What are the conditions for the terms of a three-term geometric progression to be considered the side lengths from which a triangle can be constructed?

Can any of these triangles be right-angled?
Let the sides of a triangle be $a, b, c$ in increasing order, and let $b=a q, c=a q^{2}(q \geq 1)$. Among these, the triangle inequalities are satisfied, so in particular

$$
c<a+b \quad \text { i.e. } \quad a q^{2}<a + a q
$$

(The other two inequalities are trivially satisfied.) Since $a$ is positive, for this to hold,

$$
q^{2}-q-1<0
$$

must be true. This expression holds between the two roots of the left-hand side, and since $q$ must also be greater than 1, the geometric sequences that satisfy this condition have a ratio

$$
1 \leq q<\frac{\sqrt{5}+1}{2}
$$

For a right-angled triangle,

$$
a^{2} q^{4}=a^{2}+a^{2} q^{2}, \quad \text { so } q^{4}-q^{2}-1=0
$$

must be true, whose positive real solution is $\sqrt{\frac{\sqrt{5}+1}{2}}$, so the right-angled triangles that satisfy this are similar to the one with sides $1, \sqrt{\frac{\sqrt{5}+1}{2}}, \frac{\sqrt{5}+1}{2}$.