3. An isosceles triangle with angle $\alpha$ at the base is the base of a pyramid such that all its lateral edges form equal angles $\varphi=90^{\circ}-\alpha$ with the base. The plane passing through the height of the pyramid and the vertex of the isosceles triangle at the base cuts the pyramid into a figure whose area is $p$. Determine the volume of the pyramid.
Solution. Let A, B, C be the vertices of the base of the pyramid, the apex of the pyramid be $D$, and the foot of the height of the side face $BCD$ be $E$ (see the diagram on the right). Further, let the side of the base be $b$, the radius of its circumscribed circle be $R$, and $S$ be the foot of the height dropped from the apex of the pyramid. The triangles $DAS$ and $DBS$ have corresponding right angles $\angle ASD$ and $\angle BSD$, a common side $SD$, and $\angle DAS = \angle DBS = \varphi$, so they are

![](https://cdn.mathpix.com/cropped/2024_06_05_e4cff208225cb6887089g-2.jpg?height=396&width=395&top_left_y=1388&top_left_x=1069)
congruent. Therefore, $AS = BS$ and similarly $AS = CS$, which means that $S$ is the center of the circumscribed circle of the base and $AS = BS = R = \frac{b}{2 \sin \alpha}$.

From the right triangle $ABE$ we get $AE = b \sin \alpha$, and from the right triangle $DAS$ we have

$$
SD = R \operatorname{ctg} \alpha = \frac{b}{2 \sin \alpha} \operatorname{ctg} \alpha = \frac{b \cos \alpha}{2 \sin^2 \alpha}
$$

Therefore,

$$
p = \frac{1}{2} AE \cdot SD = \frac{1}{2} b \sin \alpha \frac{b \cos \alpha}{2 \sin^2 \alpha} = \frac{1}{4} b^2 \operatorname{ctg} \alpha, \text{ i.e., } b^2 = 4 p \operatorname{tg} \alpha
$$

For the volume of the pyramid, we get

$$
V = \frac{1}{3} \cdot \frac{1}{2} b^2 \sin (180^\circ - 2 \alpha) \frac{b \cos \alpha}{2 \sin^2 \alpha} = \frac{1}{6} b^3 \frac{\cos^2 \alpha}{\sin \alpha} = \frac{4}{3} \frac{\cos^2 \alpha}{\sin \alpha} (p \operatorname{tg} \alpha)^{\frac{3}{2}}
$$