\section*{Exercise 2 - 081222}

In a three-sided pyramid, let the base be an equilateral triangle with side length \(a\), and the apex \(S\) be at a height \(h\) above the intersection point \(M\) of the medians of the base triangle. What is the value of the quotient \(\frac{h}{a}\) if the inclination angle between two side faces of the pyramid is \(90^{\circ}\)?
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The angle between two side faces lies in a plane that is perpendicular to both side faces. This perpendicular plane intersects the two side faces in two side face heights, which are perpendicular to a side length and have a common foot point.

The angle between the two side face heights is the angle between the two side faces. \(\angle A E B\) is such an angle. Its size is \(\tau\). According to the problem statement, \(\tau=90^{\circ}\). Triangle AEB is isosceles with the base \(\overline{A B}\). By the interior angle sum theorem, \(\tau+2 \epsilon=180^{\circ}\) and thus \(\epsilon=45^{\circ}\). By the cosine rule in triangle AB:

\[
|A E|^{2}=|A B|^{2}+|B E|^{2}-2|A B||A E| \cos \epsilon
\]

Since triangles \(A C S\) and \(B C S\) are congruent and isosceles, \(|A E|=|B E|\). Substituting into (*) also yields with \(\cos 45^{\circ}=\frac{1}{2} \sqrt{2}\) and \(|A B|=a\):

\[
\begin{aligned}
& |B E|^{2}=|A B|^{2}+|B E|^{2}-2|A B||A E| \cdot \frac{1}{2} \sqrt{2} \\
& |A B||A E| \cdot \sqrt{2}=|A B|^{2} \Longleftrightarrow|A E|=\frac{a}{2} \sqrt{2}
\end{aligned}
\]

Let \(\varphi=\angle B C E\). Triangle \(C B E\) is right-angled. It holds: \(\sin \varphi=\frac{\frac{a}{2} \sqrt{2}}{a} \Longleftrightarrow \varphi=45^{\circ}\).

Triangle \(C B S\) is isosceles with base \(C B\). Let \(h_{s}\) be the height on the base and \(s\) the length of the side edge. It holds \(\cos \varphi=\frac{\frac{a}{2}}{s} \Longleftrightarrow s=\frac{a}{2} \sqrt{2}\)

Triangle \(A B C\) is equilateral. The height in the equilateral triangle is \(\frac{a}{2} \sqrt{3}\) and the medians (in the triangle simultaneously heights) intersect in the ratio \(2: 1\), where the segment from the centroid to the vertex is twice as long as the segment from the centroid to the midpoint of the side.

Triangle \(H C S\) is right-angled. By the Pythagorean theorem:

\[
\left(\frac{2}{3} \frac{a}{2} \sqrt{3}\right)^{2}+h^{2}=\left(\frac{a}{2} \sqrt{2}\right)^{2} \Longleftrightarrow h=\frac{a}{6} \sqrt{6}
\]

The quotient is thus \(\frac{h}{a}=\frac{\sqrt{6}}{6}\).