China Mathematical Competition

Suppose that the common ratio of $a$, $b$, $c$ is $q$. Then $b=aq$, $c=a{q}^2$. We have $$\begin{array}{rl}\frac{\sin A\cot C+\cos A}{\sin B\cot C+\cos B}& =\frac{\sin A\cos C+\cos A\sin C}{\sin B\cos C+\cos B\sin C}\\ & =\frac{\sin (A+C)}{\sin (B+C)}=\frac{\sin (\pi -B)}{\sin (\pi -A)}\\ & =\frac{\sin B}{\sin A}=\frac{b}{a}=q.\end{array}$$ 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, $$\{\begin{array}{l}a+aq>a{q}^2,\\ aq+a{q}^2>a.\end{array}$$ It follows that $$\{\begin{array}{l}{q}^2-q-1<0,\\ q^2+q-1>0.\end{array}$$ Their solutions are $$\{\begin{array}{l}\frac{\sqrt{5}-1}{2}<q<\frac{\sqrt{5}+1}{2},\\ q>\frac{\sqrt{5}-1}{2}\text{ or }q<-\frac{\sqrt{5}+1}{2}.\end{array}$$ It is only possible that $$\frac{\sqrt{5}-1}{2}<q<\frac{\sqrt{5}+1}{2}$$