11.192. A plane passing through the vertex of a cone intersects the base along a chord, the length of which is equal to the radius of this base. Determine the ratio of the volumes of the resulting parts of the cone.
Solution.

Let $AO = r, SO = h$ (Fig. 11.79), $V$ be the volume of the cone, and $V_{1}$ and $V_{2}$ be the volumes of its parts. We will find $V_{1}$ as the difference between the volumes of the part of the cone whose base is the sector $AOB$ and the pyramid whose base is $\triangle AOB$. According to the condition, $AB = r$, i.e., $AB$ is a side of the inscribed regular hexagon, and thus $V_{1} = \frac{1}{6} \cdot \frac{1}{3} \pi r^{2} h - \frac{1}{3} \cdot \frac{r^{2} \sqrt{3}}{4} h = \frac{1}{3} S_{1} h$, where $S_{1}$ is the area of the segment $AmB$. Then $V_{2} = V - V_{1} = \frac{1}{3} S_{2} h$, where $S_{2} = \pi r^{2} - S_{1}$. Therefore, $V_{1}: V_{2} = S_{1}: S_{2}$. Next, we find $S_{1} = \frac{1}{6} \pi r^{2} - \frac{r^{2} \sqrt{3}}{4} = \frac{r^{2}(2 \pi - 3 \sqrt{3})}{12}, S_{2} = \pi r^{2} - \frac{r^{2}(2 \pi - 3 \sqrt{3})}{12} = \frac{r^{2}(10 \pi + 3 \sqrt{3})}{12} \Rightarrow V_{1}: V_{2} = \frac{2 \pi - 3 \sqrt{3}}{10 \pi + 3 \sqrt{3}}$.

Answer: $\quad \frac{2 \pi - 3 \sqrt{3}}{10 \pi + 3 \sqrt{3}}$.