2016 AMC 12B 第 18 题

Question

方程

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

的图像所围成区域的面积是多少？

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

答案：B

知识点：配方法对称性圆面积

难度评级：1990

小提示：

图像关于两条坐标轴对称；在第一象限中方程为

x^2+y^2=x+y

。

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

x^2+y^2=x+y

大提示：

配方得

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

；第一象限内的区域是一个直角三角形加一个半圆。

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

解答：

由对称性，只看第一象限，此时方程为

x^2+y^2=x+y

，即

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

。这是一个圆心在

\left(\tfrac12,\tfrac12\right)

、经过

(1,0)

和

(0,1)

的圆。由于圆心是连接

(1,0)

与

(0,1)

的弦的中点，第一象限内围成的区域由面积为

\tfrac12

的直角三角形和半径为

\dfrac{\sqrt2}{2}

、面积为

\dfrac\pi4

的半圆组成。乘以

4

个象限，总面积为

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

。

所以正确答案是 B。

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.

2016 AMC 12B 第 18 题

答案：B

难度评级：1990

小提示：

大提示：

解答：

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