2016 AMC 12B Problem 18

Question

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

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

\pi+\sqrt2

\pi+2

\pi+2\sqrt2

2\pi+\sqrt2

2\pi+2\sqrt2

The graph is symmetric about both axes; in the first quadrant the equation is

x^2+y^2=x+y

Complete the square to

\left(x-\tfrac12\right)^2+\left(y-\tfrac12\right)^2 =\tfrac12;

in that quadrant the region is a right triangle plus a semicircle

By symmetry, consider the first quadrant, where the equation is

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

or

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

This is a circle centered at

\left(\tfrac12,\tfrac12\right)

passing through

(1,0)

and

(0,1);

since the center is the midpoint of that chord, the enclosed first-quadrant region is the right triangle with legs to

(1,0)

and

(0,1)

(area

\tfrac12

) plus a semicircle of radius

\dfrac{\sqrt2}{2}

(area

\dfrac\pi4

). Multiplying by

4

for all quadrants gives

4\left(\tfrac12+\tfrac\pi4\right)=\pi+2.

Thus, the correct answer is B.

Problem 18 in other years