3. From the midpoint $O$ of the height $SE$ of a regular quadrilateral pyramid with apex $S$, a normal $OM$ is drawn to the lateral edge and a normal $OK$ is drawn to the lateral face of the pyramid. If the lengths of these normals are $\overline{OM}=p$ and $\overline{OK}=q$, calculate the volume of the pyramid (i.e., express the volume in terms of $p$ and $q$).
Solution. Let the side length of the base of the pyramid be $a$, the diagonal of the base $d$, and let $h=\overline{S E}$. We have $\triangle M O S \sim \triangle E A S$, so

$$
\frac{p}{\frac{d}{2}}=\frac{\frac{h}{2}}{\sqrt{\left(\frac{d}{2}\right)^{2}+h^{2}}}
$$

By manipulating this equation and considering that $d=a \sqrt{2}$, we get

$$
\frac{4}{h^{2}}+\frac{8}{a^{2}}=\frac{1}{p^{2}}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_74ce08a8bb971e811c89g-07.jpg?height=466&width=387&top_left_y=375&top_left_x=1092)

On the other hand, $\triangle K O S \sim \triangle E L S$ (where $L$ is the midpoint of the base side), so we have

$$
\frac{q}{\frac{a}{2}}=\frac{\frac{h}{2}}{\sqrt{h^{2}+\left(\frac{a}{2}\right)^{2}}}
$$

By manipulating this equation, we get

$$
\frac{16}{a^{2}}+\frac{4}{h^{2}}=\frac{1}{q^{2}}
$$

Subtracting (1) from (2), we get $\frac{8}{a^{2}}=\frac{1}{q^{2}}-\frac{1}{p^{2}}=\frac{p^{2}-q^{2}}{p^{2} q^{2}}$ and thus $a^{2}=\frac{8 p^{2} q^{2}}{p^{2}-q^{2}}$. Substituting into (1), we get $h^{2}=\frac{4 p^{2} q^{2}}{2 q^{2}+p^{2}}$. Now let's calculate the volume of the pyramid

$$
V=\frac{B h}{3}=\frac{a^{2} h}{3}=\frac{16 p^{3} q^{3}}{3\left(p^{2}-q^{2}\right) \sqrt{2 q^{2}+p^{2}}}
$$