Question

下图中，外面的正方形内含有无限多个正方形，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 $0 \lt k \lt 1$ 。相邻正方形之间的区域交替涂色，如图所示（图不一定按比例绘制）。

阴影部分的面积是原正方形面积的 $64\%$ 。 $k$ 等于多少？

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：1560

解答：

设外层正方形边长为 $1$ 各正方形边长为 $1, k, k^2, \ldots$ 而阴影环带交替出现，所以阴影面积为 $1 - k^2 + k^4 - k^6 + \cdots = \frac{1}{1 + k^2}$ 题目给出这等于 $64\% = \frac{16}{25}$ 所以 $1 + k^2 = \frac{25}{16}$ 因此 $k^2 = \frac{9}{16}$ 得到 $k = \frac{3}{4}$ 所以正确答案是 D。

Let the outer side be $1.$ The squares have sides $1, k, k^2, \ldots,$ and the shaded rings alternate, so the shaded area is $1 - k^2 + k^4 - k^6 + \cdots = \frac{1}{1 + k^2}.$ We're told this equals $64\% = \frac{16}{25},$ so $1 + k^2 = \frac{25}{16}.$ Then $k^2 = \frac{9}{16},$ giving $k = \frac{3}{4}.$ Thus, D is the correct answer.

2025 AMC 10A 第 13 题

答案：D

难度评级：1560

解答：

其他年份的第 13 题