Imagine three identical circles, each with radius \( r \), touching each other. Their centers form an equilateral triangle with sides of length \( 2r \).
Now, consider two more circles, X and Y, each touching all three original circles externally. Let their radii be \( R_x \) and \( R_y \), with \( R_x \) being the larger radius. We need to find the ratio \( \frac{R_x}{R_y} \).
The radius (R) of a circle tangent externally to three identical, mutually tangent circles is given by: \[ R = \frac{r(k^2 + \sqrt{3}k - 1)}{k^2 + \sqrt{3}k + 1} \] where \( k = r/R \).
For the larger circle X, we have: \[ R_x = \frac{r}{\sqrt{3} - 1} \]=\[ r(\sqrt{3} + 1) \]
For the smaller circle Y, the formula is applied in reverse:\[ R_y = \frac{r}{\sqrt{3} + 1} = r(\sqrt{3} - 1) \]
Now, let's calculate the ratio \( \frac{R_x}{R_y} \):
\[ \frac{R_x}{R_y} = \frac{r(\sqrt{3} + 1)}{r(\sqrt{3} - 1)} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \]
To rationalize the denominator, multiply the numerator and denominator by \( \sqrt{3} + 1 \):
\[ = \frac{(\sqrt{3} + 1)^2}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{3 + 2\sqrt{3} + 1}{3 - 1} = \frac{4 + 2\sqrt{3}}{2} \]
Simplifying this expression gives:
\[ = 2 + \sqrt{3} \]
Therefore, the required ratio is \( \boxed{7 + 4\sqrt{3} : 1} \).