4 $[\underline{\text { Tangent Spheres }}]$

A cone contains five equal spheres. Four of them lie on the base of the cone, each touching two others lying on the base and the lateral surface of the cone. The fifth sphere touches the lateral surface and the other four spheres. Find the volume of the cone if the radius of each sphere is $r$.
Let the spheres with centers $O_1, O_2, O_3$, and $O_4$ (Fig.1) touch the base of a cone with vertex $A$ and base radius $R$ at points $Q_1, Q_2, Q_3$, and $Q_4$ respectively, and the center $O_5$ of the fifth sphere lies on the height $AP$ of the cone, which is equal to $h$. Then $P$ is the center of the square $Q_1Q_2Q_3Q_4$ with side length $2r$, so

$$
PQ_1 = PQ_2 = PQ_3 = PQ_4 = r \sqrt{2}
$$

Consider the regular quadrilateral pyramid $O_1O_2O_3O_4O_5$ with vertex $O_5$, all of whose edges are equal to 2. The angle between its lateral edge and the plane of the base is $45^\circ$. Since the lateral edge of the pyramid $O_1O_2O_3O_4O_5$ is parallel to some generatrix of the cone, the angle at the base of the axial section of the cone is also $45^\circ$. Consider the axial section $ABC$ of the cone (Fig.2), passing through the point $O_1$. Let the point $Q_1$ lie on the leg $BP$ of the right triangle $APB$. Since $BO_1$ is the bisector of the angle $ABP$,

$$
BQ_1 = O_1Q_1 \operatorname{ctg} \angle Q_1BO_1 = r \operatorname{ctg} \frac{45^\circ}{2} = r(1 + \sqrt{2}).
$$

Therefore,

$$
\begin{gathered}
R = PB = PQ_1 + Q_1B = r \sqrt{2} + r(1 + \sqrt{2}) = r(1 + 2\sqrt{2}) \\
h = R \operatorname{tg} 45^\circ = R = r(1 + 2\sqrt{2}).
\end{gathered}
$$

If $V$ is the volume of the cone, then

$$
V = \frac{1}{3} \pi R^2 h = \frac{1}{3} \pi R^3 = \frac{1}{3} \pi r^3 (1 + 2\sqrt{2})^3.
$$

## Answer

$\frac{1}{3}(1 + 2\sqrt{2})^3 \pi r^3$