1. (2003 Beijing Competition Question) The three sides of a triangle form a geometric sequence. Then, the range of the common ratio $q$ is
1. Fill in $\frac{\sqrt{5}-1}{2}0, q \geqslant 1$. Then $a + aq > aq^2$, i.e., $q^2 - q - 1 < 0$, and $a + aq > 0$, i.e., $q^2 + q - 1 > 0$. Solving these, we get $\frac{\sqrt{5}-1}{2} < q < 1$.
In summary, $\frac{\sqrt{5}-1}{2} < q < \frac{\sqrt{5}+1}{2}$.