## Task B-3.2.

A right circular cone is inscribed in a cylinder such that its base coincides with one base of the cylinder. The vertex of the cone is located at the center of the other base of the cylinder. If the measure of the central angle of the cone's lateral surface is $120^{\circ}$, determine the ratio of the surface areas of the cone and the cylinder.
## Solution.

The surface area of a cylinder with radius $R$ and height $H$ is $O_{V}=2 R \pi(R+H)$.

The surface area of a cone with the same base, height $H$, and slant height $s$ is $O_{S}=R \pi(R+s)$.

Then, $\frac{O_{S}}{O_{V}}=\frac{R+s}{2(R+H)}$.

Since the central angle of the lateral surface $\alpha=120^{\circ}$, we have:

$\frac{360^{\circ}}{120^{\circ}}=\frac{s}{R}$, so $R=\frac{1}{3} s$.

$H=\sqrt{s^{2}-R^{2}}$, so $H=\frac{2 \sqrt{2}}{3} s$.

1 point

Therefore:

$\frac{O_{S}}{O_{V}}=\frac{R+s}{2(R+H)}=\frac{2}{1+2 \sqrt{2}}=\frac{-2+4 \sqrt{2}}{7}$.

Note: The student does not need to rationalize the denominator in the final solution to receive full marks.