8. (Iceland 2) A plane passing through the midpoint of a circular cone's height and tangent to the circumference of the base circle divides the cone into two parts. Find the ratio of the smaller part to the entire cone.

保留源文本的换行和格式如下：

8. (Iceland 2) A plane passing through the midpoint of a circular cone's height and tangent to the circumference of the base circle divides the cone into two parts. 
Find the ratio of the smaller part to the entire cone.
Let's assume
the base of the cone is 1,
the height PN = h.
(Figure 5). Denote
$A, C$, and let $A$

Let $B$ be the antipodal point of $A$ on the circumference of the base circle, $M$ be the midpoint of the height $PN$, $O$ be the midpoint of $AC$, i.e., the center of the circle. Also, let $E$ be a point on the ellipse such that $OE$ is the semi-major axis of the ellipse. Extend $PE$ and $PO$ to intersect the base of the cone at $F$ and $G$, respectively. Clearly, $EG \perp AB$, $OE \perp CA$.

Observe $\triangle APB$ (Figure 6), connect $N, O$, and extend $NO$ to intersect $AP$ at $D$. Since $N, O$ are the midpoints of $AB$ and $AC$, respectively, $D$ is the midpoint of $AP$, hence $O$ is the centroid of $\triangle ANP$. From this, we know that $G$ is the midpoint of $AN$, and
$$
PC: PB = DO: DN = 1: 3.
$$

Let the distance from point $P$ to the line $AC$ be $d$, the volume of the cone be $V$, and the volume of the smaller part be $V_1$, then
$$
\begin{array}{l}
V = \frac{1}{3} \pi h, \\
V_1 = \frac{1}{3} \pi d \cdot OA \cdot OE. \text{ [Solution] }
\end{array}
$$

But $d \cdot OA = S_{\triangle PCA} = \frac{1}{3} S_{\triangle PAB} = \frac{1}{3} h$.
Thus, $V_1 = \frac{1}{9} \pi h \cdot OE = \frac{1}{3} V \cdot OE$.
Also, since $FG^2 = AG \cdot GB = \frac{1}{2} \cdot \frac{3}{2} = \frac{3}{4}$, we have $FG = \frac{\sqrt{3}}{2}$. Therefore, from
$$
OE: FG = PO: PG = \frac{2}{3}
$$

we get $OE = \frac{1}{\sqrt{3}} \cdot$ Finally, we have
$$
\frac{V_1}{V} = \frac{1}{3} OE = \frac{\sqrt{3}}{9}.
$$