10. (ICE 2) A plane cuts a right circular cone into two parts. The plane is tangent to the circumference of the base of the cone and passes through the midpoint of the altitude. Find the ratio of the volume of the smaller part to the volume of the whole cone.
10. Let $r$ be the radius of the base and $h$ the height of the cone. We may assume w.l.o.g. that $r=1$. Let $A$ be the top of the cone, $B C$ the diameter of the circumference of the base such that the plane touches the circumference at $B, O$ the center of the base, and $H$ the midpoint of $O A$ (also belonging to the plane). Let $B H$ cut the sheet of the cone at $D$. By applying Menelaus's theorem to $\triangle A O C$ and $\triangle B H O$, we conclude that $\frac{\Delta D}{D D}=\frac{C R}{B O} \cdot \frac{O H}{H A}=\frac{1}{2}$ and $\frac{H R}{D B}=\frac{H A}{A O} \cdot \frac{O C}{C B}=\frac{1}{4}$.
The plane cuts the cone in an ellipse whose major axis is $B D$. Let $E$ be the center of this ellipse and $F G$ its minor axis. We have $\frac{A E}{E D}=\frac{1}{2}$. Let $E^{\prime}, F^{\prime}, G^{\prime}$ be radial projections of $E, F, G$ from $A$ onto the base of the cone. Then $E$ sits on $B C$. Let $h(X)$ denote the height of a point $X$ with respect to the base of the cone. We have $h(E)=h(D) / 2=h / 3$.

Hence $E F=2 E^{r} F^{v} / 3$. Applying Menelaus's theorem to $\triangle B H O$ we get $\frac{O E^{\prime}}{E^{\top} B}=\frac{B E}{E H} \cdot \frac{H A}{A O}=1$. Hence $E F=\frac{2}{3} \frac{\sqrt{3}}{2}=\frac{1}{\sqrt{3}}$.
Let $d$ denote the distance from $A$ to the plane. Let $V_{1}$ and $V$ denote the volume of the cone above the plane (on the same side of the plane as $A$ ) and the total volume of the cone. We have
$$
\begin{aligned}
\frac{V_{1}}{V} & =\frac{B E \cdot E F \cdot d}{h}=\frac{(2 B H / 3)(1 / \sqrt{3})\left(2 S_{A H B} / B H\right)}{h} \\
& =\frac{(2 / 3)(1 / \sqrt{3})(h / 2)}{h}=\frac{1}{3 \sqrt{3}} .
\end{aligned}
$$

Since this ratio is smaller than $1 / 2$, we have indeed selected the correct volume for our ratio.