Three circles touch one another externally. The tangents at their points of contact meet at a point whose distance from the point of contact is 4. The ratio of product of their radii to the sum of radii of the circles is

$$\begin{gathered} \mathrm{Let} {\mathrm{r}}_1,{\mathrm{r}}_2,{\mathrm{r}}_3 \mathrm{are} \mathrm{radii} \mathrm{of} \mathrm{three} \mathrm{circles} \mathrm{let} \mathrm{O} \\ \mathrm{is} \mathrm{the} \mathrm{radical} \mathrm{centre}, \mathrm{OL}=4 \mathrm{Let} \mathrm{A},\mathrm{B},\mathrm{C} \mathrm{are} \mathrm{centres} \\ \mathrm{AB}={\mathrm{r}}_1+{\mathrm{r}}_2=\mathrm{c} \\ \mathrm{BC}={\mathrm{r}}_2+{\mathrm{r}}_3=\mathrm{a} \\ \mathrm{CA}={\mathrm{r}}_1+{\mathrm{r}}_3=\mathrm{b} \\ \mathrm{s}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}={\mathrm{r}}_1+{\mathrm{r}}_2+{\mathrm{r}}_3 \\ ∆=\sqrt{\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})} \\ \frac{∆(∆\mathrm{ABC})}{\mathrm{semi}-\mathrm{perimeter} ∆\mathrm{ABC}}=\mathrm{radius} \\ =\frac{{\mathrm{r}}_1{\mathrm{r}}_2{\mathrm{r}}_3}{{\mathrm{r}}_1+{\mathrm{r}}_2+{\mathrm{r}}_3}=\frac{16}{1}=16:1 \end{gathered}$$