3. The base of the pyramid is a square. The angles of inclination of the lateral faces of the pyramid to the base are in the ratio $1: 2: 4: 2$. Determine these angles.
Solution. Let $ABCD$ be the base, and $S$ be the apex of the pyramid. The slant angles of the lateral faces $SAB$, $SBC$, $SCD$, $SDA$ are $\varphi, 2\varphi, 4\varphi, 2\varphi$. Let $a$ be the side of the base, and $H$ be the height of the pyramid (see the diagram on the right). First, let us assume that $\frac{\pi}{2} < 4\varphi < \pi$ (we will later prove that it cannot be $4\varphi \leq \frac{\pi}{2}$). The plane that contains the apex of the pyramid and is perpendicular to the lines $BC$ and $DA$ intersects these lines outside the segments $BC$ and $DA$; let these intersections be denoted by $E$ and $F$. The intersections of the planes through $S$ perpendicular to $AB$ and $CD$ with these lines are denoted by $M$ and $N$ (these points lie on $AB$ and $CD$, respectively). These planes contain the height of the pyramid, whose foot is denoted by $O$. By assumption, we have

$$
\measuredangle SMN = \varphi, \measuredangle SEF = \measuredangle SFE = 2\varphi, \quad \measuredangle SNM = 4\varphi.
$$

From the corresponding right triangles, we get

$$
\operatorname{ctg} \varphi = \frac{OM}{H}, \operatorname{ctg} 2\varphi = \frac{OE}{H} = \frac{a}{2H}, \operatorname{ctg} 4\varphi = -\frac{ON}{H}
$$

which implies

$$
\operatorname{ctg} \varphi + \operatorname{ctg} 4\varphi = \frac{OM - ON}{H} = \frac{MN}{H} = \frac{a}{H} = 2 \operatorname{ctg} 2\varphi
$$

This equation for $\frac{\pi}{8} < \varphi < \frac{\pi}{4}$ is sequentially transformed into the following equivalent equations

$$
\begin{aligned}
& \operatorname{ctg} 4\varphi - \operatorname{ctg} 2\varphi = \operatorname{ctg} 2\varphi - \operatorname{ctg} \varphi \\
& \frac{\sin (4\varphi - 2\varphi)}{\sin 4\varphi \sin 2\varphi} = \frac{\sin (2\varphi - \varphi)}{\sin 2\varphi \sin \varphi}, \\
& \sin 2\varphi = 2 \sin 2\varphi \cos 2\varphi
\end{aligned}
$$

so $\varphi = \frac{\pi}{6}$. Therefore, the required angles are $\frac{\pi}{6}, \frac{\pi}{3}, \frac{2\pi}{3}, \frac{\pi}{3}$.

We will prove that it cannot be $4\varphi \leq \frac{\pi}{2}$. Indeed, if this is the case, a similar procedure as above yields that $\varphi$ must satisfy the equation $\operatorname{ctg} \varphi + \operatorname{ctg} 4\varphi = 2 \operatorname{ctg} 2\varphi$, which has no solutions for $0 < \varphi \leq \frac{\pi}{8}$.