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

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

解答：

其他年份的第 7 题