Bulgarian Mathematical Olympiad

Solution:
Denote the circles by $k_{1}(O_{1}, r)$, $k_{2}(O_{2}, R)$ and $k(O, x)$, where $r < x < R$. Let $A$, $B$ and $C$ be the feet of the perpendiculars from $O_{1}$, $O$ and $O_{2}$, respectively, to the arm $p$ of the given angle $O p q$. Let $l$ be the line through $O_{1}$ parallel to $p$ and let $l$ meet $O B$ and $O_{2} C$ at points $M$ and $N$, respectively. Then

$\triangle O_{1} O M \sim \triangle O_{1} O_{2} N$ and therefore
$$\frac{O O_{1}}{O_{1} O_{2}} = \frac{O M}{O_{2} N}$$
We have $O M = O B - B M = O B - O_{1} A = x - r$, $O_{1} O_{2} = r + R$ and $O_{2} N = O_{2} C - C N = O_{2} C - O_{1} A = R - r$.

If $k$ passes through $O_{1}$, then $O O_{1} = x$ and we get the equation
$$\frac{x}{R + r} = \frac{x - r}{R - r}$$
whence $x = \frac{r + R}{2}$.

If $k$ passes through $O_{2}$, then $O O_{1} = R + r - x$ and
$$\frac{R + r - x}{R + r} = \frac{x - r}{R - r}$$
whence we have again $x = \frac{r + R}{2}$.

In both cases $O$ is the midpoint of $O_{1} O_{2}$ and $k$ passes through $O_{1}$ and $O_{2}$.