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Thread: constructible angle

  1. #1
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    constructible angle

    cosθ=3/7, θ is an acute angle.
    prove θ cannot be trisected with straightedge and compass?




    my approach: angle θ can't be constructed with straightedge and compass if cosθ is transcendental, but cosθ=3/7 is algebraic and so it is not transcendental?

    Please help me with this question, thanks.
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by elmo View Post
    cosθ=3/7, θ is an acute angle.
    prove θ cannot be trisected with straightedge and compass?




    my approach: angle θ can't be constructed with straightedge and compass if cosθ is transcendental, but cosθ=3/7 is algebraic and so it is not transcendental?

    Please help me with this question, thanks.
    The question you're asking is if the angle $\displaystyle \frac\theta3 $ is constructible.

    The angle $\displaystyle \frac\theta3 $ is constructible $\displaystyle \iff \cos\left(\frac\theta3\right) $ and $\displaystyle \sin\left(\frac\theta3\right) $ are both constructible numbers.

    Let's compute $\displaystyle \cos\left(\frac\theta3\right) $ by the triple angle formula:

    $\displaystyle \cos(\theta) = 4\cos\left(\frac\theta3\right)^3-3\cos\left(\frac\theta3\right) $

    So we see $\displaystyle \cos\left(\frac\theta3\right) $ is a root of $\displaystyle f(x)=28x^3-21x-3 $.

    $\displaystyle f(x) $ has no linear factors and thus is irreducible. Hence $\displaystyle f(x) = m_{\cos\left(\frac\theta3\right),\mathbb{Q}}(x) $.

    Since a number $\displaystyle \alpha $ is constructible $\displaystyle \iff \text{deg}(m_{\alpha,\mathbb{Q}}(x)) = 2^n $, we see that $\displaystyle \cos\left(\frac\theta3\right) $ is not constructible $\displaystyle \implies $ the angle $\displaystyle \frac\theta3 $ isn't either.
    Last edited by chiph588@; Apr 2nd 2010 at 07:48 AM.
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