1. A bottle in the shape of a cone rests on its base. It is filled with water until the liquid level reaches 8 centimeters measured vertically below the vertex of the cone. If the bottle is now inverted, without changing the amount of water inside, the empty space in the upper part of the inverted cone is 2 centimeters high.

How tall is the bottle?
Solution: Let $h$ be the height in centimeters of the bottle and $r$ the radius of the base of the cone. The volume of the bottle is equal to $\frac{\pi r^{2} h}{3}$. While the bottle is resting on its base, the water occupies a volume equal to the difference between the volume of the bottle and that of the cone of height 8 centimeters obtained by slicing the bottle at the level of the liquid. Since the plane is parallel to the base of the bottle, this latter cone is similar to the bottle. The ratio of similarity is equal to the ratio of the two heights, which is $8 / h$. Therefore, the volume occupied by the water is

$$
\frac{\pi r^{2} h}{3}-\frac{\pi\left(\frac{8 r}{h}\right)^{2} \cdot 8}{3}
$$

After the bottle is turned upside down, the water occupies the volume of a cone of height $h-2$ similar to the bottle. Therefore, the volume occupied by the water is

$$
\frac{\pi\left(\frac{(h-2) r}{h}\right)^{2}(h-2)}{3} .
$$

By equating the two expressions for the volume occupied by the water after multiplying them by $\frac{3 h^{2}}{\pi r^{2}}$, we obtain

$$
h^{3}-8^{3}=(h-2)^{3} \text {, }
$$

which is equivalent to

$$
h^{2}-2 h-84=0 .
$$

The last equality is a quadratic equation in $h$ with a single positive solution, which is $h=1+\sqrt{85}$.

Second Solution: Note that the following three solids are similar: the cone formed by the entire bottle, the cone without water when the bottle is resting on its base, and the cone occupied by the water when the bottle is inverted. The ratios of similarity between the three cones are given by the ratios of their heights: if $h$ is the height in centimeters of the bottle, they are (in order) $h: 8: h-2$. Since the ratio of the volumes of two similar solids is given by the cube of the ratio of similarity, the volumes are in the ratio $h^{3}: 8^{3}:(h-2)^{3}$. Let $V$ be the volume of the bottle; imposing that the volume of the water is the same in the two cases discussed is equivalent to considering the equation $V\left(1-\frac{8^{3}}{h^{3}}\right)=V \frac{(h-2)^{3}}{h^{3}}$; this, multiplying by $h^{3}$, becomes $h^{3}-512=h^{3}-6 h^{2}+12 h-8$, or $h^{2}-2 h-84=0$, which has as its only positive solution $h=1+\sqrt{85}$.