A regular hexagonal prism has a base $A_{1}, \ldots, A_{6}$ and a top face $B_{1}, \ldots, B_{6}$, with the center of the latter being O. The side edge $A_{1} B_{1}$ is equal to the diagonal $A_{1} A_{3}$. Take points $C_{1}, C_{3}, C_{5}$ on the side edges $A_{1} B_{1}, A_{3} B_{3}, A_{5} B_{5}$ respectively, such that $A_{1} C_{1}=A_{3} C_{3}=A_{5} C_{5}=A_{1} A_{2}$. Consider the part of the prism below the planes $C_{1} O C_{3}, C_{3} O C_{5}, C_{5} O C_{1}$. Determine the surface area and volume of this part if $A_{1} A_{2}=1$.
1. Denoting the center of the base by $O^{\prime}$, the symmetries of the prism are: a $60^{\circ}$ rotation about the $O O^{\prime}$ axis, reflections in the planes $A_{i} O A_{i+3}$ $(i=1,2,3)$, and 4 additional mirror planes which we will not need. After selecting the points $C_{j} (j=1,3,5)$ and cutting the body with the planes $C_{j} O C_{j+2}$, the remaining part is symmetric with respect to the planes $A_{i} O A_{i+3}$, but only $120^{\circ}$ rotations about the $O O^{\prime}$ axis map the body onto itself (cyclically permuting $C_{1}, C_{3}, C_{5}$), because the plane sloping from $O$ to $C_{1}$ and $C_{3}$ intersects the line of the edge $A_{2} B_{2}$ at a point deeper than $C_{1}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_01bbfccb7cf9c85179e1g-1.jpg?height=647&width=593&top_left_y=392&top_left_x=755)

Denoting this intersection point by $D_{2}$, the quadrilateral $O C_{1} D_{2} C_{3}$ is a rhombus. Indeed, the side face $A_{2} B_{2} B_{3}$ is parallel to the plane $A_{1} C_{1} O$ (i.e., $A_{1} O A_{4}$), so the mirror images of these planes in the plane $A_{2} B_{2} O$, the planes $A_{2} B_{2} B_{1}$ and $A_{3} C_{3} O$, are also parallel. Therefore, the intersection lines of the plane $O C_{1} C_{3}$ are $D_{2} C_{3} \| C_{1} O$ and $D_{2} C_{1} \| C_{3} O$, and clearly $O C_{3} = O C_{1}$. Denoting the center of this rhombus by $M$, $D_{2}$ is twice as far below the plane of the top face as $M, C_{1}$, and $C_{3}$, i.e., $B_{2} D_{2} = 2 B_{1} C_{1}$. And since

$$
B_{1} C_{1} = B_{1} A_{1} - C_{1} A_{1} = A_{3} A_{1} - A_{2} A_{1} < A_{2} A_{3} = C_{1} A_{1}
$$

i.e., $C_{1}$ is higher than half the height of the prism, $D_{2}$ is an internal point of the edge $A_{2} B_{2}$.

Thus, the rhombus mentioned is a face of the remaining part, and from the side faces of the prism, trapezoids congruent to $A_{2} D_{2} C_{1} A_{1}$ remain.

2. For the purpose of surface area calculation, based on the above,

$$
\begin{gathered}
B_{2} D_{2} = 2(\sqrt{3} - 1), \quad O D_{2} = \sqrt{1 + 4(\sqrt{3} - 1)^{2}} = \sqrt{17 - 8 \sqrt{3}}, \\
C_{1} C_{3} = \sqrt{3}, \quad A_{2} D_{2} = 2 - \sqrt{3}
\end{gathered}
$$

and the combined area of the 3 rhombuses, 6 trapezoids, and the base hexagon is

$$
F = 3 \cdot \frac{\sqrt{3} \cdot \sqrt{17 - 8 \sqrt{3}}}{2} + 6 \cdot \frac{3 - \sqrt{3}}{2} + 6 \cdot \frac{\sqrt{3}}{4} = 9 + 1.5 \sqrt{51 - 24 \sqrt{3}} - 1.5 \sqrt{3} = 11.01 \text{ area units. }
$$

3. Cut the examined body with the plane $S$ passing through $C_{1}$ and parallel to the base, removing the pyramid $O C_{1} C_{3} C_{5}$, and denote the points of $O O^{\prime}$ and $A_{2} B_{2}$ in $S$ by $O^{\prime \prime}$ and $C_{2}$, respectively. A $180^{\circ}$ rotation about the $C_{1} C_{3}$ axis maps $O^{\prime \prime}$ to $C_{2}$ and therefore $O$ to $D_{2}$, so the pyramid $O O^{\prime \prime} C_{1} C_{3}$ is suitable for filling the pyramid-shaped space $D_{2} C_{2} C_{1} C_{3}$ that forms above the examined body and below $S$. It is evident from this that the volume of the examined body is the same as that of the prism with base $A_{1} \ldots A_{6}$ and side edge $A_{1} C_{1}$, which is $1.5 \sqrt{3} = 2.60$ volume units.