, 2013

Solution:

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 $\omega_{1}$. $O_{2} B = O_{2} A = r_{2}$ because they are both radii of $\omega_{2}$.

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,
$$\begin{gathered} \frac{r_{2}}{r_{1}} = \frac{r_{1}}{r_{2} - r_{1}} \\ r_{1}^{2} - r_{2}^{2} + r_{1} r_{2} = 0 \end{gathered}$$

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 $\omega_{1}$. $O_{2} B = O_{2} A = r_{2}$ because they are both radii of $\omega_{2}$.

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 $\triangle A B O_{1} \sim \triangle O_{2} A B$.

$$\begin{gathered} \frac{r_{2}}{r_{1}} = \frac{r_{1}}{r_{2} + r_{1}} \\ r_{1}^{2} - r_{2}^{2} - r_{1} r_{2} = 0 \end{gathered}$$

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