4. Let $\omega_{1}$ and $\omega_{2}$ be circles with centers $O_{1}$ and $O_{2}$, respectively, and radii $r_{1}$ and $r_{2}$, respectively. Suppose that $O_{2}$ is on $\omega_{1}$. Let $A$ be one of the intersections of $\omega_{1}$ and $\omega_{2}$, and $B$ be one of the two intersections of line $O_{1} O_{2}$ with $\omega_{2}$. If $A B=O_{1} A$, find all possible values of $\frac{r_{1}}{r_{2}}$.
Answer: $\frac{-1+\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}$ There are two configurations to this problem, namely, $B$ in between the segment $O_{1} O_{2}$ and $B$ on the ray $O_{1} O_{2}$ passing through the side of $O_{2}$ Case 1: Let us only consider the triangle $A B O_{2} . A B=A O_{1}=O_{1} O_{2}=r_{1}$ because of the hypothesis and $A O_{1}$ and $O_{1} O_{2}$ are radii of $w_{1} . O_{2} B=O_{2} A=r_{2}$ because they are both radii of $w_{2}$.
Then by the isosceles triangles, $\angle A O_{1} B=\angle A B O_{1}=\angle A B O_{2}=\angle O_{2} A B$. Thus can establish that $\triangle A B O_{1} \sim \triangle O_{2} A B$.
Thus,
$$
\begin{array}{c}
\frac{r_{2}}{r_{1}}=\frac{r_{1}}{r_{2}-r_{1}} \\
r_{1}^{2}-r_{2}^{2}+r_{1} r_{2}=0
\end{array}
$$

By straightforward quadratic equation computation and discarding the negative solution,
$$
\frac{r_{1}}{r_{2}}=\frac{-1+\sqrt{5}}{2}
$$

Case 2: Similar to case 1, let us only consider the triangle $A B O_{1}$. $A B=A O_{1}=O_{1} O_{2}=r_{1}$ because of the hypothesis and $A O_{1}$ and $O_{1} O_{2}$ are radii of $w_{1} . O_{2} B=O_{2} A=r_{2}$ because they are both radii of $w_{2}$.
Then by the isosceles triangles, $\angle A O_{1} B=\angle A B O_{1}=\angle A B O_{2}=\angle O_{2} A B$. Thus can establish that $\triangle A B O_{1} \sim \triangle O_{2} A B$.
Now,
$$
\begin{array}{c}
\frac{r_{2}}{r_{1}}=\frac{r_{1}}{r_{2}+r_{1}} \\
r_{1}^{2}-r_{2}^{2}-r_{1} r_{2}=0
\end{array}
$$

By straightforward quadratic equation computation and discarding the negative solution,
$$
\frac{r_{1}}{r_{2}}=\frac{1+\sqrt{5}}{2}
$$