1965 AMC 12 第 36 题

先试着解答 1965 AMC 12 第 36 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 1965 AMC 12 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

36.

给定两条不同的直线 OAOA 和 OBOB。从 OAOA 上一点向 OBOB 作垂线,再从该垂线的垂足向 OAOA 作垂线。从第二条垂线的垂足再向 OBOB 作垂线,如此无限继续。第一、第二条垂线段的长度分别为 aa 和 bb;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 OAOA and OB.OB. From a point in OAOA a perpendicular is drawn to OB;OB; from the foot of this perpendicular a line is drawn perpendicular to OA.OA. From the foot of this second perpendicular a line is drawn perpendicular to OB;OB; and so on indefinitely. The lengths of the first and second perpendiculars are aa and b,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:
ba−b\dfrac b{a-b}
aa−b\dfrac a{a-b}
aba−b\dfrac{ab}{a-b}
b2a−b\dfrac{b^2}{a-b}
a2a−b\dfrac{a^2}{a-b}
答案:E
知识点:等比数列相似求和
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 aa、公比为 ba\frac{b}{a} 的等比数列
The perpendicular lengths form a geometric sequence with first term aa and ratio ba\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,ba,b, 公比为 ba\frac{b}{a}。 收敛意味着 0<ba<10\lt \frac{b}{a}\lt1, 因而总和为 a1−ba=a2a−b。\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,a,b, the common ratio is ba.\frac{b}{a}. Convergence implies 0<ba<1,0\lt \frac{b}{a}\lt1, and the sum is a1−ba=a2a−b.\frac{a}{1-\frac{b}{a}}=\frac{a^2}{a-b}.
Therefore, the correct answer is E.
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