AMC12 2025 A

AMC12 2025 A · Q5

AMC12 2025 A · Q5. It mainly tests Exponents & radicals, Area & perimeter.

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 $0 < k < 1.$ 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 $k?$

下图中，外部正方形包含无限多个正方形，每个正方形有相同的中心且边与外部正方形平行。相邻正方形的边长比为 $k$，其中 $0 < k < 1$。正方形之间的空间交替着色，如图所示（图未按比例绘制）。着色部分的面积是原正方形面积的64%。$k$ 等于多少？

(A) \frac 35 \frac 35

(B) \frac {16}{25} \frac {16}{25}

(D) \frac 34 \frac 34

(E) \frac 45 \frac 45

Answer

Correct choice: (D)

正确答案：（D）

Solution

Let the side length of the largest square be $a,$ so it has area $a^2.$ Hence, the second-largest square has area $a^2k^2,$ the third-largest has $a^2k^4,$ and so on. It follows that the total shaded area is \[a^2-a^2k^2+a^2k^4-a^2k^6+...=a^2(1-k^2+k^4-k^6+...)=a^2\dfrac{1}{1+k^2}.\] The ratio of the area of the shaded region to that of the original square is then \[\dfrac{a^2\frac{1}{1+k^2}}{a^2}=\dfrac{1}{1+k^2}=\dfrac{64}{100}\] \[\implies 64+64k^2=100\implies k^2=\dfrac{36}{64}\implies k=\boxed{\text{(D) }\dfrac{3}{4}}.\] We can just let $a=1$ because the question deals with ratios, meaning that there wouldn't be a loss of generality if we let the side length equal some value, getting the same answer $\boxed{D}$. This was done in solution 3.

设最大正方形的边长为 $a$，面积为 $a^2$。因此，第二大正方形面积为 $a^2k^2$，第三大为 $a^2k^4$，依此类推。着色总面积为 \[a^2-a^2k^2+a^2k^4-a^2k^6+...=a^2(1-k^2+k^4-k^6+...)=a^2\dfrac{1}{1+k^2}.\] 着色区域面积与原正方形面积之比为 \[\dfrac{a^2\frac{1}{1+k^2}}{a^2}=\dfrac{1}{1+k^2}=\dfrac{64}{100}\] \[\implies 64+64k^2=100\implies k^2=\dfrac{36}{64}\implies k=\boxed{\text{(D) }\dfrac{3}{4}}.\] 我们可以令 $a=1$，因为问题是关于比率的，令边长为某值不会丢失一般性，得到相同答案 $\boxed{D}$。这是解决方案3中的做法。

Topics