Question

Let a, b, c be the three sides of a triangle. Suppose that $\frac{a}{b}=\frac{b}{c}=q.$

Prove that $\frac{\sqrt{5}-1}{2}<q<\frac{\sqrt{5}+1}{2}$

Answer 1

Sum of any two sides is greater than third side
$a+b>c$......(2)
We have a - b < c < a + b
$\frac{a}{b}-1<\frac{c}{b}<\frac{a}{b}+1$
$q-1<\frac{1}{q}<q+1$
$q^2-q<1<q^2+q(sinceq>0)$
$q^2-q-1<0,q^2+q+1>0$
$\frac{1-\sqrt{5}}{2}<q<\frac{1+\sqrt{5}}{2}$;
$q<\frac{-1-\sqrt{5}}{2}$ or $q>\frac{-1+\sqrt{5}}{2}$
$\Rightarrow \frac{\sqrt{5}-1}{2}<q<\frac{\sqrt{5}+1}{2}$