10.248. To two externally tangent circles with radii $R$ and $r$, a secant is constructed such that the circles intercept three equal segments on it. Find the lengths of these segments.
Solution.

Let the desired length be $2x$. Then $AB = 4x$, $AO_1 = \sqrt{R^2 - x^2}$, $BO_2 = \sqrt{r^2 - x^2}$ (Fig. 10.55). Draw $O_2C \parallel AB$. In $\triangle O_1O_2C$, we have $O_1C = \sqrt{O_1O_2^2 - O_2C^2} \Rightarrow \sqrt{R^2 - x^2} - \sqrt{r^2 - x^2} = \sqrt{(R + r)^2 - 16x^2}^{*}$.

Multiplying both sides of equation (*) by $\sqrt{R^2 - x^2} + \sqrt{r^2 - x^2}$, we get

\[
\begin{aligned}
& R^2 - r^2 = \sqrt{(R + r)^2 - 16x^2} \left( \sqrt{R^2 - x^2} + \sqrt{r^2 - x^2} \right) \Rightarrow \\
& \Rightarrow \sqrt{R^2 - x^2} + \sqrt{r^2 - x^2} = \frac{R^2 - r^2}{\sqrt{(R + r)^2 - 16x^2}}(* *)
\end{aligned}
\]

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0747.jpg?height=440&width=670&top_left_y=81&top_left_x=113)

Fig. 10.55

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0747.jpg?height=478&width=454&top_left_y=84&top_left_x=790)

Fig. 10.56

Adding equations (*) and (**), we have

\[
\begin{aligned}
& 2 \sqrt{R^2 - x^2} = \sqrt{(R + r)^2 - 16x^2} + \frac{R^2 - r^2}{\sqrt{(R + r)^2 - 16x^2}} \Rightarrow \\
\Rightarrow & \frac{(R + r)^2 - 16x^2 + R^2 - r^2}{\sqrt{(R + r)^2 - 16x^2}} = \frac{2(R^2 + Rr - 8x^2)}{\sqrt{(R + r)^2 - 16x^2}} = \frac{2(R(R + r) - 8x^2)}{\sqrt{(R + r)^2 - 16x^2}} \Rightarrow
\end{aligned}
\]

\[
\Rightarrow \sqrt{R^2 - x^2} \cdot \sqrt{(R + r)^2 - 16x^2} = R(R + r) - 8x^2 \Rightarrow
\]

\[
\Rightarrow (R^2 - x^2)((R + r)^2 - 16x^2) = (R(R + r) - 8x^2)^2 \Rightarrow
\]

\[
\Rightarrow R^2(R + r)^2 - 16R^2x^2 - x^2(R + r)^2 + 16x^4 = R^2(R + r)^2 - 16R(R + r)x^2 + 64x^4.
\]

Since $x \neq 0$, we arrive at the equation $48x^2 = 14Rr - R^2 - r^2$, from which $x = \frac{1}{4} \sqrt{\frac{14Rr - R^2 - r^2}{3}}$ (the second root of the equation is not suitable). The desired length is $2x = \frac{1}{2} \sqrt{\frac{14Rr - R^2 - r^2}{3}}$.

\[
\text{Answer: } \frac{1}{2} \sqrt{\frac{14Rr - R^2 - r^2}{3}}
\]