## Task A-3.2.

Four spheres of radius $R$ lie on the base of a cone such that each touches two of the other spheres and the lateral surface of the cone. A fifth sphere of the same radius touches the first four spheres and the lateral surface of the cone. Determine the volume of this cone.
## Solution.

Let $A, B, C$ and $D$ be the centers of the spheres that lie on the base of the cone, and $E$ be the center of the fifth sphere.

![](https://cdn.mathpix.com/cropped/2024_05_30_a630fb9dab34a5225850g-14.jpg?height=771&width=1205&top_left_y=1274&top_left_x=357)

Then $A B C D E$ is a regular square pyramid with all edges of length $2 R$. Therefore, $A B C D$ is a square and $|A C|=2 R \sqrt{2}$.

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Consider the cross-section of the cone by the plane passing through points $A, C$, and $E$.

Let $P$ be the vertex of the cone, and let $Q$ and $R$ be the other two vertices of this cross-section.

![](https://cdn.mathpix.com/cropped/2024_05_30_a630fb9dab34a5225850g-15.jpg?height=571&width=981&top_left_y=177&top_left_x=469)

Since the spheres with centers $A, C$, and $E$ touch the lateral surface of the cone, it follows that $A E \| Q P$ and $C E \| R P$, so $\varangle Q P R=90^{\circ}$.

Thus, triangle $Q P R$ is an isosceles right triangle, and the radius of the base of the cone is equal to its height.

Let $S$ and $T$ be the points of tangency of the fifth sphere with the generators $\overline{Q P}$ and $\overline{R P}$, respectively. Then the quadrilateral $P S E T$ is a square with side length $r$, and we conclude that $|P E|=R \sqrt{2}$.

Let $O$ be the center of the square $A B C D$, i.e., the midpoint of the segment $\overline{A C}$. Then

$$
|O E|=\frac{1}{2}|A C|=R \sqrt{2}
$$

Finally, let $N$ be the foot of the height of the cone, i.e., the midpoint of the segment $\overline{Q R}$. Then $|O N|=R$ and we conclude that the height of the cone is

$$
\begin{aligned}
v & =|P N|=|P E|+|E O|+|O N| \\
& =R \sqrt{2}+R \sqrt{2}+R \\
& =(1+2 \sqrt{2}) R
\end{aligned}
$$

The volume of the cone is

$$
\begin{aligned}
V & =\frac{1}{3} B v \\
& =\frac{1}{3}(1+2 \sqrt{2})^{2} R^{2} \pi \cdot(1+2 \sqrt{2}) R \\
& =\frac{(1+2 \sqrt{2})^{3}}{3} \cdot R^{3} \pi \\
& =\frac{25+22 \sqrt{2}}{3} \cdot R^{3} \pi
\end{aligned}
$$