proof

**anime_mania** · March 30th 2008, 06:41 PM

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

**roy_zhang** · March 30th 2008, 10:24 PM

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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