2. (15 points) Find the range of the similarity coefficient of triangles with side lengths $x, y, z$ and $y, z, p$. In your answer, specify the nearest integers between which the found interval lies.
Solution: Let's write down the similarity condition for the triangles.

$\frac{x}{y}=\frac{y}{z}=\frac{z}{p}=k ; y^{2}=x z ; z^{2}=y p \quad$ or $\quad x=y k, y=z k$. Since the triangles are similar, it is sufficient to write the triangle existence conditions (triangle inequality) for only one of them $\left\{\begin{array}{l}k^{2}+1>k \\ 1+k>k^{2} \\ k^{2}+k>1\end{array}\right.$. Solving this system of conditions, we get that $k \in\left(\frac{\sqrt{5}}{2}-\frac{1}{2} ; \frac{\sqrt{5}}{2}+\frac{1}{2}\right)$. All found values of $k$ lie to the right of 0 but to the left of 2.

Answer: 0 and 2.

Grading Criteria:

| 15 pts | Fully justified solution |
| :--- | :--- |
| 10 pts | There is an arithmetic error in solving the system of conditions. |
| 5 pts | Not all triangle existence conditions are written, or there is an enumeration of integer values of the similarity coefficient. |