2023 AMC 8 第 12 题

Question

下图显示一个大的未涂色圆，内部有若干较小的未涂色圆和阴影圆。大未涂色圆内部的面积中，阴影部分占几分之几？

The figure below shows a large unshaded circle with a number of smaller unshaded and shaded circles in its interior. What fraction of the interior of the large unshaded circle is shaded?

\dfrac{1}{4}

\dfrac{11}{36}

\dfrac{1}{3}

\dfrac{19}{36}

\dfrac{5}{9}

答案：B

知识点：圆面积面积比

难度评级：1370

小提示：

圆面积与半径的平方成正比

Circle areas are proportional to the square of the radius

大提示：

用最小圆的面积作为单位来数阴影面积

Count shaded area in units of the smallest circle’s area

视频讲解：

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文字解答：

不妨设每个最小圆的半径为

1

。

这样有

3

个阴影单位圆，总面积为

3 \cdot 1^2 \pi = 3 \pi

。

另外还有一个半径为

4

的阴影圆，其中挖去了两个半径为

2

的未涂色圆。

这部分额外阴影面积为

\begin{align*} 4^2 \pi - 2 \cdot 2^2 \pi &= 16 \pi - 8 \pi \\ &= 8 \pi\text{。} \end{align*}

阴影总面积为

8\pi + 3\pi = 11\pi\text{。}

大未涂色圆的面积为

6^2\pi = 36\pi\text{。}

所以所求分数为

\dfrac{11\pi}{36\pi} = \dfrac{11}{36}\text{。}

所以正确答案是 B。

Let each smallest circle have radius

1.

This means that there are

3

shaded unit circles, which total to

3 \cdot 1^2 \pi = 3 \pi

area.

There is also a shaded circle with radius

4

with two unshaded circles of radius

2

inside.

This gives us an extra shaded area of

\begin{align*} 4^2 \pi - 2 \cdot 2^2 \pi &= 16 \pi - 8 \pi \\ &= 8 \pi. \end{align*}

The total shaded area is

8\pi + 3\pi = 11\pi.

The area of the large unshaded circle is

6^2\pi = 36\pi.

Therefore, the desired fraction is

\dfrac{11\pi}{36\pi} = \dfrac{11}{36}.

Thus, B is the correct answer.