## Task A-4.7. (20 points)

The angle at the vertex of the axial section of a right circular cone is $2 \alpha$, and the radius of the base is $r$. A regular hexagonal prism is inscribed in the cone, with all its edges of equal length (one base of the prism lies in the plane of the base of the cone, and the remaining vertices are on the lateral surface of the cone). Calculate the surface area of the prism using $\alpha$ and $r$.
## Solution.

Let all edges of the inscribed prism be of length $a$ and let $A B V$ be the axial section of the given cone, and $C D E F$ the section of the inscribed prism by the same plane (see figure). The shape $C D E F$ is a rectangle with side lengths $|D E|=a$ and $|E F|=2 a$ (since $\overline{E F}$ is the longer diagonal of the upper base of the prism, a regular hexagon).

![](https://cdn.mathpix.com/cropped/2024_05_30_59925bf190fc107f19b7g-20.jpg?height=586&width=551&top_left_y=875&top_left_x=750)

(5 points)

Since $\varangle D E B = \varangle O V B = \alpha$ (angles with parallel legs), from the right triangle $D E B$ we get $|D B| = |D E| \operatorname{tg} \alpha$, or $r - a = a \operatorname{tg} \alpha$.

From this, we express $a = \frac{r}{1 + \operatorname{tg} \alpha}$.

(5 points)

The surface area of the given prism is $O = 2 \cdot \frac{6 a^{2} \sqrt{3}}{4} + 6 a^{2} = a^{2} (3 \sqrt{3} + 6)$.

(5 points)

By substituting the expression for $a$, we finally get $O = \frac{r^{2} (3 \sqrt{3} + 6)}{(1 + \operatorname{tg} \alpha)^{2}}$.

(5 points)