1. (10 points) The lengths of the sides of a scalene triangle turned out to be consecutive terms of a certain geometric progression. The common ratio of this progression can be
1) 1.7 ;
2) 0.5 ;
3) 2.0 ;
4) another answer.
Answer: 4).

Solution: Let the sides of the triangle be $b, q b$, and $q^{2} b$, where $q$ is the sought common ratio of the geometric progression. For each side, the triangle inequality must hold:

$$
\left\{\begin{array}{l}
b+q b>b q^{2} \\
q b+q^{2} b>b \\
b+q^{2} b>b q .
\end{array}\right.
$$

From this, we have the following system of inequalities:

$$
\left\{\begin{array}{l}
q^{2}-q-1<0 \\
q^{2}-q+1>0
\end{array}\right.
$$

Notice that the last inequality is true for any value of the common ratio $q$. When $q=2$, the first inequality is not satisfied:

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

Solving the first two inequalities together, we find that the common ratio of the progression can be a number from the interval

$$
\left(\frac{-1+\sqrt{5}}{2} ; \frac{1+\sqrt{5}}{2}\right) \simeq(0.618 ; 1.618)
$$