An Egyptian pyramid (a regular quadrilateral pyramid) has a base perimeter equal to the circumference of a circle whose radius is the height of the pyramid. What angle does one of the triangular faces of the pyramid form with the base plane?
Let the base edge of the pyramid be $a$, and its height be $m$. According to the condition,

$$
\begin{aligned}
4 a & =2 \pi m \\
\text { hence } \quad m & =\frac{2 a}{\pi}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_50d8cf99e780520e27e0g-1.jpg?height=485&width=726&top_left_y=457&top_left_x=686)

The lateral faces of the pyramid are isosceles triangles, where the median to the base is also the altitude. We know that the angle of inclination between a line and a plane is the angle it forms with its orthogonal projection on the plane. Using the notation in the diagram and (1),

$$
\operatorname{tg} \alpha=\frac{m}{a / 2}=\frac{4 a}{a \pi}=\frac{4}{\pi}, \quad \text { hence } \quad \alpha \approx 51^{\circ} 51^{\prime}
$$

The other two medians form the same angle with the base plane due to symmetry, so it is sufficient to determine one of them.

The projection of the median $A F$ is $A F^{\prime}$; here $F$ bisects $E B$, and $F^{\prime}$ bisects $K B$, and therefore $F F^{\prime}$ is parallel to $E K$. From the right triangle $A F F^{\prime}$,

$$
\operatorname{tg} \beta=\frac{F F^{\prime}}{F^{\prime} A}
$$

From (1), $F F^{\prime}=\frac{m}{2}=\frac{a}{\pi}$, and $A F^{\prime}$ can be determined using the Pythagorean theorem:

$$
A F^{\prime}=\sqrt{A K^{2}+K F^{\prime 2}}=\sqrt{\left(\frac{a \sqrt{2}}{2}\right)^{2}+\left(\frac{a \sqrt{2}}{4}\right)^{2}}=a \sqrt{\frac{5}{8}}
$$

Substituting the obtained values:

$$
\operatorname{tg} \beta=\frac{a / \pi}{a \sqrt{5 / 8}}=\frac{\sqrt{8}}{\pi \sqrt{5}}, \quad \text { hence } \quad \beta \approx 21^{\circ} 55^{\prime}
$$