Angles of a triangle - are in a Geometric Progression, possible values for the common ratio other than 1 [closed]

Let us assume that there exists a triangle with measures of its angles in a Geometric Progression (G.P.) with a common ratio other than 1.

Then what are the possible ranges of (that is starting set and ending set) of measures of angles - when calculated to 2 decimal places of precision?

Is there a point of inflexion / break-even in this case? (By this I mean - one set of solutions become complement of the other set - thereby no need for us to find by calculating but we can infer by inspection) If so, how to find it?

Solution 1:

The general form for three numbers in geometric progression is $a, ar, ar^2$. Here of course $a > 0$. We can assume $r \geq 1$, or else the angles are just in decreasing rather than increasing order. We know the sum of the angles must be $\pi$ radians or $180^\circ$.

None of these restrictions put any upper bound on $r$. So choosing any $r \geq 1$, there's a triangle with angles

$$ \frac{180^\circ}{r^2+r+1}, \frac{180^\circ r}{r^2+r+1}, \frac{180^\circ r^2}{r^2+r+1} $$

If for example $r=1000$, the angles can be approximately $0.00017982^\circ, 0.17982^\circ, 179.82^\circ$, close to a degenerate linear triangle. Of course the $r=1$ limit gives the equiangular $60^\circ, 60^\circ, 60^\circ$.