Question

给定两条不同的直线 $OA$ 和 $OB$ 。从 $OA$ 上一点向 $OB$ 作垂线，再从该垂线的垂足向 $OA$ 作垂线。从第二条垂线的垂足再向 $OB$ 作垂线，如此无限继续。第一、第二条垂线段的长度分别为 $a$ 和 $b$ ；当垂线条数无限增加时，各垂线段长度之和趋于一个极限。此极限为：

Given distinct straight lines $OA$ and $OB.$ From a point in $OA$ a perpendicular is drawn to $OB;$ from the foot of this perpendicular a line is drawn perpendicular to $OA.$ From the foot of this second perpendicular a line is drawn perpendicular to $OB;$ and so on indefinitely. The lengths of the first and second perpendiculars are $a$ and $b,$ respectively. Then the sum of the lengths of the perpendiculars approaches a limit as the number of perpendiculars grows beyond all bounds. This limit is:

$\dfrac b{a-b}$

$\dfrac a{a-b}$

$\dfrac{ab}{a-b}$

$\dfrac{b^2}{a-b}$

$\dfrac{a^2}{a-b}$

答案：E

知识点：等比数列相似求和

难度评级：2190

小提示：

两条固定直线依次形成的直角三角形相似

Successive right triangles formed by the two fixed lines are similar

大提示：

垂线段长度构成首项为 $a$ 、公比为 $\frac{b}{a}$ 的等比数列

The perpendicular lengths form a geometric sequence with first term $a$ and ratio $\frac{b}{a}$

解答：

每个新的直角三角形都有相同的锐角，所以垂线段长度构成等比数列。由于前两项为 $a,b$ ，公比为 $\frac{b}{a}$ 。收敛意味着 $0\lt \frac{b}{a}\lt1$ ，因而总和为 $\frac{a}{1-\frac{b}{a}}=\frac{a^2}{a-b}\text{。}$

因此，正确答案是 E。

Each new right triangle has the same acute angle, so the perpendicular lengths form a geometric sequence. Since the first two lengths are $a,b,$ the common ratio is $\frac{b}{a}.$ Convergence implies $0\lt \frac{b}{a}\lt1,$ and the sum is $\frac{a}{1-\frac{b}{a}}=\frac{a^2}{a-b}.$

Therefore, the correct answer is E.

1965 AMC 12 第 36 题

答案：E

难度评级：2190

小提示：

大提示：

解答：

其他年份的第 36 题