Re: Exact Trisecting



bassam king karzeddin wrote:
> Re: Exact Trisecting
> Posted: Jul 26, 2005 2:25 AM Plain Text Reply
>
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> quasi wrote:
> > Bassam Karzeddin wrote:
> >
> > > An arbitrary angle and its exact trisection angle
> > > fits exactly in the following symbolic triangle with the following sides:
> > > a^3 , a*(b^2-a^2) , b*(b^2-2*a^2)
> > > Where : 2 >= b/a >= sqrt(2)
> > > (a,b):are positive real numbers
> >
>
> And,iff, 2>= b/a >= F,Then: a*(b^2-a^2) >= b*(b^2-2*a^2)

If b=2 and a=1, a*(b^2-a^2) = 3 and b*(b^2-2*a^2) = 4...

> Where : F is a Fibonacci Number (F=1.618033989...)

Let t = b/a s.t. a>0, b>0, t in [sqrt(2), 2]

then you want to know for which t, the inequality
t^2 - 1 >= t*(t^2-2) holds.

To solve the inequation t^3 - t^2 - 2*t + 1 <= 0,
just find the roots of the cubic, Maple gives the
approximate values:

[1.801937737, -1.246979604, 0.4450418679]

Hence the cubic is negative for t in [1.414, 1.802].

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