\section*{Exercise 1 - 061211}

The Cheops Pyramid in Egypt has the shape of a pyramid with the square base \(A B C D\). The apex \(S\) is \(140 \mathrm{~m}\) above the center \(M\) of the base. The side length of the base is \(231 \mathrm{~m}\). Let's assume for a moment that the following is possible:

A tourist climbs the pyramid in such a way that he starts from \(A\) and travels in a straight line perpendicular to the edge \(B S\) until he reaches this edge at point \(B_{1}\). After reaching this edge at point \(B_{1}\), he continues in a straight line perpendicular to the edge \(C S\) until he reaches this edge at point \(C_{1}\), and from there, he proceeds similarly to point \(D_{1}\) on edge \(D S\) and to point \(A_{1}\) on edge \(A S\).

a) How long would the path taken by him from \(A\) to point \(A_{1}\) be?

b) At what height above the base would the tourist be at point \(A_{1}\)?

c) What angles would the straight segments of the path form with the plane of the base?
}
![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1817.jpg?height=504&width=1218&top_left_y=998&top_left_x=424)
- From the height \(h:=\overline{M S}=140 \mathrm{~m}\) and the length of the base \(a:=\overline{A B}=\overline{B C}=\overline{C D}=\) \(\overline{A D}=231 \mathrm{~m}\), the length of the edges \(k:=\overline{A S}=\overline{B S}=\overline{C S}=\overline{D S}\) can be easily calculated (the triangles \(A M S\) and \(A B C\) are right-angled; thus, the Pythagorean theorem can be used in each case):

\[
\begin{aligned}
k=\overline{A S} & =\sqrt{\overline{A M}^{2}+\overline{M S}^{2}} \\
& =\sqrt{\left(\frac{1}{2} \cdot \overline{A C}\right)^{2}+h^{2}} \\
& =\sqrt{\frac{1}{4} \cdot\left(\overline{A B}^{2}+\overline{B C}^{2}\right)+h^{2}} \\
& =\sqrt{\frac{1}{4} \cdot\left(a^{2}+a^{2}\right)+h^{2}} \\
& =\sqrt{\frac{1}{2} a^{2}+h^{2}}=215.13 \mathrm{~m}
\end{aligned}
\]

The path length \(s\) (where \(\gamma=\angle A S B\)) is now sought:

\[
\begin{aligned}
s & =\overline{A B_{1}}+\overline{B_{1} C_{1}}+\overline{C_{1} D_{1}}+\overline{D_{1} A_{1}} \\
& =\overline{A S} \sin \gamma+\overline{B_{1}^{*} S} \sin \gamma+\overline{C_{1}^{*} S} \sin \gamma+\overline{D_{1}^{*} S} \sin \gamma \\
& =\sin \gamma \cdot\left(k+\overline{B_{1} S}+\overline{C_{1} S}+\overline{D_{1} S}\right)
\end{aligned}
\]

with

\[
\begin{aligned}
& \overline{B_{1} S}=k \cdot \cos \gamma \\
& \overline{C_{1} S}=\overline{B_{1} S} \cdot \cos \gamma=k \cdot \cos ^{2} \gamma \\
& \overline{D_{1} S}=\overline{C_{1} S} \cdot \cos \gamma=k \cdot \cos ^{3} \gamma \\
& \overline{A_{1} S}=\overline{D_{1} S} \cdot \cos \gamma=k \cdot \cos ^{4} \gamma
\end{aligned}
\]

Substituting these into each other, we get:

\[
s=k \cdot \sin \gamma \cdot\left(1+\cos \gamma+\cos ^{2} \gamma+\cos ^{3} \gamma\right)
\]

We observe that the triangles \(A B B_{1}\) and \(A S^{*} S\) are similar (both are right-angled and share the angle \(\beta\): \(\beta=\angle B A S=\angle A B S\)). Thus, \(\sin \alpha\) can be calculated as follows:

\[
\sin \alpha=\overline{A S^{*}}: \overline{A S}=a: 2 k
\]

\(\left(\Rightarrow \alpha=32.47^{\circ}\right)\). And for \(\cos \alpha\), using \(\sin ^{2} \alpha+\cos ^{2} \alpha=1\), we get:

\[
\cos \alpha=\sqrt{1-a^{2}: 4 k^{2}}
\]

\(\sin \gamma\) can be calculated from \(\overline{A B}\) and \(\overline{A S}\) (using the sine rule): \(\sin \gamma: \sin \beta=\overline{A B}: \overline{A S}=a: k\), where \(\sin \beta=\cos \alpha\) holds, thus:

\[
\sin \gamma=\frac{a}{k} \cdot \sqrt{1-a^{2}: 4 k^{2}}
\]

Similarly, for \(\cos \gamma\):

\[
\cos \gamma=\sqrt{1-\frac{a^{2}}{k^{2}} \cdot\left(1-\frac{a^{2}}{4 k^{2}}\right)}=\sqrt{1-\frac{a^{2}}{k^{2}}+\frac{a^{4}}{4 k^{4}}}
\]

Substituting (1), (3), and (4) into (2), we get the result that the tourist must travel \(327 \mathrm{~m}\). For verification, the individual segments are also given: \(\overline{A B_{1}}=195 \mathrm{~m}, \overline{B_{1} C_{1}}=83 \mathrm{~m}, \overline{C_{1} D_{1}}=\) \(35 \mathrm{~m}, \overline{D_{1} A_{1}}=15 \mathrm{~m}\).
- For the sought height \(h_{a}\), with the notations from the first image: \(h_{a}: h=\overline{A_{1} A}: \overline{A S}\). From this, we get:

\[
h_{a}=h \cdot \frac{\overline{A_{1} A}}{k}=\frac{h \cdot\left(k-\overline{A_{1} S}\right)}{k}=\frac{h \cdot\left(k-k \cdot \cos ^{4} \gamma\right)}{k}=h \cdot\left(1-\cos ^{4} \gamma\right)=135.50 \mathrm{~m}
\]
- The straight partial paths form the same angle with the base. For this, the height is needed analogously to the previous subtask with respect to \(B_{1}\):

\[
h_{b}=h \cdot \frac{\overline{B_{1} B}}{k}=\frac{h \cdot\left(k-\overline{B_{1} S}\right)}{k}=\frac{h \cdot(k-k \cdot \cos \gamma)}{k}=h \cdot(1-\cos \gamma)=80.70 \mathrm{~m}
\]

The sine of the sought angle can now be expressed as follows:

\[
\epsilon=\arcsin \left(h_{b}: \overline{A B_{1}}\right)=\arcsin \left(\frac{h \cdot(1-\cos \gamma)}{k \cdot \sin \gamma}\right)
\]

Substituting the values, we get: \(\epsilon=65.5^{\circ}\).