IV OBM

The left-hand diagram shows the four spheres on the base. Evidently $AC=2r\sqrt{2}$.

The right-hand diagram shows a vertical section through $A$, $C$ and the center $O$ of the top sphere. $P$ is the apex of the cone and $QR$ is a diameter of its base. Evidently $PQR$ is similar to $OAC$ and its sides are parallel and at distance $r$ outside the corresponding sides of $OAC$.



Also $AOC$ is congruent to $ABC$, so $\angle AOC=90^{\circ }$. Hence $\angle QPR=90^{\circ }$ also and so $OP=r\sqrt{2}$. The altitude from $O$ in $AOC$ has length $AC/2=r\sqrt{2}$. Hence the altitude from $P$ in $PQR$ has length $OP+r\sqrt{2}+r=(2\sqrt{2}+1)r$. Thus the radius of the base of the cone is also $(2\sqrt{2}+1)r$ and its volume is $\frac{\pi r^3(2\sqrt{2}+1{)}^3}{3}$.