2. Two circles touch each other externally and each touches a larger circle internally. The radius of one is half, and that of the other is one-third of the radius of the largest circle. Find the ratio of the length of the segment of the common internal tangent to the smaller circles, contained within the largest, to its diameter.

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# Solution

Let the radii of the smaller circles be $2r$ and $3r$. Then the radius of the largest (outer) circle is $6r$.

![](https://cdn.mathpix.com/cropped/2024_05_06_56dfd6ceffa3415f7b2dg-2.jpg?height=627&width=621&top_left_y=150&top_left_x=730)

Consider $\triangle O_{1} O O_{2}$. Its sides are $3r, 4r$, and $5r$, so it is a right triangle.

Let the point of intersection 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 $O_{1} O O_{2}, K F O_{2}, K D O$ 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 $O D=\frac{6}{5} r$.

From the right triangle $O D M$, we find half of the desired chord $M D^{2}=(6 r)^{2}-\left(\frac{6}{5} r\right)^{2}=\frac{36 \cdot 24}{25} r$.

The desired ratio of the chord to the diameter is $\frac{M D}{6 r}=\frac{\sqrt{24}}{5}=\frac{2 \sqrt{6}}{5}$.

Answer: $\frac{2 \sqrt{6}}{5}$.