A sphere with a radius of 1 is inscribed in a cone, the height of which is twice the diameter of the sphere. Find the ratio of the total surface area of the cone to the surface area of the sphere.
Consider the axial section of a cone with vertex $A$. We obtain an isosceles triangle $A B C$ with vertex $A$, in which a circle is inscribed, the center $O$ of which coincides with the center of the sphere inscribed in the cone. Let $M$ and $K$ be the points of tangency of the circle with the sides $B C$ and $A B$, respectively, and $P$ be the point of intersection of the circle with the height $A M$. Clearly, $M$ is the center of the base of the cone. Additionally,

$$
O K=O M=1, A P=P M=2, A O=A P+O P=2+1=3
$$

Let $\angle A B M=\angle A O K=\alpha$. Then

![](https://cdn.mathpix.com/cropped/2024_05_06_acc13965d0370bfc03dcg-05.jpg?height=98&width=989&top_left_y=2698&top_left_x=535)

$$
\begin{gathered}
B M=\frac{A M}{\operatorname{kg} \angle A B M}=\frac{1}{B_{B 2}}=\frac{1}{2 \sqrt{2}}=\sqrt{2} \\
A B=\frac{B M}{m \angle A B M}=\frac{\sqrt{2}}{m \times B \Delta}=\frac{\sqrt{2}}{3}=3 \sqrt{2}
\end{gathered}
$$

Let $S$ be the total surface area of the cone, $l$ be the slant height of the cone, $r$ be the radius of the base, $R$ be the radius of the sphere, and $s$ be the surface area of the sphere. Then

$$
\begin{gathered}
S=\pi r l+\pi r 2=\pi r(l+r)=\pi \cdot B M \cdot(A B+B M)=\pi \cdot \sqrt{2}(3 \sqrt{2}+\sqrt{2})=8 \pi \\
s=4 \pi R 2=4 \pi \cdot O M 2=4 \pi
\end{gathered}
$$

Therefore, $\frac{\stackrel{5}{\Delta}}{\Delta}=\frac{8 \pi}{1 \pi}=2$.

## Answer

2.00