Triangle $\Delta ABC$ , $a,b,c$ are in G.P.

Question

If in a triangle $\Delta ABC$ the sides $a,b,c $ are in Geometric Progression.Find out the range of common ratio of the Geometric Progression.

I understood that the twist is that we are bound under these conditions: $$a+b>c;\\a+c>b;\\b+c>a;$$ Please help . How to solve these Inequalities?

Answer 1

Let the common ratio be denoted by $\;\dfrac{a_n}{a_{n+1}} =\dfrac 1r,\;$ or $\dfrac {a_{n+1}}{a_n} = r.\;$ Then, WLOG, let $$a = 1,\; b = r, \;c = r^2\tag{1}$$

Given the constraints you list (abiding by the triangle inequality), we indeed have

$$1 + r - r^2 \gt 0, \;1 + r^2 - r \gt 0, \; \text{ and} \;r^2 + r - 1 \gt 0\tag{2}$$.

Solving for the values of $\,r\,$ that satisfyeach inequality in $(2)$, by using the number line (plot by WolframAlpha) to indicate which interval(s) of values of $r$ for which each inequality holds, we see the intersection of those intervals, within which $r$ satisfies all three inequalities:

There is one integer solution: $r = 1$, but that's not terribly interesting. So pick say, $r = \frac 34$, and solve for $a, b, c$, and confirm. Indeed, any value of $r$ within the interval of solutions given by $$\frac 12\left(\sqrt 5 - 1\right) \;<\; r\; < \;\frac 12(\sqrt 5 + 1)$$

is a solution for $\;r$.

Answer 2

Call the numbers $1$, $r$, $r^2$ (without loss of generality)

Then $1 + r - r^2 > 0$, $1 + r^2 - r > 0$, $r^2 + r - 1 > 0$

Solve for the possible regions of $r$ in each case and see where they intersect (Use quadratic equation I suppose)

Edit: How interesting, the solution space ends up being $(\phi - 1, \phi)$ where $\phi$ is the golden ratio. If we are only counting $r \geq 1$ which I suppose is reasonable, then we get $[1, \phi)$.