How to minimise the area of the folded part of a piece of paper when it touches the other side I am studying maths purely out of interest and have come across this question in my text book:
A rectangular piece of paper ABCD is folded about the line joining points P on AB and Q on AD so that the new position of A is on CD. If AB = a and AD = b, where $a \ge\frac{2b}{\sqrt3}$, show that the least possible area of the triangle APQ is obtained when the angle AQP is equal to $\frac{\pi}{3}$. What is the significance of the condition $a \ge\frac{2b}{\sqrt3}$?
I realise I need to get an equation for the area of the fold, which I can then differentiate. Even then, I am not sure how to relate this to angle AQP.I have looked at solutions on the internet but they tend to look at the length of the crease, not the area of the fold.
I have taken A in my diagram to represent its position before the fold. I could be mistaken.
This is how I visualise it:

I have said:
$x^2 = m^2 + (b - x)^2 = m^2 + b^2 - 2bx +x^2 \implies m^2 = 2b(x - \frac{b}{2})$
$L^2 = (L - m)^2 + b^2 \implies L = \frac{m^2 + b^2}{2m}$
$y^2 = x^2 + L^2 = x^2 + \frac{(m^2 + b^2)^2}{4m^2} = x^2 + \frac{(2bx - b^2 + b^2)^2}{4m^2} = x^2 + \frac{4b^2x^2}{8b(x - \frac{b}{2})} = \frac{bx^2}{2x - b}$
After this I am not sure how to proceed.
 A: If $\angle AQP = \theta$, $\angle APQ = 90^0 - \theta$.
Draw a perp from $A_1$ to line segment $AP$ and say it is point $A_2$ on line segement $AP$.
Then $A_1A_2 = b$ and $\angle A_2PA_1 = 2 \ \angle APQ = 180^0 - 2\theta \ $.
So $AP = L = \frac{A_1 A_2}{\sin(180^0 - 2 \theta)} = \frac{b}{\sin (180^0 - 2 \theta)} = \frac{b}{\sin 2 \theta} \ $ (in $\triangle A_1PA_2$)
$AQ = x = L \cot \theta = \frac{b}{2 \sin^2 \theta}$
Area of $\triangle APQ = \frac{1}{2} \ L \ x = \frac{b^2}{8 \sin^3 \theta \cos \theta}$
So to minimize area of $\triangle APQ$, we need to maximize $f(\theta) = \sin ^3 \theta \ \cos \theta$.
$f'(\theta) = 3 \sin^2 \theta \cos^2 \theta - \sin^4 \theta = 0$
$\implies \sin \theta = 0, \tan \theta = \pm \sqrt3$. This leads to $\theta = \frac{\pi}{3}$.
Which also tells that $L = \frac{b}{\sin 2 \theta} = \frac{2b}{\sqrt3}$
Also note that for us to be able to fold such that we obtain minimum area,
$a \geq L =  \frac{2b}{\sqrt3}$.
A: I think you were almost there. You don't need $y$ to compute the area, so you can omit that part of your calculation. So you have two variables, $x$ and $m$, but these are related. You've solved for $m^2$ in terms of $x$, but you can instead solve for $x$ in terms of $m^2$. Then you can minimize $A=\frac{1}{2}xL$, which is a function of the single variable $m$.
Notice that there are natural endpoints for this minimization. Start by assuming that the piece of paper is semi-infinite. That is, let $a$ go to infinity so you don't have to worry about not getting a triangle when you make the fold. You should be able to see that $m$ must be greater than $0$. (As $m$ approaches $0$, the area of the triangle approaches infinity.) On the other hand, $m$ can be at most $b$ (since $m\le x\le b$), at which point the triangle area is $\frac{1}{2}b^2$. So you are minimizing on the interval $(0,b]$.
