Folded Page Ⅰ

Label the figure as shown. Then the width of the leaf $CE=1$ . Let the width of the part folded over $DE = BD = x$ and the angle between the crease $AD$ and the right edge of the page $AE$ , $\angle DAE = \theta$ . There are many triangles similar to $\triangle ADE$ . We note that $BE = 2x \cos \theta$ and that

$\begin{aligned} \cos \theta & = \frac {CE}{BE} = \frac 1{2x\cos \theta} \\ \implies x & = \frac 1{2\cos^2 \theta} \end{aligned}$

The length of the crease $AD = \dfrac x{\sin \theta} = \dfrac 1{2\sin \theta \cos^2 \theta}$ . $AD$ is minimum when $f(\theta) = \sin \theta \cos^2 \theta$ is maximum.

$\begin{aligned} \frac {d f(\theta)}{d\theta} & = \cos^3 \theta - 2\sin^2 \theta \cos \theta & \small \color{#3D99F6} \text{Putting }\frac {d f(\theta)}{d\theta} = 0 \\ \implies \cos^2 \theta & = 2 \sin^2 \theta \\ \tan \theta & = \frac 1{\sqrt 2} & \small \color{#3D99F6} \implies \sin \theta = \frac 1{\sqrt 3}, \cos \theta = \sqrt{\frac 23} \\ \frac {d^2 f(\theta)}{d\theta^2} & = - 3\sin \theta \cos^2 \theta - 4\sin \theta \cos^2 \theta + 2 \sin^3 \theta & \small \color{#3D99F6} \text{Putting }\theta = \tan^{-1} \frac 1{\sqrt 2} \\ & = - \frac 4{\sqrt 3} < 0 \end{aligned}$

Therefore, $f(\theta)$ is maximum and $AD$ is minimum when $\theta = \tan^{-1} \frac 1{\sqrt 2}$ or $x = \dfrac 1{2\cos^2 \theta} = \dfrac 34 = \boxed{0.75}$ .