Question

如图，在一组等距平行线上画出一个角。阴影区域 $\mathcal{C}$ 的面积与阴影区域 $\mathcal{B}$ 的面积之比为 $\frac{11}{5}$ 。求阴影区域 $\mathcal{D}$ 的面积与阴影区域 $\mathcal{A}$ 的面积之比。

An angle is drawn on a set of equally spaced parallel lines as shown. The ratio of the area of shaded region $\mathcal{C}$ to the area of shaded region $\mathcal{B}$ is $\frac{11}{5}.$ Find the ratio of the area of shaded region $\mathcal{D}$ to the area of shaded region $\mathcal{A}.$

答案

答案：408

知识点：相似面积比平方差

难度评级：2500

解答：

取相邻平行线的间距为单位长度，令 $x$ 为角的顶点到第一条线的距离。第 $j$ 条线截出的三角形与 $\mathcal{A}$ 相似，相似比为 $\frac{x + j - 1}{x}$ ，所以其面积与 $(x + j - 1)^2$ 成正比，而第 $j$ 条线与第 $j + 1$ 条线之间的条带面积与 $(x + j)^2$ $- (x + j - 1)^2 = 2x + 2j - 1$ 成正比。

区域 $\mathcal{B}$ 、 $\mathcal{C}$ 、 $\mathcal{D}$ 分别是从第 $2$ 、 $4$ 、 $6$ 条线开始的条带，面积分别与 $2x + 3$ 、 $2x + 7$ 、 $2x + 11$ 成正比。已知比值得 $\frac{2x+7}{2x+3} = \frac{11}{5}$ ，所以 $10x + 35 = 22x + 33$ ，从而 $x = \frac{1}{6}$ 。

因为 $\mathcal{A}$ 的面积与 $x^2$ 成正比，所求比值为 $\frac{2x + 11}{x^2} = \frac{34/3}{1/36} = 408$ 。

Take the spacing between consecutive lines as the unit, and let $x$ be the distance from the vertex of the angle to the first line. The triangle cut off by the $j$ th line is similar to triangle $\mathcal{A}$ with ratio $\frac{x + j - 1}{x},$ so its area is proportional to $(x + j - 1)^2,$ and the strip between lines $j$ and $j + 1$ has area proportional to $(x + j)^2$ $- (x + j - 1)^2 = 2x + 2j - 1.$

Regions $\mathcal{B},$ $\mathcal{C},$ and $\mathcal{D}$ are the strips beginning at lines $2,$ $4,$ and $6,$ with areas proportional to $2x + 3,$ $2x + 7,$ and $2x + 11.$ The given ratio yields $\frac{2x+7}{2x+3} = \frac{11}{5},$ so $10x + 35 = 22x + 33$ and $x = \frac{1}{6}.$

Since $\mathcal{A}$ has area proportional to $x^2,$ the requested ratio is $\frac{2x + 11}{x^2} = \frac{34/3}{1/36} = 408.$

2006 AIME I 第 7 题

答案：408

难度评级：2500

解答：

其他年份的第 7 题