Problem 10. A cylinder of volume 9 is inscribed in a cone. The plane of the upper base of this cylinder cuts off a frustum of volume 63 from the original cone. Find the volume of the original cone.
Answer: 64.

Solution. Let the height and radius of the original cone be $H$ and $R$, and the height and radius of the cylinder be $h$ and $r$. We use the formula for the volume of a frustum of a cone: $\frac{1}{3} \pi\left(R^{2}+R r+r^{2}\right) h=63$. We also know that $\pi r^{2} h=9$. Dividing the corresponding parts of the equations, we get $\left(\frac{R}{r}\right)^{2}+\left(\frac{R}{r}\right)+1=$ $\frac{63 \cdot 3}{9}=21$. Solving the quadratic equation, we get the roots 4 and -5, only the positive root has geometric meaning. $R / r=4, \frac{H-h}{H}=4, \frac{h}{H}=\frac{3}{4}$, from which we obtain for the original cone:

$$
V=\frac{1}{3} \pi R^{2} H=\frac{1}{3}\left(\pi r^{2} h\right)\left(\frac{R}{r}\right)^{2} \frac{H}{h}=\frac{1}{3} \cdot 9 \cdot 4^{2} \cdot \frac{4}{3}=64
$$

## Variant II