x+y,

or

(x-\tfrac12)^2+(y-\tfrac12)^2=\tfrac12.

In the first quadrant, the enclosed region is the triangle under

x+y=1

, with area

\frac12

, plus a semicircle of radius

\sqrt{\frac{1}{2}}

, with area

\frac\pi4

.

Multiplying by

4

, the total area is

4\left(\frac12+\frac\pi4\right)=2+\pi.

Thus, the correct answer is B.

Problem 21 in other years