We inscribe a sphere in a certain right circular cone, the surface of which is $\frac{2}{3}$ of the cone's lateral surface. What is the apex angle of the cone? (By the apex angle of the cone, we mean: the angle subtended by two opposite generators of the cone.)
Consider the cross-section of a cone and an inscribed sphere along an axis. The diagram is shown below.

![](https://cdn.mathpix.com/cropped/2024_05_02_2ebd6e4dd3fae443e9a1g-1.jpg?height=586&width=458&top_left_y=192&top_left_x=839)

According to the problem,

$$
\frac{2 r \pi a}{3}=4 \varrho^{2} \pi
$$

which simplifies to

$$
a r=6 \varrho^{2}
$$

It is known that the area of the triangle can be expressed using the semiperimeter $(2 s)$ and the radius of the inscribed circle as $t=s \varrho$. Additionally, the area of the triangle is $A O_{1} \cdot O_{1} C=r \sqrt{a^{2}-r^{2}}$, and since in this case $s=a+r$,

$$
(a+r) \varrho=r \sqrt{a^{2}-r^{2}}
$$

which simplifies to

$$
\varrho=r \frac{\sqrt{a^{2}-r^{2}}}{a+r}=r \sqrt{\frac{(a-r)(a+r)}{(a+r)^{2}}}=r \frac{\sqrt{a-r}}{a+r}
$$

Substituting these values into (1),

$$
a r=\frac{6 r^{2}(a-r)}{a+r}
$$

which simplifies to

$$
a^{2}+a r=6 a r-6 r^{2}
$$

or

$$
6 r^{2}-5 a r+a^{2}=0
$$

from which

$$
r_{1}=\frac{a}{2}, \quad r_{2}=\frac{a}{3}
$$

The half-angle of the cone is $\frac{\gamma}{2}$, so

$$
\sin \frac{\gamma}{2}=\frac{r}{a}
$$

which gives

$$
\begin{array}{ll}
\sin \frac{\gamma_{1}}{2}=\frac{r_{1}}{a}=\frac{1}{2}, & \text { from which } \quad \frac{\gamma_{1}}{2}=30^{\circ}, \text { and thus } \gamma_{1}=60^{\circ}, \\
\sin \frac{\gamma_{2}}{2}=\frac{r_{2}}{a}=\frac{1}{3}, & \text { from which } \quad \frac{\gamma_{2}}{2}=19^{\circ} 28^{\prime}, \text { and thus } \gamma_{2}=38^{\circ} 56^{\prime}
\end{array}
$$