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Math Help - proof

  1. #1
    Junior Member
    Joined
    Jan 2007
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    proof

    hey this is a herd proof, cant figure out
    The lower right-hand corner of a long piece of rectangular paper 6 in. wide is folded over to the left-hand edge as shown. The length L of the fold depends on the angle Show: on attachment



    (Remember, it doesn’t work to show it’s true for a few values of ; you must prove it in general.)

    thanks
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  2. #2
    Junior Member roy_zhang's Avatar
    Joined
    Mar 2008
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    I think there must be multiple approaches to this proof. Below is what I did. The whole idea is to find a connection (equation) between the length L and the given angle \angle\theta.

    Please see the attached the figure, I draw a line AB from the vertex of \angle\theta and parallel to lower side of the rectangle. According to your given information, we have AB=6, \angle CAD=\theta and \angle BAC=90^{\circ}-2\theta.

    In right triangle \triangle ABC, we have \cos(\angle BAC)=\frac{AB}{AC}=\frac{6}{AC}. Realize that \cos(\angle BAC)=\cos(90^{\circ}-2\theta)=\sin(2\theta). So we have \sin(2\theta)=\frac{6}{AC}\Rightarrow AC=\frac{6}{\sin(2\theta)}.

    In right triangle \triangle ACD, we have \cos(\angle CAD)=\cos(\theta)=\frac{AC}{L}\Rightarrow L=\frac{AC}{\cos(\theta)}. Substitute AC=\frac{6}{\sin(2\theta)} obtained earlier, we have

    L=\frac{6}{\sin(2\theta)\cos(\theta)}=\frac{6}{2\s  in(\theta)\cos(\theta)\cos(\theta)}=\frac{3}{\sin(  \theta)\cos^{2}(\theta)}


    Roy
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