A regular square pyramid has a base edge of $a$, and the adjacent side faces form an angle of $\alpha$ with each other. What is the height of the pyramid? (This is a generalization of Problem 3 from the first round of the Rákosi Mátyás competition.)
I. Solution: The lettering is shown in the following figure.

![](https://cdn.mathpix.com/cropped/2024_05_02_0b9f78cc2b4ad143a7c1g-1.jpg?height=501&width=466&top_left_y=205&top_left_x=819)

Let the height we are looking for be $O M = m$. Since the pyramid is regular, $B D \perp M C$, and thus a plane can be drawn through $B D$ that is perpendicular to the side edge $M C$. Let this plane intersect $M C$ at $N$, then $B N D_{\triangle}$ is an isosceles triangle, with the angle at $N$ being the given $\alpha$.

In $\triangle M N O \approx \triangle M O C$, because these two triangles are right-angled and share the angle at $M$, and thus

$$
m : O N = M C : O C
$$

which gives

$$
m = \frac{O N \cdot M C}{O C}
$$

Since

$$
O B = O D = \frac{a \sqrt{2}}{2}, \quad \text{therefore} \quad O N = \frac{a \sqrt{2}}{2} \operatorname{cotg} \frac{\alpha}{2}.
$$

By the Pythagorean theorem,

$$
M C = \sqrt{\frac{a^{2}}{2} + m^{2}}
$$

and

$$
O C = \frac{a \sqrt{2}}{2}
$$

Substituting the values from (2), (3), and (4) into (1),

$$
m = \sqrt{\frac{a^{2}}{2} + m^{2}} \cdot \operatorname{cotg} \frac{\alpha}{2}
$$

which simplifies to

$$
m^{2} = \frac{a^{2}}{2} \operatorname{cotg}^{2} \frac{\alpha}{2} + m^{2} \operatorname{cotg}^{2} \frac{\alpha}{2}
$$

and thus

$$
m = \sqrt{\frac{a^{2} \operatorname{cotg}^{2} \frac{\alpha}{2}}{2\left(1 - \operatorname{cotg}^{2} \frac{\alpha}{2}\right)}} = \frac{a}{\sqrt{2 \operatorname{tg}^{2} \frac{\alpha}{2} - 2}}
$$

Remark: The right-hand side of (5) is equivalent to the following expressions:

$$
\frac{a \cos \frac{\alpha}{2}}{\sqrt{2 \cos (180 - \alpha)}} = a \sqrt{\frac{1 + \cos \alpha}{4 \cos (180 - \alpha)}} = \frac{a \cos \frac{\alpha}{2}}{\sqrt{2 - 4 \cos^{2} \frac{\alpha}{2}}} = \frac{\alpha}{2} \sqrt{\operatorname{cotg} \frac{\alpha}{2} \operatorname{tg}(180^{\circ} - \alpha)}
$$

and so on. Substituting $\alpha = 120^{\circ}$ into these expressions, we obtain the special value $m = \frac{a}{2}$ mentioned in the Rákosi competition problem. We get a real and finite value for $m$ only if $90^{\circ} < \alpha$. The upper limit $\alpha < 180^{\circ}$ follows from the interpretation of $\alpha$.

II. Solution: Consider the two pyramids whose common base is $\triangle B N D$ and whose heights are $N M$ and $N C$, respectively. The sum of the volumes of these two pyramids is equal to half the volume of the original pyramid, i.e., (denoting the area of the common base by $t$),

$$
t \frac{N M}{3} + t \frac{N C}{3} = \frac{a^{2}}{2} \cdot \frac{m}{3}
$$

which simplifies to

$$
t \cdot M C = \frac{a^{2}}{2} \cdot m
$$

Thus,

$$
m = \frac{2 t \cdot M C}{a^{2}} = \frac{a \sqrt{2} \cdot O N \cdot M C}{a^{2}} = \frac{O N \cdot M C \cdot \sqrt{2}}{a}
$$

which is identical to the relationship (1) in the first solution.