99. Given a regular triangular pyramid $S A B C$ (with $S$ as its vertex). The edge $S C$ of this pyramid coincides with the lateral edge of a regular triangular prism $A_{1} B_{1} C A_{2} B_{2} S\left(A_{1} A_{2}, B_{1} B_{2}\right.$ and $C S$ are lateral edges, and $A_{1} B_{1} C$ is one of the bases). The vertices $A_{1}$ and $B_{1}$ lie in the plane of the face $S A B$ of the pyramid. What fraction of the volume of the entire pyramid is the volume of the part of the pyramid that lies inside the prism, if the ratio of the length of the lateral edge of the pyramid to the side of its base is $2 / \sqrt{3}$?
99. Let $K$ be the midpoint of $AB$, and $P$ be the foot of the perpendicular dropped from $K$ to $CS$. Take points $M$ and $N$ on $AB$ such that $\triangle PMN$ is equilateral (Fig. 11). The pyramid $SPMN$ can be extended to a regular prism $PMNSM_1N_1$ such that $PMN$ and $SM_1N_1$ are its bases, and $PS, MM_1, NN_1$ are its lateral edges. The prism $A_1B_1CA_2B_2S$ will be homothetic to the prism $PMNSM_1N_1$ with the center at $S$ and the coefficient $|CS| / |PS|$. It is easy to see that the sought fraction of the volume of the pyramid $SABC$ that lies inside the prism $\boldsymbol{A}_1B_1CA_2B_2S$ is equal to the ratio $|MN| / |AB|$. Let $|AB| = a \sqrt{3}, |CS| = 2a$. Find:

$$
\begin{aligned}
& |SK| = \frac{\sqrt{13}}{2} a, \quad |CK| = \frac{3}{2} a, \quad |PS| = \frac{5}{4} a, \quad |PK| = \frac{3 \sqrt{3}}{4} a \\
& \quad |MN| = |PK| \frac{2}{\sqrt{3}} = \frac{3}{2} a, \quad |MN| / |AB| = \frac{\sqrt{3}}{2} \\
& \text{Answer: } \frac{\sqrt{3}}{2} .
\end{aligned}
$$