A regular three-sided pyramid's lateral face forms an angle $\varphi$ with the base, and the angle between two lateral edges is $\varepsilon$. When will $\varphi=\varepsilon$?
Let the base of the regular triangular pyramid be $ABC$, its apex $S$, and the base edge $a$. The center of the base is denoted by $O$, so $SO$, the height of the pyramid, is perpendicular to the base plane, hence $\angle DOS = 90^\circ$ (see the figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_651f91734049773cca39g-1.jpg?height=482&width=501&top_left_y=266&top_left_x=801)

The side face height $SD = m_0$ bisects the dihedral angle $\varepsilon$ and forms the dihedral angle $\varphi$ with the base height $CD$. From the right triangle $ADS$,

$$
m_0 = \frac{a}{2} \operatorname{ctg} \frac{\varepsilon}{2}
$$

From the right triangle $DOS$,

$$
m_0 = \frac{OD}{\cos \varphi} = \frac{1}{3} \cdot DC \cdot \frac{1}{\cos \varphi} = \frac{a \sqrt{3}}{6 \cos \varphi}
$$

By equating (1) and (2), we get

$$
\operatorname{ctg} \frac{\varepsilon}{2} = \frac{\sqrt{3}}{3 \cos \varphi}, \quad \text{or} \quad 3 \cos \varphi = \sqrt{3} \cdot \operatorname{tg} \frac{\varepsilon}{2}
$$

Considering the condition $\varphi = \varepsilon$, squaring both sides yields

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

From this, we obtain the cubic equation for $\cos \varepsilon$,

$$
3 \cos^3 \varepsilon + 3 \cos^2 \varepsilon + \cos \varepsilon = 1
$$

Multiplying both sides by 9,

$$
27 \cos^3 \varepsilon + 27 \cos^2 \varepsilon + 9 \cos \varepsilon = 9
$$

Notice that the left side is the first three terms of the cube of $3 \cos \varepsilon + 1$, so adding 1 to both sides,

$$
(3 \cos \varepsilon + 1)^3 = 10
$$

Thus,

$$
\cos \varepsilon = \frac{\sqrt[3]{10} - 1}{3} \sim 0.3848
$$

and therefore,

$$
\varepsilon = \varphi = 67^\circ 22'
$$