12.340. The side of the base of a regular quadrilateral pyramid is equal to $a$, and the plane angle at the vertex of the pyramid is $\alpha$. Find the distance from the center of the base of the pyramid to its lateral edge.
## Solution.

Let $EO$ be the height of the regular pyramid $EABCD$ (Fig. 12.201), $OF$ be the distance from point $O$ to the lateral edge $EC$, and $\angle ECO = \beta$. Since

$$
\angle CED = \alpha, \text{ then } \angle ECD = 90^\circ - \frac{\alpha}{2}.
$$

$OC$ is the projection of $EC$ onto the plane of the base of the pyramid, and $CD$ is a line lying in this plane.

Then $\cos \angle ECD = \cos \angle ECO \cdot \cos \angle OCD$;

$$
\cos \left(90^\circ - \frac{\alpha}{2}\right) = \cos \beta \cdot \cos 45^\circ \Leftrightarrow \cos \beta = \frac{\sin \frac{\alpha}{2}}{\cos 45^\circ} = \sqrt{2} \sin \frac{\alpha}{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1099.jpg?height=668&width=765&top_left_y=78&top_left_x=320)

Fig. 12.201

In $\triangle OFC \left(\angle OFC = 90^\circ\right)$:

$$
OF = OC \cdot \sin \beta = \frac{a \sqrt{2}}{2} \sqrt{1 - \cos^2 \beta} = \frac{a \sqrt{2}}{2} \sqrt{1 - 2 \sin^2 \frac{\alpha}{2}} = \frac{a}{2} \sqrt{2 \cos \alpha}
$$

Answer: $\frac{a}{2} \sqrt{2 \cos \alpha}$.