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At the base of the right prism $A B C A 1 B 1 C 1$ lies a right triangle with legs $A B=8$ and $B C=6$. The hypotenuse AC is the diameter of the base of a cone, the vertex of which is located on the edge $A 1 B 1$. The lateral surface of the cone intersects the edge $A B$ at point $M$ such that $A M=5$. Find the volume of the cone.
Let $O$ be the midpoint of $AC$ (the center of the base of the cone), $S$ - the vertex of the cone, $SK$ - the generatrix of the cone passing through point $M$. Drop a perpendicular $ON$ from point $O$ to edge $AB$. The planes of the faces $ABC$ and $AA_1B_1B$ are perpendicular because the prism is a right prism. Therefore, the line $ON$ is perpendicular to the plane $AA_1B_1B$. Then the segment $NS$ is the orthogonal projection of the height $OS$ of the cone onto the plane $AA_1B_1B$. At the same time, point $N$ is the projection of the center of the base of the cone onto the plane passing through the generatrices $SA$ and $SK$, so $N$ lies on the bisector $SE$ of the isosceles triangle $ASK$. The segment $ON$ is the midline of the right triangle $ABC$, so $ON = \frac{1}{2} BC = 3$, and since $N$ is the midpoint of $AB$, then

$$
MN = BN - BM = \frac{1}{2} AB - MB = 4 - (8 - 5) = 1.
$$

Consider the isosceles triangle $ASK$ located in the plane $AA_1B_1B$. Its median $SE$ intersects the segment $AM$ at point $N$ and at the same time $AN = 4$ and $MN = 1$. Draw a line through point $E$ parallel to $AM$. Let this line intersect $SK$ at point $F$. Then $EF$ is the midline of triangle $AMK$, so $EF = \frac{1}{2} AM = \frac{5}{2}$. From the similarity of triangles $SNM$ and $SEF$ we find that

$$
\frac{SN}{SE'} = \frac{MN}{EF'} = \frac{1}{5} = \frac{2}{5}
$$

Therefore, $\frac{SN}{NE} = \frac{2}{3}$. Let $SN = 2t, NE = 3t$. Consider the right triangle $SOE$. Its height $ON$, drawn from the vertex of the right angle, divides the hypotenuse $SE$ into segments $SN = 2t$ and $NE = 3t$, so $SN \cdot NE = ON^2$, or $6t^2 = 9$, from which we find that $t^2 = \frac{3}{2}$. Then

$$
SO^2 = SN \cdot SE = 2t(2t + 3t) = 10t^2 = 10 \cdot \frac{3}{2} = 15
$$

Let $V$ be the volume of the cone. Then

$$
V = \frac{1}{3} \pi OA^2 \cdot SO = \frac{1}{3} \pi \cdot 25 \cdot \sqrt{15} = 25 \pi \sqrt{\frac{5}{3}}.
$$

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