3-3. Find the volume of a regular quadrilateral pyramid, the sides of the base of which are $a$, and the planar angles at the vertex are equal to the angles of inclination of the lateral edges to the base plane.
Solution 3. Let the length of the lateral edge be $b$, and the height of the pyramid be $h$. The desired volume $V$ is equal to $a^{2} h / 3$. Let the planar angles at the vertex and the angles of inclination of the lateral edges to the base plane be $\alpha$. Then $b \sin \alpha=h, b \cos \alpha=\frac{\sqrt{2}}{2} a$ and $b \sin \frac{\alpha}{2}=\frac{a}{2}$. Therefore, $h=b \sin \alpha=\frac{a \sin \alpha}{2 \sin (\alpha / 2)}=a \cos \frac{\alpha}{2}$, and thus $V=\frac{1}{3} a^{3} \cos \frac{\alpha}{2}$. In addition, $\sin \frac{\alpha}{2}=\frac{1}{\sqrt{2}} \cos \alpha$. Let $\cos \frac{\alpha}{2}=x$ Then $\sqrt{1-x^{2}}=\frac{1}{\sqrt{2}}\left(2 x^{2}-1\right)$. Solving this equation and keeping only the positive root, we get $x=\frac{1}{2} \sqrt{1+\sqrt{5}}$.

## 3rd Option