The Egyptian Cheops pyramid is a regular quadrilateral pyramid $A B C D S$, with a base edge of $a=231 \mathrm{~m}$ and a height of $m=140 \mathrm{~m}$. A tourist travels in such a way that starting from $A$, they proceed in a direction perpendicular to the side edge $S B$. Upon reaching the side edge $S B$ at point $B_{1}$, they continue their journey perpendicularly to the side edge $D C$ to point $C_{1}$, and so on. At what height will the tourist be when they reach the side edge $S A$ at point $A_{1}$ again? What is the angle of elevation of the tourist's path? What is the distance traveled by the tourist to point $A_{1}$?
I. Solution: The labeling is shown in the figure. First, let's calculate the distance $u$ traveled up to point $A_{1}$, which consists of the following four segments: $A B_{1}=u_{1}, B_{1} C_{1}=u_{2}, C_{1} D_{1}=u_{3}$, and $D_{1} A_{1}=u_{4}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_9b70f4b52cc5ab8e104eg-1.jpg?height=418&width=593&top_left_y=230&top_left_x=755)

If we denote the slant height of the pyramid by $b$ and the angle at vertex $S$ by $\varepsilon$, then from the right triangle $A B_{1} S$,

$$
u_{1}=b \sin \varepsilon, \quad \text { and } \quad S B_{1}=b \cos \varepsilon
$$

Similarly, from the right triangle $B_{1} C_{1} S$,

$$
u_{2}=S B_{1} \sin \varepsilon=b \cos \varepsilon \cdot \sin \varepsilon=u_{1} \cos \varepsilon
$$

The planes passing through points $B_{1}, C_{1}, D_{1}$ and parallel to the base cut off smaller pyramids from the original pyramid, all of which are similar to the original pyramid. Based on this similarity, we can state that the angles of elevation of the four segments are equal, and furthermore, $u_{3}=u_{2} \cos \varepsilon$ and $u_{4}=u_{3}$, meaning the four segments form a geometric sequence with the first term $u_{1}=b \sin \varepsilon$ and common ratio $\cos \varepsilon$. Therefore,

$$
u=u_{1}+u_{2}+u_{3}+u_{4}=b \sin \varepsilon \frac{1-\cos ^{4} \varepsilon}{1-\cos \varepsilon}=b \sin \varepsilon(1+\cos \varepsilon)\left(1+\cos ^{2} \varepsilon\right)
$$

Since $A O=\frac{a \sqrt{2}}{2}$, or $A O^{2}=\frac{a^{2}}{2}$, the slant height of the pyramid is

$$
S A=b=\sqrt{m^{2}+A O^{2}}=\sqrt{140^{2}+\frac{231^{2}}{2}}=215.1
$$

From the right triangle $A E S$, $\sin \frac{\varepsilon}{2}=\frac{a}{2 b}=\frac{231}{430.2}$, hence

$$
\frac{\varepsilon}{2}=32^{\circ} 28^{\prime}, \quad \varepsilon=64^{\circ} 56^{\prime}
$$

Substituting these values into (I):

$$
u=215.1 \cdot \sin 64^{\circ} 56^{\prime} \cdot 1.424 \cdot 1.180
$$

Using a school 4-digit logarithmic table, the distance traveled to point $A_{1}$ is

$$
u \approx 327.4 \text { meters. }
$$

Let the height of point $A_{1}$ be $x$, and the height of point $B_{1}$ be $y$.

$$
x: u=y: u_{1}
$$

Furthermore, in the right triangle $B O S$,

$$
y: m=B_{1} B: b
$$

where

$$
B_{1} B=b-S B_{1}=b-b \cos \varepsilon=b(1-\cos \varepsilon)=2 b \sin ^{2} \frac{\varepsilon}{2}
$$

Thus, from (3),

$$
y=\frac{m \cdot B_{1} B}{b}=\frac{m \cdot 2 b \sin ^{2} \frac{\varepsilon}{2}}{b}=2 m \sin ^{2} \frac{\varepsilon}{2}
$$

and from (2),

$$
\begin{gathered}
x=\frac{u y}{u_{1}}=\frac{u \cdot 2 m \sin ^{2} \frac{\varepsilon}{2}}{b \sin \varepsilon}=\frac{u \cdot 2 m \sin ^{2} \frac{\varepsilon}{2}}{b \cdot 2 \sin \frac{\varepsilon}{2} \cos \frac{\varepsilon}{2}}=\frac{u m \sin \frac{\varepsilon}{2}}{b \cos \frac{\varepsilon}{2}}=\frac{u m \operatorname{tg} \frac{\varepsilon}{2}}{b}= \\
=\frac{327.4 \cdot 140 \cdot \operatorname{tg} 32^{\circ} 28^{\prime}}{215.1}
\end{gathered}
$$

Thus, the height of point $A_{1}$ is

$$
x \approx 135.6 \text { meters. }
$$

The angle of elevation, denoted by $\omega$ in $\angle B_{1} A B_{1}^{\prime}$,

$$
\sin \omega=\frac{y}{u_{1}}=\frac{x}{u}=\frac{135.6}{327.4}
$$

hence

$$
\omega=24^{\circ} 28^{\prime}
$$

II. Solution: After calculating $b$ and $\varepsilon$, we can first calculate the height $x$ of point $A_{1}$.

Indeed, $x: m=A A_{1}: b$, where $A A_{1}=b-S A_{1} S B_{1}=b \cos \varepsilon, S C_{1}=S B_{1} \cos \varepsilon=b \cos ^{2} \varepsilon, S D_{1}=S C_{1} \cos \varepsilon=b \cos ^{3} \varepsilon$ and $S A_{1}=b \cos ^{4} \varepsilon$.

Thus,

$$
x=\frac{m \cdot A A_{1}}{b}=\frac{m\left(b-b \cos ^{4} \varepsilon\right)}{b}=m\left(1-\cos ^{4} \varepsilon\right)(\approx 140 \cdot 0.9678 \approx 135.7)
$$

According to the first solution,

$$
u_{1}=b \sin \varepsilon, \quad \text { and } \quad y=2 m \sin ^{2} \frac{\varepsilon}{2}
$$

thus,

$$
\sin \omega=\frac{y}{u_{1}}=\frac{2 m \sin ^{2} \frac{\varepsilon}{2}}{2 b \sin \frac{\varepsilon}{2} \cos \frac{\varepsilon}{2}}=\frac{m \operatorname{tg} \frac{\varepsilon}{2}}{b}\left(\approx \frac{140 \operatorname{tg} 32^{\circ} 28^{\prime}}{215.1}\right)
$$

hence,

$$
\omega=24^{\circ} 28^{\prime}
$$

Since $\sin \omega=\frac{x}{u}$, then

$$
u=\frac{x}{\sin \omega}\left(\approx \frac{135.7}{\sin 24^{\circ} 28^{\prime}}=327.6\right)
$$