8,9

A cylinder is circumscribed around a sphere of radius $R$. Point $P$ is located inside the cylinder on its axis and is $\frac{3}{4} R$ away from the bottom base. Through this point, a plane $\alpha$ is drawn, which has only one common point with the circle of the base. A cone is inscribed in the sphere, with its base lying in the plane $\alpha$ and its vertex located above this plane. Find the volume of the cone.
Consider the section of the cylinder by a plane passing through the center of the sphere and point $P$ (Fig. 2). We obtain a square $A B C D$, in which a circle of radius $R$ with center $O$ is inscribed. Let the intersecting plane intersect the plane $\alpha$ along the line $A K$ ($K$ is a point on $B C$), and $M$ and $N$ are the points of intersection of the circle with the line $A K$. Then $M N$ is the diameter of the base of the specified cone, the midpoint $Q$ of the segment $M N$ is the center of the base of the cone, $E$ is the point of tangency of the circle with the side $A B$, and the point $L$ of intersection of the extension of the segment $Q O$ beyond point $O$ with the circle inscribed in the square $A B C D$ is the vertex of the cone. Let $Q M=Q N=r, L Q=h, \angle K A B=\phi$. Then

\[
\begin{gathered}
\operatorname{tg} \phi=\frac{P E}{A E}=\frac{\frac{3}{R} R}{R}=\frac{3}{4}, \cos \phi=\frac{4}{5} \\
O Q=O P \cos \angle P O Q=\frac{1}{4} R \cos \phi=\frac{1}{5} R
\end{gathered}
\]

From the right triangle $O Q N$ we find that

\[
r=Q N=\sqrt{O N^{2}-O Q^{2}}=\sqrt{R^{2}-\left(\frac{1}{5} R\right)^{2}}=\frac{2}{5} R \sqrt{6}
\]

and since

\[
h=Q L=O L+O Q=R+\frac{1}{5} R=\frac{6}{5} R
\]

then, denoting the volume of the cone by $V$, we get

\[
V=\frac{1}{3} \pi r 2 h=\frac{1}{3} \pi\left(\frac{2}{5} R \sqrt{6}\right) 2 \cdot \frac{6}{5} R=\frac{45}{125} \pi R 3 .
\]

## Answer

$\frac{45}{125 \pi} R 3$.