Solution: Draw a rectangle with longer side

\[l\]

and shorter side \[s\]

. Draw a diagonal \[AC\]

then the crease \[PQ\]

must bisect this crease at right angles.The length of the diagonal

\[AC\]

is \[\sqrt{l^2-s^2}\]

by Pythagoras Theorem and the distance \[AO\]

is half of this: \[AO= \frac{\sqrt{l^2+s^2}}{2}\]

Also since the crease

\[PQ\]

is of length \[l\]

and the diagonal bisects the crease, \[OP=\frac{l}{2}\]

\[AOP\]

is a right angled triangle since \[PQ\]

bisects \[AC\]

at right angles. We can find two expressions for \[AP\]

and equate them.Drop a perpendicular from P to the base of the rectangle then

\[PMQ\]

is a right angled triangle and \[PM=s\]

.By Pythagoras Theorem

\[QM= \sqrt{l^2 - s^2}\]

and \[PO\]

is half of this: \[PM=\frac{\sqrt{l^2-s^2}}{2}\]

.Now apply Pythagoras Theorem to the triangle

\[AOP\]

.\[ ( \frac{l}{2} + \frac{\sqrt{l^2-s^2}}{2})^2 = (\frac{l}{2})^2 + (\frac{\sqrt{l^2+s^2}}{2})^2\]

\[\frac{l^2}{4} + \frac{l \sqrt{l^2-s^2}}{2} +\frac{l^2-s^2}{4}=\frac{l^2}{4} + \frac{l^2+s^2}{4} \]

\[\frac{l \sqrt{l^2-s^2}}{2}= \frac{s^2}{2}\]

\[l^2(l^2-s^2) =s^4\]

\[ l^4-l^2s^2-s^4=0\]

We can treat this as a quadratic in

\[l^2\]

with solutions \[l^2 = \frac{s^2 \pm \sqrt{s^2+4s^2}}{2} = s^2 (\frac{1 \pm \sqrt{5}}{2}) \]

We only need the positive option since the negative one mean

\[l^2\]

is negative.Hence

\[l= s \sqrt{\frac{1+ \sqrt{5}}{2}}\]