12.305. In a regular triangular prism, a plane passing through the center of the base and the centers of symmetry of two lateral faces forms an acute angle $\alpha$ with the base plane. Find the area of the section formed by this plane if the side of the base is equal to $a$.
## Solution.

Let $O$ be the center of the base $ABC$ of the regular prism $ABC A_1 B_1 C_1$. Points $P$ and $Q$ are the centers of symmetry of the diagonals of the faces $AA_1 C_1 C$ and $AA_1 B_1 B$ (Fig. 12.167). The intersecting plane passing through points $O$, $P$, and $Q$ intersects the parallel faces $ABC$ and $A_1 B_1 C_1$ along segments $EF$ and $MN$ respectively, with $K$ being the midpoint of $MN$.

Draw perpendiculars $M M_1$, $N N_1$, and $K K_1$ to the plane $ABC$. The trapezoid $E M_1 N_1 F$ is the projection of the trapezoid $E M N F$ onto the base plane.

Since $O$ is the center of $\triangle ABC$, we have $EC = \frac{1}{3} AC = \frac{a}{3}$, and $EF = \frac{2a}{3}$.

$\triangle A_1 M P = \triangle C E P$. Therefore, $A M_1 = A_1 M = E C = \frac{a}{3}$, and $M_1 N_1 = \frac{a}{3}$.

\[
\begin{aligned}
& K_1 O = A K_1 = \frac{1}{2} A O = \frac{a \sqrt{3}}{6} \\
& S_{E M_1 N_1 F} = \frac{E F + M_1 N_1}{2} \cdot K_1 O = \frac{a}{2} \cdot \frac{a \sqrt{3}}{6} = \frac{a^2 \sqrt{3}}{12}
\end{aligned}
\]

Then, $S_{E M N F} = \frac{S_{E M_1 N_1 F}}{\cos \alpha} = \frac{a^2 \sqrt{3}}{12 \cos \alpha}$.

Omiem: $\frac{a^2 \sqrt{3}}{12 \cos \alpha}$.