6. A sphere is inscribed in a right circular cone, the radius of the base of which is equal to 2. Find the volume of this sphere if it is three times smaller than the volume of the cone.
Answer: $\frac{16 \pi}{9} \sqrt{9+5 \sqrt{3}}$ or $\frac{16 \pi}{9} \sqrt{9-5 \sqrt{3}}$.

Solution. Let $r$ be the radius of the inscribed sphere, $R=2$ be the radius of the base, $h$ be the height of the cone, $2 \varphi$ be the angle between the slant height of the cone and the plane of its base, $0<\varphi<\pi / 4, t=\operatorname{tg} \varphi, 0<t<1$. Then $r=R t$, $h=R \operatorname{tg} 2 \varphi=R \frac{2 t}{1-t^{2}}$. By the condition

$$
k V_{\text {sphere }}=V_{\text {cone }} \quad \Leftrightarrow \quad k \cdot \frac{4}{3} \pi r^{3}=\frac{1}{3} \pi R^{2} h \quad \Leftrightarrow \quad 4 k t^{3}=\frac{2 t}{1-t^{2}} \quad \Leftrightarrow \quad 2 k t^{4}-2 k t^{2}+1=0
$$

For $k=3$, this equation has two solutions $t_{1,2}=\sqrt{\frac{k \pm \sqrt{k^{2}-2 k}}{2 k}}=\frac{\sqrt{3 \pm \sqrt{3}}}{\sqrt{6}}$, both of which satisfy the condition $0<t<1$. Therefore, there are two possible cases, in one of which the volume of the sphere is $V=\frac{4}{3} \pi R^{3} t^{3}=\frac{16 \pi}{9} \sqrt{9+5 \sqrt{3}}$, and in the other $V=\frac{16 \pi}{9} \sqrt{9-5 \sqrt{3}}$.

Answer to variant 2: $18 \pi \sqrt{20+14 \sqrt{2}}$ or $18 \pi \sqrt{20-14 \sqrt{2}}$.

Answer to variant 3: $\frac{4 \pi}{3} \sqrt{20+14 \sqrt{2}}$ or $\frac{4 \pi}{3} \sqrt{20-14 \sqrt{2}}$.

Answer to variant 4: $18 \pi \sqrt{9+5 \sqrt{3}}$ or $18 \pi \sqrt{9-5 \sqrt{3}}$.