\section*{Problem 5 - 161035}

For a right prism and a right pyramid, the following assumptions are given: Both bodies have the same base; this is an equilateral triangle with a given side length \(a\). The apex of the pyramid lies in the top face of the prism.

Determine the values for the (common) height \(h\) of the prism (and the pyramid) for which, under the given assumptions, the lateral surface area of the prism is equal to the lateral surface area of the pyramid!
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Since the pyramid is straight, its apex lies at the center of gravity of the prism's top face. The mantle of the pyramid consists of three congruent isosceles triangles with base length \(a\). If the length of the corresponding height in these triangles is denoted by \(H\), then according to the Pythagorean theorem,

\[
H^{2}=h^{2}+\left(\frac{1}{6} a \sqrt{3}\right)^{2}=h^{2}+\frac{1}{12} a^{2}
\]

Thus, the mantle of the pyramid has the area

\[
M_{1}=\frac{3}{2} a \sqrt{h^{2}+\frac{1}{12} a^{2}}
\]

The mantle of the prism has the area \(M_{2}=3 a h\). Therefore, due to \(a>0\) and \(h>0\), the requirement \(M_1=M_{2}\) is successively equivalent to

\[
\begin{aligned}
3 a h & =\frac{3}{2} a \sqrt{h^{2}+\frac{1}{12} a^{2}} \\
2 h & =\sqrt{h^{2}+\frac{1}{12} a^{2}} \\
4 h^{2} & =h^{2}+\frac{1}{12} a^{2} \\
h & =\frac{1}{6} a
\end{aligned}
\]