We painted the lateral surface of a truncated cone up to half of its slant height blue, and the rest red. The blue surface area is twice as large as the red one. How many times larger is the radius of the base circle compared to the radius of the top circle?
The lateral surface area of a frustum of a cone: $P_{f}=(R+r) a \pi$, where $r$ and $R$ are the radii of the top and bottom circles, respectively, and $a$ is the slant height of the frustum. If the cone were cut exactly at half its height, the radius of the cut circle would be the arithmetic mean of the radii of the base and top circles, that is, $\frac{R+r}{2}$. Thus, we can write the following equation that satisfies the condition:

$$
\left(\frac{R+r}{2}+R\right) \frac{a}{2} \pi=2\left(\frac{R+r}{2}+r\right) \frac{a}{2} \pi
$$

from which, after simplification, $R=5 r$.