TITLE: Minimize the Area
QUESTION [8 upvotes]: The lower corner of a page is to folded to reach the opposite inner edge. 
We have to find the width of the folded part if the Area of the folded part is minimum. 
Now how I proceeded:
Let the width of the page be $1$.
Let the folded part be $x$. 
And the angle of the part folded be $\theta$.
Now I found a relation between $x$ and $\theta$ and thus wrote the area entirely as a function  of tan($\theta$) and minimized it. 
What I want to know is of  other different  ways to do this. 
Here's a sketch :

Just in case someone  wants to check their answers,  the width x comes out to be 2/3.

REPLY [2 votes]: I suggest working with $\theta':={\pi\over2}-\theta$ instead, and putting $\tan\theta'=:\tau$. You then have
$${1-x\over x}=\cos(\pi-2\theta)=\cos(2\theta')={1-\tau^2\over 1+\tau^2}$$
and therefore ${\displaystyle x={1+\tau^2\over2}}$.
The area in question then comes to
$$A={1\over2}x\cdot{x\over\tau}={(1+\tau^2)^2\over 8\tau}\ .$$
It turns out that $A$ is minimal when $\tau^2={1\over3}$, or $\theta'=30^\circ$, so that $x={2\over3}$. 
The simplicity of the result opens up a chance that it can be obtained also using some geometric reasoning.