\[a, \: ar, \: ar^2\]
. 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 \[r^2=\frac{1+ \sqrt{5}}{2} \rightarrow r=\sqrt{\frac{1+\sqrt{5}}{2}}\]