## Ratio of Lengths of Sides of Rectangle When Bisector of a Diagonal Equals Longer Side

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?
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
$l= s \sqrt{\frac{1+ \sqrt{5}}{2}}$