A regular triangular pyramid's lateral face forms an angle $\varphi$ with the base, and the angle between two adjacent lateral edges is $\varepsilon$. Express some trigonometric function of $\varepsilon$ as a function of $\varphi$ functions. a) For what value of $\varphi$ will $\varepsilon=45^{\circ}$? b) For what value of $\varepsilon$ will $\varphi=45^{\circ}$?
I. Solution: The labeling is shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_02_71c0483923fd273b5979g-1.jpg?height=407&width=523&top_left_y=187&top_left_x=779)

$$
O A_{1}=\varrho=\frac{1}{3} \cdot \frac{a \sqrt{3}}{2}=\frac{a \sqrt{3}}{6}, \quad S A_{1}=m_{0}=\frac{\varrho}{\cos \varphi}=\frac{a \sqrt{3}}{6 \cos \varphi}
$$

and thus

$$
\operatorname{tg} \frac{\varepsilon}{2}=\frac{a}{2 m_{0}}=\frac{a}{2} \cdot \frac{6 \cos \varphi}{a \sqrt{3}}=\sqrt{3} \cdot \cos \varphi
$$

But

$$
\operatorname{tg} \frac{\varepsilon}{2}=\sqrt{\frac{1-\cos \varepsilon}{1+\cos \varepsilon}}
$$

so

$$
\frac{1-\cos \varepsilon}{1+\cos \varepsilon}=\operatorname{tg}^{2} \frac{\varepsilon}{2}=3 \cos ^{2} \varphi
$$

from which

$$
\cos \varepsilon=\frac{1-3 \cos ^{2} \varphi}{1+3 \cos ^{2} \varphi}
$$

a) If $\varepsilon=45^{\circ}$ then $\cos \varepsilon=\frac{\sqrt{2}}{2}$, so

$$
\frac{\sqrt{2}}{2}\left(1+3 \cos ^{2} \varphi\right)=1-3 \cos ^{2} \varphi
$$

thus

$$
\sqrt{2}+3 \sqrt{2} \cos ^{2} \varphi=2-6 \cos ^{2} \varphi
$$

and hence

$$
(6+3 \sqrt{2}) \cos ^{2} \varphi=2-\sqrt{2}
$$

from which

$$
\begin{gathered}
\cos ^{2} \varphi=\frac{2-\sqrt{2}}{6+3 \sqrt{2}}=\frac{(2-\sqrt{2})(6-3 \sqrt{2})}{36-18}=\frac{18-12 \sqrt{2}}{18}=\frac{3-\sqrt{2}}{3} \approx \\
\approx \frac{0.1716}{3}=0.0572
\end{gathered}
$$

Using the 4-digit logarithmic table from school,

$$
\varphi=76^{\circ} 10^{\prime}
$$

so

b) If $\varphi=45^{\circ}$, then $\cos \varphi=\frac{\sqrt{2}}{2}$ and $\cos ^{2} \varphi=\frac{1}{2}$,

$$
\cos \varepsilon=\frac{1-3 \cdot \frac{1}{2}}{1+3 \cdot \frac{1}{2}}=\frac{2-3}{2+3}=-\frac{1}{5}
$$

so

$$
\cos \left(180^{\circ}-\varepsilon\right)=\frac{1}{5}=0.2
$$

from which

$$
180^{\circ}-\varepsilon=78^{\circ} 28^{\prime}
$$

and thus

$$
\varepsilon=101^{\circ} 32^{\prime}
$$

Keresztély Sándor (Miskolc, Földes Ferenc g. IV. o. t.)

II. Solution: If in the relationship (1) of the first solution we do not substitute the value (2) for $\operatorname{tg} \frac{\varepsilon}{2}$, but use the following formula, then

$$
\operatorname{tg} \varepsilon=\frac{2 \operatorname{tg} \frac{\varepsilon}{2}}{1-\operatorname{tg}^{2} \frac{\varepsilon}{2}}=\frac{2 \sqrt{2} \cdot \cos \varphi}{1-3 \cdot \cos \varphi}
$$

The numerical results naturally agree with those obtained in the first solution.

Kovács László (Debrecen, Ref. g. III. o. t.)

III. Solution: In the right triangle $A_{1} O S$,

$$
\operatorname{tg} \varphi=\frac{m}{\varrho}=\frac{6 m}{a \sqrt{3}}=\frac{2 \sqrt{3} m}{a}
$$

Denoting the distance $O B=\frac{2}{3} \cdot \frac{a}{2} \sqrt{3}=\frac{a}{3} \sqrt{3}$ by $r$, in the right triangle $B O S$,

$$
\sin \frac{\varepsilon}{2}=\frac{a}{2 S B}=\frac{a}{2 \sqrt{m^{2}+r^{2}}}
$$

and thus

$$
\sin ^{2} \frac{\varepsilon}{2}=\frac{c^{2}}{4\left(m^{2}+\frac{a^{2}}{3}\right)}=\frac{3 a^{2}}{12 m^{2}+4 a^{2}}
$$

From (1),

$$
m^{2}=\frac{a^{2} \operatorname{tg} \varphi}{12}
$$

Substituting this value into (2) and simplifying by $a^{2}$,

$$
\sin ^{2} \frac{\varepsilon}{2}=\frac{1-\cos \varepsilon}{2}=\frac{3}{\operatorname{tg}^{2} \varphi+4}
$$

from which

$$
\cos \varepsilon=1-\frac{6}{\operatorname{tg}^{2} \varphi+4}=\frac{\operatorname{tg}^{2} \varphi-2}{\operatorname{tg}^{2} \varphi+4}
$$

Mohos Béla (Pannonhalma, Benczés g. III. o. t.)