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Thread: proof

  1. #1
    Junior Member
    Jan 2007


    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.)

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  2. #2
    Junior Member roy_zhang's Avatar
    Mar 2008
    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 $\displaystyle L$ and the given angle $\displaystyle \angle\theta$.

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

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

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

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

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