The length of the external tangent of circles with radii $r$ and $R$ is twice the length of the internal tangent. Find the distance between the centers of these circles.

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Let $O_1$ and $O_2$ be the centers of circles with radii $r$ and $R$ respectively, $AB$ be a common external tangent to these circles, and $CD$ be a common internal tangent (points $A$ and $C$ lie on the first circle, $B$ and $D$ on the second), $AB = 2CD$. Drop perpendiculars $O_1F$ and $O_2P$ from the centers $O_1$ and $O_2$ of the circles to the lines $O_2B$ and $O_1C$ respectively. Then

$$
AB = O_1F, \quad CD = O_2P, \quad O_2F = R - r, \quad O_1P = R + r.
$$

From the right triangles $O_1FO_2$ and $O_1PO_2$, we find that

$$
\begin{aligned}
& AB^2 = O_1F^2 = O_1O_2^2 - O_2F^2 = O_1O_2^2 - (R - r)^2 \\
& CD^2 = O_2P^2 = O_1O_2^2 - O_1P^2 = O_1O_2^2 - (R + r)^2
\end{aligned}
$$

Since $AB^2 = 4CD^2$, we get the equation

$$
O_1O_2^2 - (R - r)^2 = 4O_1O_2^2 - 4(R + r)^2,
$$

from which we find that

$$
O_1O_2^2 = \frac{1}{3} \left( 4(R + r)^2 - (R - r)^2 \right) = R^2 + \frac{10}{3} rR + r^2
$$

Therefore,

$$
O_1O_2 = \sqrt{R^2 + r^2 + \frac{10}{3} rR}
$$