Re: Trisecting an arbitrary angle




Le 19/07/05 15:27, dans
8693294.1121779705730.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx, « bassam king
karzeddin » <bassam@xxxxxxxxxx> a écrit :

>> 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.
> Here is an example:
> A Triangle that have an angle & its seventh multisection
> (that is THETA & 7*THETA) will be given in the following symbolic (A,B,C)
> triangle .Where the sides are:
>
> A = 1
> B = X^3-5*X^2+6*X-1
> C = SQRT(X)*(X^3-6*X^2+10*X-4)
>
> Where : 4 >= X >= y
> y = some fixed value I will define later


Looks good.

I put Maple result at http://jcarbaut.free.fr/7section.txt

Here are some of the results:

64*cA^7 - 112*cA^5 + 56*cA^3 - 7*cA = cB

So you should have 7*A = B. But that needs a check on A.
We must have 0 < A < pi/8 (so that A+B<pi)

cos(A) = t/2

So, valid A are found for t in
2 * cos(0) = 2
2 * cos(pi/2) = sqrt(2+sqrt(2)) ~ 1.84

So valid t are in ]sqrt(2+sqrt(2)), 2[

Simple check:

t=19/10

then A,B,C are (in radians): [0.3175604293, 2.222923006, 0.6011092193]

.

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