Question

If one corner of a long rectangular sheet of paper of width $1$ unit is folded over, so as to reach the opposite edge of the sheet, then

• A. minimum length of the crease is $\frac{3\sqrt{3}}{4}$
• B. minimum length of the crease is $\frac{\sqrt{3}}{4}$
• C. reduced width of the paper is $\frac{1}{2}$
• D. reduced width of the paper is $\frac{1}{4}$

Answer 1

Answer: (A), (D)

The correct options are
A minimum length of the crease is $\frac{3\sqrt{3}}{4}$
D reduced width of the paper is $\frac{1}{4}$

Let $EF=x$,


$E{C}^{\prime }=EC=x\cos \theta$
In $△C^{\prime }BE$,
$BE=x\cos \theta \cos (\pi -2\theta )\Rightarrow BE=-x\cos \theta \cos 2\theta BC=1\Rightarrow x\cos \theta -x\cos \theta \cos 2\theta =1\Rightarrow x=\frac{1}{\cos \theta (1-\cos 2\theta )}$
Let
$z=\cos \theta (1-\cos 2\theta )\Rightarrow \frac{dz}{d\theta }=\cos \theta \left(2\sin 2\theta \right)-\sin \theta (1-\cos 2\theta )\Rightarrow \cos \theta \left(2\sin 2\theta \right)-\sin \theta (1-\cos 2\theta )=0\Rightarrow 2\sin \theta (2{\cos }^2\theta -{\sin }^2\theta )=0\Rightarrow 2-3{\sin }^2\theta =0[\because \sin \theta \ne 0]\Rightarrow \sin \theta =\sqrt{\frac{2}{3}}[\because \theta \text{is acute angle}]\frac{dz}{d\theta }=2\sin \theta (2-3{\sin }^2\theta )\Rightarrow \frac{d^{2}z}{d{\theta }^2}{|}_{\sin \theta =\sqrt{\frac{2}{3}}}<0$
So, at $\sin \theta =\sqrt{\frac{2}{3}},z$ is maximum.
Hence, $x$ is minimum.
$x_{min}=\frac{1}{\cos \theta (1-\cos 2\theta )}\Rightarrow x_{min}=\frac{1}{\cos \theta \left(2{\sin }^2\theta \right)}\Rightarrow x_{min}=\frac{1}{\frac{1}{\sqrt{3}}\times 2\times \frac{2}{3}}=\frac{3\sqrt{3}}{4}$

Reduced width,
$=BE=1-EC=1-x_{min}\cos \theta =1-\frac{3\sqrt{3}}{4}\times \frac{1}{\sqrt{3}}=\frac{1}{4}$