Stereometry problem(difficult) Three identical spheres, each with radius $=a$ are inserted into a cylinder in such a way, that:


*

*Each sphere is tangent to the lateral surface;

*Each sphere is tangent to the other $2$ spheres;

*$2$ spheres touch the lower base while the third one touches the upper base;

*The height of the cylinder is $3a$.


Question

Find the radius of the cylinder's base.

 A: The centers of the three spheres form an equilateral triangle $\Delta$ of sidelength $2a$ and height $\sqrt{3}a$. The vertices $A$ and $B$ of $\Delta$ lie at level $a$ over the base and $C$ at level $2a$. Denote the projection of $C$ onto level $a$ by $C'$ and the midpoint of $AB$ by $M$. As $\angle(CC'M)=90^\circ$ one computes $|C'M|=\sqrt{2}a$.
The $C$-centered sphere might as well have its center at $C'$. Therefore we have to compute the radius $R$ of the smallest circle containing the three disks at level $a$ with radius $a$ and centers $A$, $B$, $C'$. Below I shall prove that $$R=r+a\ ,\tag{1}$$ where $r$ is  the circumradius  of the isosceles triangle $\Delta':=(A,B,C')$. 
Looking at the rectangular triangle $C'MA$ and drawing the median of its hypotenuse $C'A$ one  computes $r={3\sqrt{2}\over 4} a$. Therefore
$$R=\left(1+{3\sqrt{2}\over 4}\right)a\ .$$
Proof of $(1)$: Let $Q$ be the center of the circumcircle of $\Delta'$. Draw rays from $Q$ through $A$, $B$, $C'$ of length $r+a$, having endpoints $A''$, $B''$, $C''$. Any circle containing the three mentioned disks must contain the points $A''$, $B''$, $C''$. The circle of radius $r+a$ with center $Q$ is the smallest such circle, and this circle does in fact contain our three disks.
