2025 AMC 12A 第 5 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

5.

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 kk,其中 0<k<10 \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%64\%kk 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,k, where 0<k<1.0 \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%64\% of the area of the original square. What is k?k?

35\dfrac{3}{5}

1625\dfrac{16}{25}

23\dfrac{2}{3}

34\dfrac{3}{4}

45\dfrac{4}{5}

答案:D
知识点:等比数列面积比
难度评级:1270
小提示:

设外层正方形面积为 11;这些正方形的面积为 1,k2,k4,1, k^2, k^4, \ldots

Take the outer square to have area 11; the squares have areas 1,k2,k4,1, k^2, k^4, \ldots

大提示:

阴影环带构成一个等比级数,其和为 11+k2\dfrac{1}{1+k^2}

The shaded rings form a geometric series that sums to 11+k2\dfrac{1}{1+k^2}

解答:

设外层正方形面积为 11。嵌套正方形的面积为 1,k2,k4,1, k^2, k^4, \ldots,因此第 nn 个正方形与第 (n+1)(n+1) 个正方形之间的环带面积为 k2n(1k2)k^{2n}(1-k^2)

阴影环带是交替的那些,即 n=0,2,4,n = 0, 2, 4, \ldots,总面积为 j=0k4j(1k2)=1k21k4=11+k2 \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}\text{。}

11+k2=1625\dfrac{1}{1+k^2} = \dfrac{16}{25},得 1+k2=25161 + k^2 = \dfrac{25}{16},所以 k2=916k^2 = \dfrac{9}{16}k=34k = \dfrac{3}{4}

因此,正确答案是 D

Let the outer square have area 1.1. The nested squares have areas 1,k2,k4,,1, k^2, k^4, \ldots, so the ring between the nnth and (n+1)(n+1)th squares has area k2n(1k2).k^{2n}(1-k^2).

The shaded rings are the alternate ones n=0,2,4,,n = 0, 2, 4, \ldots, with total area j=0k4j(1k2)=1k21k4=11+k2. \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 11+k2=1625\dfrac{1}{1+k^2} = \dfrac{16}{25} gives 1+k2=2516,1 + k^2 = \dfrac{25}{16}, so k2=916k^2 = \dfrac{9}{16} and k=34.k = \dfrac{3}{4}.

Thus, the correct answer is D.

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