Folded Page Ⅱ

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 folded-over area $[ABD] = \frac 12 AB \cdot BD = \frac 12 \cdot \dfrac x{\tan \theta} \cdot x = \frac 1{8\sin \theta \cos^3 \theta}$ . $[ABD]$ is minimum when $f(\theta) = \sin \theta \cos^3 \theta$ is maximum.

$\begin{aligned} \frac {d f(\theta)}{d\theta} & = \cos^4 \theta - 2\sin^2 \theta \cos^2 \theta & \small \color{#3D99F6} \text{Putting }\frac {d f(\theta)}{d\theta} = 0 \\ \implies \cos^2 \theta & = 3 \sin^2 \theta \\ \tan \theta & = \frac 1{\sqrt 3} \\ \implies \theta & = 30^\circ \\ \frac {d^2 f(\theta)}{d\theta^2} & = - 4\sin \theta \cos^3 \theta - 4\sin \theta \cos^3 \theta + 4 \sin^3 \theta \cos \theta & \small \color{#3D99F6} \text{Putting }\theta = 30^\circ \\ & = - \frac {5\sqrt 3}4 < 0 \end{aligned}$

Therefore, $f(\theta)$ is maximum and $[ABD]$ is minimum when $\theta = 30^\circ$ or $x = \dfrac 1{2\cos^2 \theta} = \dfrac 23 = \boxed{0.667}$ .