AMC10 2018 B

AMC10 2018 B · Q7

AMC10 2018 B · Q7. It mainly tests Ratios & proportions, Area & perimeter.

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 $N$?

如下图，沿着一个大半圆的直径画了$N$个全等的半圆，它们的直径覆盖了大半圆的直径，没有重叠。设$A$为小半圆的总面积，$B$为大半圆内部但小半圆外部的区域面积。比例$A:B=1:18$。$N$是多少？

(A) 16 16

(B) 17 17

(D) 19 19

(E) 36 36

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Suppose without loss of generality that each small semicircle has radius 1; then the large semicircle has radius $N$. The area of each small semicircle is $\frac{\pi}{2}$, and the area of the large semicircle is $N^2\cdot\frac{\pi}{2}$. The combined area $A$ of the $N$ small semicircles is $N\cdot\frac{\pi}{2}$, and the area $B$ inside the large semicircle but outside the small semicircles is $N^2\cdot\frac{\pi}{2}-N\cdot\frac{\pi}{2}=(N^2-N)\cdot\frac{\pi}{2}$. Thus the ratio $A:B$ of the areas is $N:(N^2-N)$, which is $1:(N-1)$. Because this ratio is given to be $1:18$, it follows that $N-1=18$ and $N=19$.

答案（D）：不失一般性，设每个小半圆的半径为 $1$，则大半圆的半径为 $N$。每个小半圆的面积为 $\frac{\pi}{2}$，大半圆的面积为 $N^2\cdot\frac{\pi}{2}$。$N$ 个小半圆的总面积 $A$ 为 $N\cdot\frac{\pi}{2}$，而大半圆内但在小半圆外的面积 $B$ 为 $N^2\cdot\frac{\pi}{2}-N\cdot\frac{\pi}{2}=(N^2-N)\cdot\frac{\pi}{2}$。因此面积之比 $A:B$ 为 $N:(N^2-N)$，可化为 $1:(N-1)$。由于该比值给定为 $1:18$，可得 $N-1=18$，所以 $N=19$。

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