1. Find the ratio $k=\frac{M}{B}$, where $M$ is the lateral surface area of the cone, and $B$ is the area of its base, given that the volume of the cone is twice the volume of the inscribed sphere in it.
Solution. Let $H, s$ and $r$ be the height, slant height, and radius of the base of the cone, respectively, and let $R$ be the radius of the inscribed sphere. From the similarity of the triangles in the diagram, we have:

$$
\frac{r}{s}=\frac{R}{H-R}
$$

From this, since $H=\sqrt{s^{2}-r^{2}}$, we get $R=r \sqrt{\frac{s-r}{s+r}}$. Now, if we denote the volume of the cone and the sphere by $V_{k}$ and $V_{t}$, respectively, we have:

![](https://cdn.mathpix.com/cropped/2024_06_05_4826f81c3e0b08e4acb7g-5.jpg?height=450&width=413&top_left_y=871&top_left_x=1067)

$$
\frac{V_{k}}{V_{t}}=\frac{\frac{\pi r^{2} H}{3}}{\frac{4 \pi r^{3}}{3}}=\frac{r^{2} \sqrt{s^{2}-r^{2}}}{4 r^{3} \frac{s-r}{s+r} \sqrt{\frac{s-r}{s+r}}}=\frac{(s+r)^{2}}{4 r(s-r)}
$$

From the condition of the problem $\frac{V_{k}}{V_{t}}=2$, using the first and last terms of the above equality, we get:

$$
\begin{gathered}
8 r(s-r)=(s+r)^{2} \\
9 r^{2}-6 r s+s^{2}=0 \\
\quad(3 r-s)^{2}=0
\end{gathered}
$$

From this, $s=3 r$, so $k=\frac{M}{B}=\frac{\pi r s}{\pi r^{2}}=\frac{s}{r}=3$.