[ $\underline{\text { Cone }}$

Through the edge $B C$ of the triangular pyramid $P A B C$ and the point $M$, the midpoint of the edge $P A$, a section $B C M$ is drawn. The vertex of the cone coincides with the vertex $P$ of the pyramid, and the circle of the base is inscribed in the triangle $B C M$, touching the side $B C$ at its midpoint. The points of tangency of the circle with the segments $B M$ and $C M$ are the points of intersection of the medians of the faces $A P B$ and $A P C$. The height of the cone is twice the radius of the base. Find the ratio of the lateral surface area of the pyramid to the area of the base of the pyramid.

#
Let $E$ and $D$ be the points of intersection of the medians of triangles $A P B$ and $A P C$, $K$ be the midpoint of $B C$, $O$ be the center of the circle inscribed in triangle $B M C$, and $r$ be the radius of the base of the cone, with $2 r$ being its height. Then

$$
P D=P E=P K=\sqrt{r^{2}+4 r^{2}}=r \sqrt{5}
$$

Let $M E=M D=x$. Then $B K=B E=2 x$, $C K=D C=2 x$, so triangle $B M C$ is isosceles. Then

$$
M K=\sqrt{9 x^{2}-4 x^{2}}=x \sqrt{5}
$$

If $p$ is the semiperimeter of the triangle and $S$ is its area, then

$$
S=p r=5 x r, S=\frac{1}{2} B C \cdot M K=2 \times 2 \sqrt{5}
$$

from which we find that $x=\frac{\sqrt{5}}{2}$. Let $P F$ be the median of triangle $A P B$. Since $O E \perp B M$, by the theorem of three perpendiculars, $P F \perp B M$, so

$$
S_{\triangle A P B}=2 S_{\triangle B M P}=2 \cdot \frac{1}{2} M B \cdot P E=3 x \cdot r \sqrt{5}=3 r \cdot \frac{\sqrt{5}}{2} \cdot r \sqrt{5}=\frac{15 r^{2}}{2}
$$

Similarly, $S_{\triangle A P C}=\frac{15 r^{2}}{2}$. Additionally,

$$
S_{\triangle B P C}=\frac{1}{2} B C \cdot P K=2 x \cdot r \sqrt{5}=r \sqrt{5} \cdot r \sqrt{5}=5 r 2
$$

Therefore, the lateral surface area of the pyramid $P A B C$ is $20 r 2$. From the right triangle $B E F$, we find that

$$
\begin{aligned}
B F & =\sqrt{B E^{2}+F E^{2}}=\sqrt{B E^{2}+\left(\frac{1}{2} P E\right)^{2}}= \\
& =\sqrt{4 x^{2}+\left(\frac{r \sqrt{5}}{2}\right)^{2}}=\sqrt{5 r^{2}+\frac{5 r^{2}}{4}}=\frac{5}{2} r .
\end{aligned}
$$

$A B=2 B F=5 r$. Similarly, $A C=5 r$. Then

$$
\begin{aligned}
& A K=\sqrt{A B^{2}-B K^{2}}=\sqrt{25 r^{2}-5 r^{2}}=2 r \sqrt{5} \\
& S_{\triangle A B C}=\frac{1}{2} B C \cdot A K=2 x \cdot 2 r \sqrt{5}=r \sqrt{5} \cdot 2 r \sqrt{5}=10 r 2
\end{aligned}
$$

Therefore, the desired ratio is 2.

## Answer

2.00