12.229. The plane angle at the vertex of a regular n-sided pyramid is equal to $\alpha$. The segment of the line connecting the center of the base of the pyramid with the midpoint of a lateral edge is equal to $a$. Find the total surface area of the pyramid.
Solution.

Let $CO$ be the height of the regular pyramid $CA_1A_2\ldots A_n$ (Fig. 12.96), $\angle A_1CA_2 = \alpha, B$ be the midpoint of $CA_2$, and $OB = a$.

$OB$ is the median of the right $\angle COA_2$, drawn to the hypotenuse, hence $CA_2 = 2a, S_{\triangle A_1CA_2} = \frac{1}{2} A_1C^2 \sin \alpha = 2a^2 \sin \alpha$.

The lateral surface area of the pyramid

$$
S_1 = n S_{\triangle A_2CA_2} = 2na^2 \sin \alpha
$$

If $\angle CA_2A_1$ is the angle between the line $A_1A_2$, lying in the plane of the base of the pyramid, and the inclined line $CA_2$ to this plane, $OA_2$ is the projection of $CA_2$ onto this plane, then $\cos \angle CA_2A_1 = \cos \angle OA_2A_1 \cdot \cos \angle CA_2O \Rightarrow$

$$
\Rightarrow \cos \left(\frac{\pi}{2} - \frac{\alpha}{2}\right) = \cos \left(\frac{\pi}{2} - \frac{\pi}{n}\right) \cos \angle CA_2O \Rightarrow \cos \angle CA_2O = \frac{\sin \frac{\alpha}{2}}{\sin \frac{\pi}{n}}
$$

In $\triangle A_2OC \left(\angle A_2OC = \frac{\pi}{2}\right)$:

$$
\begin{aligned}
& OA_2 = CA_2 \cos \angle CA_2O = \frac{2a \sin \frac{\alpha}{2}}{\sin \frac{\pi}{n}} \Rightarrow S_{\triangle A_1OA_2} = \frac{1}{2} OA_2^2 \sin \angle A_1OA_2 = \\
& = \frac{1}{2} \cdot \frac{4a^2 \sin^2 \frac{\alpha}{2}}{\sin^2 \frac{\pi}{n}} \cdot \sin \frac{2\pi}{n} = \frac{4a^2 \sin^2 \frac{\alpha}{2} \sin \frac{\pi}{n} \cos \frac{\pi}{n}}{\sin^2 \frac{\pi}{n}} = 4a^2 \sin^2 \frac{\alpha}{2} \operatorname{ctg} \frac{\pi}{n}
\end{aligned}
$$

The area of the base of the pyramid

$$
S_2 = n S_{\Delta A_1OA_2} = 4a^2 n \sin^2 \frac{\alpha}{2} \operatorname{ctg} \frac{\pi}{n}
$$

Then the total surface area of the pyramid

$$
\begin{aligned}
& S = S_1 + S_2 = 2na^2 \sin \alpha + 4na^2 \sin^2 \frac{\alpha}{2} \operatorname{ctg} \frac{\pi}{n} = \\
& = 4na^2 \sin \frac{\alpha}{2} \left(\cos \frac{\alpha}{2} + \frac{\sin \frac{\alpha}{2} \cos \frac{\pi}{n}}{\sin \frac{\pi}{n}}\right) = \frac{4na^2 \sin \frac{\alpha}{2} \sin \left(\frac{\alpha}{2} + \frac{\pi}{n}\right)}{\sin \frac{\pi}{n}} \\
& \text{Answer: } \frac{4na^2 \sin \frac{\alpha}{2} \sin \left(\frac{\alpha}{2} + \frac{\pi}{n}\right)}{\sin \frac{\pi}{n}}
\end{aligned}
$$