The dihedral angle at the vertex of a regular triangular pyramid $A B C D$ with base $A B C$ is $\alpha$. A regular truncated pyramid $A B C A_1 B_1 C_1$ is cut along five edges: $A_1 B_1, B_1 C_1, C_1 C, C A$, and $A B$. After that, this pyramid is unfolded onto a plane. For what values of $\alpha$ will the resulting net necessarily overlap itself?
The surface of the given truncated pyramid $A B C A 1 B 1 C 1$ consists of three isosceles trapezoids and two equilateral triangles (Fig.1). Consider the part of the net consisting of triangle $A 1 B 1 C 1$, trapezoid $A A 1 C 1 C$, trapezoid $A A 1 B 1 B$, and trapezoid $B 1^{\prime} B^{\prime} C^{\prime} C 1^{\prime}$ (Fig.2). The net will be self-intersecting if segments $A 1 B 1$ and $B 1^{\prime} C 1$ intersect. Let $M$ be their point of intersection. Then $A 1 M \angle A 1 B 1^{\prime} M$. We have:

$$
\begin{gathered}
\angle B 1^{\prime} A 1 C 1=\angle A 1 B 1^{\prime} M=\angle A 1 B 1^{\prime} C 1^{\prime}=360^{\circ}-\angle B B^{\prime} C^{\prime}-\angle B B^{\prime} A 1= \\
=360^{\circ}-2\left(180^{\circ}-\frac{180^{\circ}-\alpha}{2}\right)=360^{\circ}-360^{\circ}+180^{\circ}-\alpha=180^{\circ}-\alpha, \\
\angle A 1 M B 1^{\prime}=180^{\circ}-\angle B 1^{\prime} A 1 M-\angle A 1 B 1^{\prime} C 1^{\prime}= \\
=180^{\circ}-\left(180^{\circ}-\alpha-60^{\circ}\right)-\left(180^{\circ}-\alpha\right)=2 \alpha-120^{\circ} .
\end{gathered}
$$

From the inequality

$$
180^{\circ}-\alpha<2 \alpha-120^{\circ},
$$

we get

$$
300^{\circ}<3 \alpha,
$$

which implies

$$
100^{\circ}<\alpha.
$$

Since the sum of the plane angles of a trihedral angle is less than $360^{\circ}$, we have $3 \alpha<360^{\circ}$, hence $\alpha<120^{\circ}$.

## Answer

The problem