A cone and a cylinder have equal bases and equal heights. Their bases lie in the same plane and touch each other. Two spheres with radii equal to the radii of the bases of the cone and cylinder touch each other, the lateral surfaces of the cone and the cylinder, as well as the plane containing the other base of the cylinder and the vertex of the cone. Find the angle at the vertex of the axial section of the cone.
Let the radii of the bases of the cylinder, cone, and spheres be $r$ (Fig.1), and the angle to be found be $\alpha$. Consider the orthogonal projection of the cone, cylinder, and spheres onto the plane $\phi$, which contains the vertex of the cone and the base of the cylinder (Fig.2). We obtain four circles with centers $A, B, C$, and $D$ of radius $r$, where the circles with centers $A, C$, and $D$ touch each other pairwise, and the circle with center $B$ (the orthogonal projection of the cone's axis) touches the circle with center $A$ and intersects the circles with centers at points $C$ and $D$. Since triangle $A C D$ is equilateral, $\angle C A D=60^{\circ}$. Therefore,

$$
\angle C A B=\angle D A B=\frac{1}{2} \angle C A D=30^{\circ}
$$

From the isosceles triangle $A B C$, we find that

$$
B C=2 A B \sin \frac{1}{2} \angle C A B=4 r \sin 15^{\circ}=r(\sqrt{6}-\sqrt{2}) .
$$

Consider the section of the cone and one of the spheres by a plane passing through the axis of the cone and the center of this sphere (Fig.3). We obtain an isosceles triangle $K B L$ (the axial section of the cone) with base $K L=2 r$ and a circle with center $O$ (the center of the sphere) of radius $r$, inscribed in the angle $L B C$, where $C$ is the point of tangency of the sphere with the plane $\phi$. From the right triangle $O B C$, we find that

$$
\operatorname{tg} \angle O B C=\frac{O C}{B C}=\frac{r}{r(\sqrt{6}-\sqrt{2})}=\frac{1}{\sqrt{6}-\sqrt{2}}=\frac{\sqrt{6}+\sqrt{2}}{4} .
$$

Therefore,

$$
\begin{aligned}
\alpha= & \angle K B L=180^{\circ}-2 \angle L B C=180^{\circ}-2 \cdot 2 \angle O B C= \\
& =180^{\circ}-4 \angle O B C=180^{\circ}-4 \operatorname{arctg} \frac{\sqrt{6}+\sqrt{2}}{4} .
\end{aligned}
$$

## Answer

$\pi-4 \operatorname{arctg} \frac{\sqrt{6}+\sqrt{2}}{4}$.