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标题: (1981 年美国奥赛题 ) 已知一角大小为 [打印本页]

作者: 野菜花 时间: 2005-2-18 21:53
标题: (1981 年美国奥赛题 ) 已知一角大小为

180/n, 其中 n 为不能被 3 整除的正整数。证明：这个角可以用欧几里德的作图工具 (园规和直尺 )三等分。

作者: fzy 时间: 2005-2-18 22:11
标题: 回复：(1981 年美国奥赛题 ) 已知一角大小为

If n = 3k + 1, then (180/n) * (2k+1) - 120 = 60/n.

If n = 3k + 2, then (180/n) * (k+1) - 60 = 60/n.

作者: 野菜花 时间: 2005-2-18 22:24
标题: So quick! PF! [:B][@};-]

So quick! PF!

作者: y 时间: 2005-2-19 00:48
标题: How to determine an angle can be trisected?

How to determine an angle can be trisected by a ruler and a compass? I don't understand your solution. Thanks

作者: 野菜花 时间: 2005-2-19 01:44
标题: 回复：How to determine an angle can be trisected?

Since angle 180/n is given, to trisect it, we need draw an angle of 60/n.

fry has found the following identities:

If n = 3k + 1, then (180/n) * (2k+1) - 120 = 60/n.

If n = 3k + 2, then (180/n) * (k+1) - 60 = 60/n.

As angle 180/n is given, we may draw angles with any multiple of this given angle, and to get an 60 degree angle, we only need to draw an equilateral triangle, thus we may also get 120 degree angle.

作者: y 时间: 2005-2-19 02:34
标题: 回复：回复：How to determine an angle can be trisected?

How can you draw an angle such as 60/5 or 60/7? Thanks.

作者: 野菜花 时间: 2005-2-19 02:49
标题: 回复：回复：回复：How to determine an angle can be trisecte

for this question?

作者: y 时间: 2005-2-19 03:28
标题: 回复：回复：回复：回复：How to determine an angle can be trise

Yes.

Form the previous post, you said:

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Since angle 180/n is given, to trisect it, we need draw an angle of 60/n.

（转自：顶顶华闻 www.TopChineseNews.com ）

fry has found the following identities:

（转自：顶顶华闻 www.TopChineseNews.com ）

If n = 3k + 1, then (180/n) * (2k+1) - 120 = 60/n.

If n = 3k + 2, then (180/n) * (k+1) - 60 = 60/n.

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Therefore, when k=1, n = 4 or 5. Whenk=2, n=7 or 8. 60/n would be 60/5 and 60/7. By the way, do you mean 180/n degree or it can be radius or both?

Thanks a lot.

作者: 野菜花 时间: 2005-2-19 03:55
标题: 回复：回复：回复：回复：回复：How to determine an angle can be tr

180/n, I meant degree.

when n=5, we are given an angle of 180/5=36 degree, we need draw a 12 degree angle,

36*2-60=12.

以已知角的顶点 O为园心，任意长为半径，画弧交已知角两边 A， B，以 B 为园心 AB长为半径，在 AB弧的延长弧上截取 C，连接 OC，这样角 AOC=72度，再分别以 O， A 为园心， OA 长为半径，划弧相交于角 AOC 内 D点，连接 OD， (OAD 是等边三角形 ) 这样角 COD=12度。

其它也类似。

作者: y 时间: 2005-2-19 06:13
标题: Thanks. It is not obvious.

For other n, it will take some time to figure out how to trisect the 180/n degree of angle.

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