\section*{Problem 3 - 311033}

Let \(A B C D S\) be a pyramid; its base is a square \(A B C D\), and the perpendicular from the apex \(S\) to the base has its foot at the diagonal intersection point \(M\) of the square \(A B C D\).

Furthermore, let \(H\) be the midpoint of the segment \(M S\); the perpendicular from \(H\) to the side face \(B C S\) has the foot point \(F\), and the perpendicular from \(H\) to the edge \(C S\) has the foot point \(K\).

Under these conditions, calculate the volume of the pyramid \(A B C D S\) from the given lengths \(f=H F\) and \(k=H K\).
}

Let the midpoint of the segment \(B C\) be \(E\). The height of the pyramid be \(h\), and the edge length of the square base be \(a\). We now make a cut through \(M E S\) and through \(M C S\):
![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1599.jpg?height=360&width=664&top_left_y=1348&top_left_x=702)

The volume of the pyramid we are looking for is

\[
V=\frac{1}{3} a^{2} h
\]

We therefore need to determine \(a\) and \(h\) using the given cuts from the lengths \(f\) and \(k\). In the left cut, we have:

\[
\begin{gathered}
\frac{f}{\frac{1}{2} h}=\frac{\frac{1}{2} a}{\sqrt{h^{2}+\frac{1}{4} a^{2}}} \\
f^{2}\left(h^{2}+\frac{1}{4} a^{2}\right)=\frac{1}{16} a^{2} h^{2} \\
16 f^{2} h^{2}+4 f^{2} a^{2}=a^{2} h^{2} \\
\left(a^{2}-16 f^{2}\right) h^{2}=4 f^{2} a^{2} \\
h^{2}=\frac{4 f^{2} a^{2}}{a^{2}-16 f^{2}}
\end{gathered}
\]

From the right cut, we similarly obtain:

\[
\frac{k}{\frac{1}{2} h}=\frac{\frac{1}{2} \sqrt{2} a}{\sqrt{h^{2}+\frac{1}{2} a^{2}}}
\]

And with the same steps as above:

\[
h^{2}=\frac{4 k^{2} a^{2}}{a^{2}-8 k^{2}}
\]

We set (1) and (2) equal and solve for \(a^{2}\):

\[
\begin{gathered}
\frac{4 f^{2} a^{2}}{a^{2}-16 f^{2}}=\frac{4 k^{2} a^{2}}{a^{2}-8 k^{2}} \\
f^{2}\left(a^{2}-8 k^{2}\right)=k^{2}\left(a^{2}-16 f^{2}\right) \\
8 f^{2} k^{2}=\left(k^{2}-f^{2}\right) a^{2} \\
a^{2}=\frac{8 f^{2} k^{2}}{k^{2}-f^{2}}
\end{gathered}
\]

Substituting this into (1), we get:

\[
\begin{gathered}
h^{2}=\frac{4 f^{2} \cdot \frac{8 f^{2} k^{2}}{k^{2}-f^{2}}}{\frac{8 f^{2} k^{2}}{k^{2}-f^{2}}-16 f^{2}}=\frac{32 f^{4} k^{2}}{8 f^{2} k^{2}-16 f^{2}\left(k^{2}-f^{2}\right)}=\frac{4 f^{2} k^{2}}{2 f^{2}-k^{2}} \\
h=\frac{2 f k}{\sqrt{2 f^{2}-k^{2}}}
\end{gathered}
\]

Substituting these equations and equation (3) into the volume formula, we get:

\[
V=\frac{2 f k}{3 \sqrt{2 f^{2}-k^{2}}} \cdot \frac{8 f^{2} k^{2}}{k^{2}-f^{2}}=\frac{16 f^{3} k^{3}}{3\left(k^{2}-f^{2}\right) \sqrt{2 f^{2}-k^{2}}}
\]