# A fold is formed on a 20 cm × 30 cm rectangular sheet of paper running from the short side to the long side by placing a corner over the long side. Find the minimum possible length of the fold?

Nov 2, 2017

$L = 15 \sqrt{3} \approx 25.98 \setminus c m$

#### Explanation:

Let us set up the following variables:

 { (L, "Length of the fold", cm), (x, "DE", cm), (theta, "Angle " hat(DGE), "radians") :}

We are given that $A D = 30 c m \text{ and } A B = 20 c m$. Our aim is to find the minimum length $L$ of the fold.

By trigonometry for $\triangle D E G$ we have:

$\sin \hat{D G E} = \frac{D E}{E G} \implies \sin \theta = \frac{x}{L}$ ..... [A]

When folded the point $D$ must touch the side $B C$

$\hat{H F G} = \hat{F G D} = 2 \theta$

Now:

$\hat{H F G} + \hat{G F E} + \hat{E F C} = \pi$
$\therefore 2 \theta + \frac{\pi}{2} + \hat{E F C} = \pi$
$\therefore \hat{E F C} = \frac{\pi}{2} - 2 \theta$

By trigonometry for $\triangle C E F$ we have:

$\sin \hat{E F C} = \frac{E C}{E F}$
$\therefore \sin \left(\frac{\pi}{2} - 2 \theta\right) = \frac{C D - D E}{E F}$
$\therefore \sin \left(\frac{\pi}{2} - 2 \theta\right) = \frac{20 - x}{x}$

Using the sum of angle formula:

$\sin \left(A - B\right) \equiv \sin A \cos B - \cos A \sin B$

we have:

$\sin \left(\frac{\pi}{2}\right) \cos \left(2 \theta\right) - \cos \left(\frac{\pi}{2}\right) \sin \left(2 \theta\right) = \frac{20 - x}{x}$
$\therefore \cos \left(2 \theta\right) = \frac{20}{x} - 1$

Using the identity:

$\cos 2 A \equiv 1 - 2 {\sin}^{2} A$

we have:

$1 - 2 {\sin}^{2} \theta = \frac{20}{x} - 1$

Using $E q \left[A\right]$ this becomes:

$\setminus \setminus \setminus \setminus \setminus 1 - 2 {\left(\frac{x}{L}\right)}^{2} = \frac{20}{x} - 1$
$\therefore 2 - 2 {x}^{2} / {L}^{2} = \frac{20}{x}$
$\therefore {x}^{2} / {L}^{2} = 1 - \frac{10}{x}$
$\therefore {x}^{2} / {L}^{2} = \frac{x - 10}{x}$
$\therefore {L}^{2} / {x}^{2} = \frac{x}{x - 10}$
$\therefore {L}^{2} = {x}^{3} / \left(x - 10\right)$

As $L$ changes then $\theta$ and $x$ change accordingly, we know have ${L}^{2}$ as a function of $x$ alone, and minimizing $L$ is the same as minimizing ${L}^{2}$, so we seek a critical point of ${L}^{2}$

Differentiating wrt $x$ by applying the quotient rule we have:

$\frac{d}{\mathrm{dx}} {L}^{2} = \frac{\left(x - 10\right) \left(\frac{d}{\mathrm{dx}} \left({x}^{3}\right)\right) - \left(\frac{d}{\mathrm{dx}} \left(x - 10\right)\right) \left({x}^{3}\right)}{x - 10} ^ 2$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{\left(x - 10\right) \left(3 {x}^{2}\right) - \left(1\right) \left({x}^{3}\right)}{x - 10} ^ 2$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {x}^{2} \frac{3 \left(x - 10\right) - x}{x - 10} ^ 2$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {x}^{2} \frac{3 x - 30 - x}{x - 10} ^ 2$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {x}^{2} \frac{2 x - 30}{x - 10} ^ 2$

At a critical point this derivative vanishes, and so:

${x}^{2} \frac{2 x - 30}{x - 10} ^ 2 = 0$
${x}^{2} \left(2 x - 30\right) = 0$
$x = 0 , 15$

$x = 0$ result in no fold and so we required $x = 15$, with this value of $x$ we have:

${L}^{2} = {15}^{3} / \left(15 - 10\right)$
$\setminus \setminus \setminus \setminus = \frac{3375}{5}$
$\setminus \setminus \setminus \setminus = 675$

$\therefore L = 15 \sqrt{3} \approx 25.98 \setminus c m$