A regular hexagonal pyramid has perpendiculars erected at the midpoint of each of its six edges. By connecting the points where these consecutive perpendiculars intersect the base plane, we obtain a hexagon that is congruent to the base of the pyramid. What are the angles on the sides of the pyramid?

 untranslated portion:
Egy szabályos hatoldalú gúla hatélű csúcsában merőlegeset állítunk mindegyik oldallapjára. Az egymás utáni merőlegeseknek az alapsíkkal való metszéspontjait összekötve, a gúla alapjával egybevágó hatszöget kapunk. Mekkorák a gúla oldallapjain levő szögek? 

(Note: The untranslated portion is included to show the original text, but it should not be part of the final translation.)
We need the ratio of the base edge to the side edge of the pyramid. Let the apex of the hexagonal pyramid be $G$, one base edge be $A B$, the intersection point of the perpendicular from $G$ to the plane of the $G A B$ face with the base plane be $K$, and the projection of $G$ onto the base plane be $O$. Due to the regularity of the pyramid, the base is a regular hexagon and $O$ is its center. The rotation by $60^{\circ}$ around the $G O$ axis maps each face of the pyramid to the next, thus also mapping the constructed perpendiculars into each other, and the new hexagon into itself, making $O$ its center as well. The congruence of the two hexagons implies $O K = O A$ (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_6e6c614a8d53d68f80f3g-1.jpg?height=475&width=525&top_left_y=375&top_left_x=781)

Figure 1

Let the midpoint of the edge $A B$ be $F$, then the plane $S = G O F$ is the perpendicular bisector plane of the edge $A B$. $K$ also lies on $S$ because $G A = G B$ and $\angle K G A = \angle K G B = 90^{\circ}$, making the triangles $K G A$ and $K G B$ congruent, hence $K A = K B$. Since $G K$ is perpendicular to the line $G F$ in the face $G A B$, the triangle $G K F$ is a right triangle (Figure 2), and thus

\[
\begin{gathered}
\tan \angle G A B = \frac{G F}{A F} = \frac{2 \sqrt{F O \cdot F K}}{A B} = \frac{2 \sqrt{F O (F O + O A)}}{O A} = \\
= 2 \sqrt{\frac{F O}{O A} \left( \frac{F O}{O A} + 1 \right)} = \sqrt{\sqrt{3} (\sqrt{3} + 2)} = 2.542
\end{gathered}
\]

$\angle G A B = 68^{\circ} 32'$, and from this $\angle A G B = 42^{\circ} 56'$, since, as is known, $\frac{F O}{O A} = \frac{\sqrt{3}}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_6e6c614a8d53d68f80f3g-1.jpg?height=607&width=616&top_left_y=1412&top_left_x=749)

Figure 2 shows the solid in two projections, as well as the $A B G$ face and the $A G K$ section rotated into the base plane.

Vilmos Komornik (Budapest, Fazekas M. Gyak. Gimn., III. o. t.) Edit Jeney (Elek, Gimn., II. o. t.)

Remarks:
1. Essentially the same way,

\[
\sin \angle G A B = \frac{G F}{G A} = \frac{G F}{G K} = \sqrt{\frac{F O \cdot F K}{K O \cdot K F}} = \sqrt{\frac{F O}{K O}} = \sqrt[4]{\frac{3}{4}}
\]

2. We can also calculate from the fact that $A K G$ is an isosceles right triangle, and its hypotenuse is also the hypotenuse in the triangle $A K F$, with the legs being known.
3. On the pyramid, the angle between the side edges and the base is equal to the angle between the side edges. Indeed, based on two different expressions for the area of the side face, and from the right triangle $F K G$:

\[
\sin \angle A G B = \frac{A A'}{A G} = \frac{B G \cdot A A'}{A G^2} = \frac{A B \cdot F G}{K G^2} = \frac{K O \cdot F G}{K O \cdot K F} = \frac{F G}{K F} = \sin \angle F K G
\]