Re: Trisecting an arbitrary angle
- From: Jean-Claude Arbaut <jean-claude.arbaut@xxxxxxxxxxx>
- Date: Tue, 19 Jul 2005 14:02:16 +0200
Le 19/07/05 12:37, dans 4dd7d$42dcd769$d52f93dc$25956@xxxxxxxxxxxxxx,
« Jutta Gut » <gut.jutta.gerhard@xxxxxxxxx> a écrit :
>
> "bassam king karzeddin" <bassam@xxxxxxxxxx> schrieb im Newsbeitrag
> news:29354512.1121766911405.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxxxxx
>> That is grate,
>>
>> This,might open doors to constructible polygons
>>
>> In fact,I have deduced & proved the same thing,I have mentioned that here:
>>
>> http://mathforum.org/kb/message.jspa?messageID=3802920&tstart=0
>>
>> I will provide examples soon.
>
> If I understand correctly, you have shown that in an triangle with
> the sides a^3 , a*(b^2-a^2) , b*(b^2-2*a^2) one angle is three times
> another one.
>
> The more interesting question would be: given an angle, how to
> construct a triangle with the sides a^3 , a*(b^2-a^2) , b*(b^2-2*a^2)
> and the given angle?
It tried for some simple angles: pi/8, pi/6, pi/5, sides can be computed
with square roots. For pi/9, you have the 3rd degree equation x^3=3*x+1.
Not surprising, since pi/9 is not constructible. But it's still interesting
to know which triangles have two angles A,B such that A=3*B.
.
- References:
- Trisecting an arbitrary angle
- From: bassam king karzeddin
- Re: Trisecting an arbitrary angle
- From: bassam king karzeddin
- Re: Trisecting an arbitrary angle
- From: Jutta Gut
- Trisecting an arbitrary angle
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