2018 AMC 10B 第 7 题

Question

下图中，沿着一个大半圆的直径画了

N

个全等的小半圆，它们的直径无重叠地覆盖大半圆的直径。设

A

为这些小半圆面积之和，

B

为大半圆内部但小半圆外部区域的面积。若

A : B

为

1 : 18

，求

N

。

In the figure below,

N

congruent semicircles are drawn along a diameter of a large semicircle, with their diameters covering the diameter of the large semicircle with no overlap. Let

A

be the combined area of the small semicircles and

B

be the area of the region inside the large semicircle but outside the small semicircles. The ratio

A : B

is

1 : 18.

What is

N?

16

17

18

19

36

答案：D

知识点：圆面积比与比例

难度评级：1310

小提示：

若每个小半圆半径为

r

，则大半圆半径为

Nr

。

If each small semicircle has radius

r,

the large semicircle has radius

Nr

大提示：

证明

A : B = 1 : (N-1)

，再令它等于

1 : 18

。

Show

A : B = 1 : (N-1),

then set this equal to

1 : 18

解答：

设每个小半圆半径为

r

。因为

N

个小直径覆盖大直径，所以大半圆半径为

Nr

。小半圆面积之和为

A = N \cdot \tfrac12 \pi r^2

，大半圆面积为

\tfrac12 \pi (Nr)^2

，所以剩余区域面积为

B = \tfrac12 \pi r^2(N^2 - N)

。因此

A : B = N : N(N-1)

= 1 : (N-1)

。令

N - 1 = 18

，得

N = 19

。正确答案是 D。

Let each small semicircle have radius

r.

The

N

diameters cover the big diameter, so the large radius is

Nr.

Then

A = N \cdot \tfrac12 \pi r^2,

and the large semicircle has area

\tfrac12 \pi (Nr)^2,

so the leftover region is

B = \tfrac12 \pi r^2(N^2 - N).

This gives

A : B = N : N(N-1)

= 1 : (N-1).

Set

N - 1 = 18,

and

N = 19.

Thus, D is the correct answer.

2018 AMC 10B 第 7 题

答案：D

难度评级：1310

小提示：

大提示：

解答：

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