Question

在下图中，外面的正方形包含无限多个正方形，每个正方形都有相同的中心，且边都平行于外面的正方形。一个正方形的边长与下一个内层正方形的边长之比为 $k$ ，其中 $0 \lt k \lt 1$ 。正方形之间的区域如图所示交替涂色（图不一定按比例绘制）。

图中阴影部分的面积是原正方形面积的 $64\%$ 。 $k$ 是多少？

In the figure below, the outside square contains infinitely many squares, each of them with the same center and sides parallel to the outside square. The ratio of the side length of a square to the side length of the next inner square is $k,$ where $0 \lt k \lt 1.$ The spaces between squares are alternately shaded, as shown in the figure (which is not necessarily drawn to scale).

The area of the shaded portion of the figure is $64\%$ of the area of the original square. What is $k?$

$\dfrac{3}{5}$

$\dfrac{16}{25}$

$\dfrac{2}{3}$

$\dfrac{3}{4}$

$\dfrac{4}{5}$

答案：D

知识点：等比数列面积比

难度评级：1270

解答：

设外层正方形面积为 $1$ 。嵌套正方形的面积为 $1, k^2, k^4, \ldots$ ，因此第 $n$ 个正方形与第 $(n+1)$ 个正方形之间的环带面积为 $k^{2n}(1-k^2)$ 。

阴影环带是交替的那些，即 $n = 0, 2, 4, \ldots$ ，总面积为 $\begin{aligned} \sum_{j=0}^{\infty} k^{4j}(1-k^2) &= \frac{1-k^2}{1-k^4} \\ &= \frac{1}{1+k^2}. \end{aligned}$

令 $\dfrac{1}{1+k^2} = \dfrac{16}{25}$ ，得 $1 + k^2 = \dfrac{25}{16}$ ，所以 $k^2 = \dfrac{9}{16}$ ， $k = \dfrac{3}{4}$ 。

因此，正确答案是 D。

Let the outer square have area $1.$ The nested squares have areas $1, k^2, k^4, \ldots,$ so the ring between the $n$ th and $(n+1)$ th squares has area $k^{2n}(1-k^2).$

The shaded rings are the alternate ones $n = 0, 2, 4, \ldots,$ with total area $\begin{aligned} \sum_{j=0}^{\infty} k^{4j}(1-k^2) &= \frac{1-k^2}{1-k^4} \\ &= \frac{1}{1+k^2}. \end{aligned}$

Setting $\dfrac{1}{1+k^2} = \dfrac{16}{25}$ gives $1 + k^2 = \dfrac{25}{16},$ so $k^2 = \dfrac{9}{16}$ and $k = \dfrac{3}{4}.$

Thus, the correct answer is D.

2025 AMC 12A 第 5 题

答案：D

难度评级：1270

解答：

其他年份的第 5 题