4. Two spheres touch each other externally and each touches a larger sphere internally. The radius of one is half, and that of the other is one-third of the radius of the largest sphere. A tangent plane is constructed at the point of contact between the smaller spheres. Find the distance from this plane to the center of the largest sphere, if its radius is $R$.
# Solution

Consider the section of the described composition by a plane passing through the centers of the three spheres. The required distance will be the length of the segment $O D$ on this plane.

![](https://cdn.mathpix.com/cropped/2024_05_06_14f730f408ac40da4118g-5.jpg?height=624&width=615&top_left_y=802&top_left_x=733)

Let the radii of the smaller circles be $2r$ and $3r$. Then the radius of the largest (outer) circle is $6r$ (given: $6r = R$). Consider $\triangle O_{1} O O_{2}$. Its sides are $3r$, $4r$, and $5r$, so it is a right triangle.

Let the intersection point of the desired chord (marked in blue in the diagram) with the segment $O_{1} O_{2}$ be $F$, and with the segment $O O_{2}$ be $K$. Drop a perpendicular $O D$ from the center of the largest circle to the desired chord (the segment of the common tangent). Then the desired chord is bisected by point $D$ and is perpendicular to segments $O D$ and $O_{1} O_{2}$.

The right triangles $O_{1} O O_{2}$, $K F O_{2}$, and $K D O$ are similar. Therefore, $\frac{K O_{2}}{F O_{2}} = \frac{O_{1} O_{2}}{O O_{2}}$, from which $K O_{2} = \frac{5}{2} r$ and $K O = \frac{3}{2} r$.

Next, $\frac{O D}{K O} = \frac{O O_{2}}{O_{1} O_{2}}$, from which $O D = \frac{6}{5} r = \frac{1}{5} R$.

Answer: $\frac{1}{5} R$.