The central distance of two circles is $c=714 \mathrm{~mm}$, the external common tangents intersect at an angle of $\alpha=36^{\circ} 8^{\prime}$, and the internal ones intersect at an angle of $\beta=104^{\circ} 12^{\prime}$. What are the radii of the circles?
Let the radii of the circles be $R$ and $r$; the centers $O_{1}$ and $O_{2}$; the external tangency points $F$ and $B$; the internal tangency points $E$ and $C$; the external similarity point $A$, and the internal similarity point $D$.

From triangles $O_{1} E D$ and $O_{2} C D$:

$$
R=O_{1} D \cdot \sin \frac{\beta}{2}, r=O_{2} D \cdot \sin \frac{\beta}{2}
$$

and thus

$$
R+r=\left(O_{1} D+O_{2} D\right) \sin \frac{\beta}{2}=c \sin \frac{\beta}{2}
$$

From triangles $O_{1} F A$ and $O_{2} B A$:

$$
R=O_{1} D \cdot \sin \frac{\alpha}{2}, r=O_{2} A \cdot \sin \frac{\alpha}{2}
$$

and thus

$$
R-r=\left(O_{1} A-O_{2} A\right) \sin \frac{\alpha}{2}=c \sin \frac{\alpha}{2}
$$

From (1) and (2):

$$
\begin{aligned}
& R=\frac{c}{2}\left(\sin \frac{\beta}{2}+\sin \frac{\alpha}{2}\right)=c \sin \frac{\beta+\alpha}{4} \cos \frac{\beta-\alpha}{4} \\
& r=\frac{c}{2}\left(\sin \frac{\beta}{2}-\sin \frac{\alpha}{2}\right)=c \cos \frac{\beta+\alpha}{4} \sin \frac{\beta-\alpha}{4}
\end{aligned}
$$

Substituting the given values:

$$
R=392.42 \text{ mm}, r=170.99 \text{ mm}
$$

(Ignácz Frankl, Selmeczbánya.)

The problem was also solved by: Bálint B., Beck F., Dénes A., Devecis M., Fekete J., Freibauer E., Friedmann B., Hofbauer E., Kántor B., Kántor N., Kertész L., Kornis Ö., Laczkó E., Misángyi V., Orlowszky F., Petrogalli G., Prakatur T., Roth M., Schwartz E., Spitzer Ö., Szabó I., Szabó K., Weisz Á., Weisz J.