12.228. The dihedral angle at the base of a regular triangular pyramid is equal to $\alpha$, the lateral surface area of the pyramid is $S$. Find the distance from the center of the base to the midpoint of the slant height of a lateral face.
## Solution.

Let $FO$ be the height of the regular pyramid $FABC$ (Fig. 12.95), $D$ be the midpoint of $AC$. Then $FD \perp AC$, $OD \perp AC$, $\angle FDO = \alpha$.

If $E$ is the midpoint of the apothem $FD$, then $OE$ is the median of the right $\triangle FOD$, drawn to the hypotenuse, $OE = \frac{1}{2} FD$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0976.jpg?height=680&width=604&top_left_y=774&top_left_x=655)

Fig. 12.95

$S_{\triangle ABC} = S \cos \alpha$.

If $AC = a$, then

$$
\begin{aligned}
& S_{\triangle ABC} = \frac{a^2 \sqrt{3}}{4} = S \cos \alpha \Rightarrow a = 2 \sqrt{\frac{S \cos \alpha}{\sqrt{3}}} \\
& OD = \frac{a \sqrt{3}}{6} = 2 \sqrt{\frac{S \cos \alpha}{\sqrt{3}}} \cdot \frac{\sqrt{3}}{6} = \frac{1}{3} \sqrt{S \sqrt{3} \cos \alpha}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0977.jpg?height=625&width=617&top_left_y=79&top_left_x=81)

Fig. 12.96

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0977.jpg?height=580&width=564&top_left_y=130&top_left_x=699)

Fig. 12.97

In $\triangle FOD (\angle FOD = 90^\circ):$

$FD = \frac{OD}{\cos \angle FDO} = \frac{\sqrt{S \sqrt{3} \cos \alpha}}{3 \cos \alpha} \Rightarrow OE = \frac{\sqrt{S \sqrt{3} \cos \alpha}}{6 \cos \alpha}$.

Answer: $\frac{\sqrt{S \sqrt{3} \cos \alpha}}{6 \cos \alpha}$.