## Sides of a Right Angled Triangle With Sides in a Geometric Progression

Suppose we have a right angled triangle with sides in a geometric progression. The sides must be of the form
$a, \: ar, \: ar^2$
. Suppose
$r \gt 1$
then
$ar^2$
is the longest side, the hypotenuse, and we can use Pythagoras Theorem to write
$a^2+(ar)^2=(ar^2)^2$

$a^2+a^2r^2=a^2r^4$

$1+r^2=r^4$

$r^4-r^2-1=0$

Then
$r^2=\frac{1 \pm \sqrt{(-1)^2 - 4 \times 1 \times (-1)}}{2 \times 1} = \frac{1 \pm \sqrt{5}}{2}$

Since
$r \gt 1$
, the solution is
$r^2=\frac{1+ \sqrt{5}}{2} \rightarrow r=\sqrt{\frac{1+\sqrt{5}}{2}}$