Question

A rectangular sheet of paper is folded so that two diagonally opposite corners come together. If the crease formed is the same length as the longer side of the sheet, what is the ratio of the longer side of the sheet to the shorter side?

Answer 1

Let the sheet of paper be ABCD, and have sides AD = a, AB = b, where a b. Let the fold line be EF, of length x. Let d the length of the diagonal.
By Pythagoras' Theorem, $d^2=a^2+b^2$.
Draw straight line BD between the two corners used to make the fold. It's clear by symmetry that this diagonal intersects the fold at right angles. Further, also by symmetry, both lines meet at the center of the rectangle, X, and bisect each other.


Triangles DAB and XEB contain two common angles, and therefore are similar.
Hence $\frac{a}{b}=\frac{\left(\frac{x}{2}\right)}{\left(\frac{d}{2}\right)}=\frac{x}{d}.$
$\therefore x=\left(\frac{a}{b}\right)(a^2+b^2)$.
If x = b, as we require, then $a^2(a^2+b^2)=b^4$, and so $b^4-a^2{b}^2-a^4=0.$
Solving as a quadratic equation in $b^2$, we have $b^2=[a^2\pm \sqrt{\frac{(a^4+4a^4)}{2}}]$
$=a^2\frac{(1\pm \sqrt{5})}{2}$
Rejecting the negative roots, $\frac{b}{2}$= ratio of longer side to shorter side =$\sqrt{\frac{1+\sqrt{5}}{2}}$