AMC12 2016 B

AMC12 2016 B · Q18

AMC12 2016 B · Q18. It mainly tests Circle theorems, Area & perimeter.

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

由方程 $x^2 + y^2 = |x| + |y|$ 的图像所围成的区域面积是多少？

(A) \pi + \sqrt{2} \pi + \sqrt{2}

(B) \pi + 2 \pi + 2

(D) 2\pi + \sqrt{2} 2\pi + \sqrt{2}

(E) 2\pi + 2\sqrt{2} 2\pi + 2\sqrt{2}

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): The graph of the equation is symmetric about both axes. In the first quadrant, the equation is equivalent to $x^2+y^2-x-y=0$. Completing the square gives $(x-\frac12)^2+(y-\frac12)^2=\frac12$, so the graph in the first quadrant is an arc of the circle that is centered at $C(\frac12,\frac12)$ and contains the points $A(1,0)$ and $B(0,1)$. Because $C$ is the midpoint of $\overline{AB}$, the arc is a semicircle. The region enclosed by the graph in the first quadrant is the union of isosceles right triangle $AOB$, where $O(0,0)$ is the origin, and a semicircle with diameter $\overline{AB}$. The triangle and the semicircle have areas $\frac12$ and $\frac12\cdot\pi\left(\frac{\sqrt2}{2}\right)^2=\frac{\pi}{4}$, respectively, so the area of the region enclosed by the graph in all quadrants is $4\left(\frac12+\frac{\pi}{4}\right)=\pi+2$.

答案（B）：该方程的图像关于两条坐标轴都对称。在第一象限中，方程等价于 $x^2+y^2-x-y=0$。配方得 $(x-\frac12)^2+(y-\frac12)^2=\frac12$，因此第一象限内的图像是一段圆弧，该圆以 $C(\frac12,\frac12)$ 为圆心，并经过点 $A(1,0)$ 与 $B(0,1)$。由于 $C$ 是线段 $\overline{AB}$ 的中点，这段圆弧为一条半圆弧。第一象限内由图像围成的区域是等腰直角三角形 $AOB$（其中 $O(0,0)$ 为原点）与以 $\overline{AB}$ 为直径的半圆的并集。三角形与半圆的面积分别为 $\frac12$ 和 $\frac12\cdot\pi\left(\frac{\sqrt2}{2}\right)^2=\frac{\pi}{4}$，所以所有象限内由该图像围成区域的面积为 $4\left(\frac12+\frac{\pi}{4}\right)=\pi+2$。

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