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

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

解答：

其他年份的第 18 题