5. A cylinder is called equilateral if its height is equal to the diameter of its base. An equilateral cylinder is inscribed in a regular tetrahedron with edge $a$ such that one of the cylinder's bases lies in one of the tetrahedron's faces, and the circle of the second base touches the three other faces. Calculate the volume and the lateral surface area of this cylinder.
5. Place a right circular cylinder inside a tetrahedron such that its base coincides with the base of the tetrahedron, and its axis coincides with the height of the tetrahedron. Now, intersect the entire figure with a plane passing through a lateral edge and the height of the tetrahedron. Then, in

![](https://cdn.mathpix.com/cropped/2024_05_21_5cdadc99c18e78f1502ag-177.jpg?height=542&width=577&top_left_y=1825&top_left_x=360)

Figure 104, the plane of intersection will yield the figure shown in Figure 104. Here, $[M P]$ is an edge of the tetrahedron, $[M N]$ is the height of the base, $[N P]$ is the height of the lateral face, $O$ is the common center of the bases of the tetrahedron and the cylinder, and $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is a square in the cross-section of the cylinder. The ray $O A^{\prime}$ intersects $(N P)$ at point $A$. Taking $O$ as the center of homothety, which maps $A^{\prime}$ to $A$, applying this homothety to the cylinder, we obtain a cylinder with the cross-section $A B C D$, and the circle of its upper base will touch all the faces of the tetrahedron - at point $A$

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and at corresponding points on the other two faces. To calculate the lateral surface area and volume of the right circular cylinder, it is sufficient to know its radius $r$. The lateral surface area is $S=4 \pi r^{2}$; the volume is $V=2 \pi r^{3}$. To determine $r$, we use the similarity of triangles $P O N$ and $A D N$. We have:

$$
\frac{|A D|}{|P O|}=\frac{|D N|}{|O N|} \text {. }
$$

Here, $|A D|=2 r ;|P O|^{2}=a^{2}-\frac{a^{2}}{3}=\frac{2 a^{2}}{3} . \quad$ Further, $\quad|P O|=\frac{a \sqrt{2}}{\sqrt{3}}$; $|D N|=|O N|-|O D|=\frac{a}{2 \sqrt{3}}-r$. From equality (1), we get:

$$
r=\frac{a(2 \sqrt{3}-\sqrt{6})}{6}
$$