China Mathematical Competition

Suppose that the common ratio of $a$, $b$, $c$ is $q$. Then $b = aq$, $c = aq^2$. We have
$$\begin{aligned} \frac{\sin A \cot C + \cos A}{\sin B \cot C + \cos B} &= \frac{\sin A \cos C + \cos A \sin C}{\sin B \cos C + \cos B \sin C} \\ &= \frac{\sin(A+C)}{\sin(B+C)} = \frac{\sin(\pi-B)}{\sin(\pi-A)} \\ &= \frac{\sin B}{\sin A} = \frac{b}{a} = q. \end{aligned}$$
So we only need to determine the range of $q$. As $a$, $b$, $c$ are the sides of a triangle, they satisfy $a+b>c$ and $b+c>a$. That is to say,
$$\begin{cases} a + aq > aq^2, \\ aq + aq^2 > a. \end{cases}$$
It follows that
$$\begin{cases} q^2 - q - 1 < 0, \\ q^2 + q - 1 > 0. \end{cases}$$
Their solutions are
$$\begin{cases} \frac{\sqrt{5}-1}{2} < q < \frac{\sqrt{5}+1}{2}, \\ q > \frac{\sqrt{5}-1}{2} \text{ or } q < -\frac{\sqrt{5}+1}{2}. \end{cases}$$
It is only possible that
$$\frac{\sqrt{5}-1}{2} < q < \frac{\sqrt{5}+1}{2}$$