11.109. Find the volume of a regular quadrilateral pyramid with the side of the base equal to $a$, and the planar angles at the vertex equal to the angles of inclination of the lateral edges to the base.
## Solution.

Draw the height of the regular pyramid $M A B C D-M O$ and the height of the lateral face $A M D$ (Fig. 11.7). Let $A D=a$. Since $\angle M C O=\angle A M D$, the right-angled triangles $M O C$ and $M K D$ are equal (by hypotenuse and acute angle) and, consequently, $M O=D K$. Let $D K=x, M E$ be the height and median of $\triangle A M D$. From $\triangle M O A\left(\angle M O A=90^{\circ}\right): A M=\sqrt{A O^{2}+M O^{2}}=\sqrt{\frac{a^{2}}{2}+x^{2}}$.

From $\triangle A E M\left(\angle A E M=90^{\circ}\right): M E=\sqrt{A M^{2}-A E^{2}}=\sqrt{\frac{a^{2}}{4}+x^{2}}$.

$S_{\triangle A M D}=\frac{1}{2} A D \cdot M E=\frac{1}{2} A M \cdot D K$. Then $a \sqrt{\frac{a^{2}}{4}+x^{2}}=\sqrt{x^{2}+\frac{a^{2}}{2}} \cdot x ;$

$\frac{a^{2}\left(a^{2}+4 x^{2}\right)}{4}=\frac{x^{2}\left(a^{2}+2 x^{2}\right)}{2} ; 4 x^{4}-2 a^{2} x^{2}-a^{4}=0 ; x=\frac{a \sqrt{\sqrt{5}+1}}{2}$.

The volume of the pyramid $V=\frac{1}{3} S_{A B C D} \cdot M O=\frac{a^{3}(\sqrt{\sqrt{5}+1})}{6}$.

Answer: $\quad \frac{a^{3} \sqrt{\sqrt{5}+1}}{6}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0825.jpg?height=467&width=489&top_left_y=689&top_left_x=125)

Fig. 11.8