3. The base edge of a square pyramid is $26 \mathrm{~cm}$, and the adjacent lateral faces form a $120^{\circ}$ angle with each other. What is the height of the pyramid?
I. solution: The labeling is shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_02_1fed8d9a491ea5c925deg-1.jpg?height=377&width=569&top_left_y=229&top_left_x=775)

Let the side of the base square be $BC = a$, and the half-diagonal of the base square be $OB = OC = \frac{a \sqrt{2}}{2}$. Let the sought $OM$ be denoted by $m$. The plane through the diagonal $BD$ perpendicular to the edge $MC$ intersects the edge at $N$. Thus, $ON \perp MC$.

According to the condition, $\angle BND = 120^\circ$. Let $\angle OCN = \alpha$. Since in the isosceles triangle $BND$ the angles opposite the equal sides are $30^\circ$, we have

$$
\sin \alpha = \frac{ON}{OC} = \frac{ON}{OB} = \tan \angle OBN = \tan 30^\circ = \frac{1}{\sqrt{3}}.
$$

Using this, from the right triangle $OCM$,

$$
m = OC \cdot \tan \alpha = OC \cdot \frac{\sin \alpha}{\sqrt{1 - \sin^2 \alpha}} = \frac{a \sqrt{2}}{2} \cdot \frac{1 / \sqrt{3}}{\sqrt{1 - 1 / 3}} = \frac{a \sqrt{2}}{2 \sqrt{2}} = \frac{a}{2}.
$$

In this case, $a = 26 \text{ cm}$ and thus $m = 13 \text{ cm}$.

The resulting numerous right triangles offer various possibilities for contestants to explore - to calculate the height without using trigonometry. One such example is the following.

# II. solution:

$BF = OF = \frac{a}{2}$. From the triangle $BON$, $BN = \frac{BO}{\sin 60^\circ} = \frac{a \sqrt{2}}{2} / \frac{\sqrt{3}}{2} = \frac{a \sqrt{2}}{\sqrt{3}} = \frac{a \sqrt{6}}{3}$.

The right triangles $MFC$ and $BNC$ are similar because the acute angle at $C$ is common, so

$$
\frac{MF}{MC} = \frac{BN}{a} = \frac{\sqrt{6}}{3},
$$

or

$$
\frac{MF^2}{MC^2} = \frac{m^2 + \frac{a^2}{4}}{m^2 + \frac{a^2}{2}} = \frac{2}{3},
$$

from which

$$
\begin{aligned}
& 3 m^2 + \frac{3 a^2}{4} = 2 m^2 + a^2 \\
& m^2 = \frac{a^2}{4}, \text{ and thus } m = \frac{a}{2}.
\end{aligned}
$$

III. solution: Let $MC = b$ and $BN = p$. Since $\angle BNO = 60^\circ$, $ON = \frac{p}{2}$. The double area of triangle $COM$ can be expressed in two ways:

$$
\begin{gathered}
2t = b \cdot \frac{p}{2} = \frac{a}{2} \sqrt{2} \cdot m \\
\text{or } b p = a m \sqrt{2}
\end{gathered}
$$

The double area of triangle $BCM$ similarly is

$$
b p = a \sqrt{m^2 + \frac{a^2}{4}}
$$

From (1) and (2), it follows that

$$
\begin{aligned}
a m \sqrt{2} & = a \sqrt{m^2 + \frac{a^2}{4}} \\
2 m^2 & = m^2 + \frac{a^2}{4}
\end{aligned}
$$

from which

$$
m = \sqrt{\frac{a^2}{4}} = \frac{a}{2}.
$$

IV. solution: Take 6 pyramids that meet the conditions and place three of them side by side so that their vertices and one edge coincide. Since the angle between the lateral faces is $120^\circ$ and three such face angles are placed side by side, the three pyramids fit together without gaps. The adjacent lateral faces of two pyramids will form an angle of $360^\circ - 2 \cdot 120^\circ = 120^\circ$. By inserting a new pyramid between each pair of adjacent pyramids so that its vertex coincides with the common vertex of the already assembled pyramids, a closed solid is formed, bounded by 6 squares (the base faces of the 6 pyramids). This solid can only be a cube. The sum of the heights of two pyramids placed with their bases opposite each other is equal to the edge of the cube, or the edge of the pyramid's base. This implies that the height of the pyramid is half the length of the base edge.

This is how Szabó László (Aszód, Petőfi g. IV. o. t.) and Tuska Ferenc (Cegléd, Kossuth g. IV. o. t.) solved the problem.

Most of the contestants determined angles using trigonometric tables, which, as the above solutions show, is completely unnecessary.