$A B C D$ is a regular tetrahedron. We know that the tangent to the conical surface with vertex at $A$ and axis $A B$ touches the edge $C D$. What is the opening angle of the cone?
Let $F$ be the midpoint of segment $CD$. Then the plane $ABF$ is a plane of symmetry for both the cone and the segment $CD$. Therefore, if the segment $CD$ touches the cone at point $P$, then the reflection of $P$ in the plane $ABF$ also touches the cone. However, the segment $CD$ has exactly one common point with the cone, so $P$ coincides with its reflection, and such a point on the segment $CD$ is only one, $F$; thus $P \equiv F$ (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_42833cc31a962b6f3880g-1.jpg?height=483&width=602&top_left_y=344&top_left_x=748)

Figure 1

![](https://cdn.mathpix.com/cropped/2024_05_02_42833cc31a962b6f3880g-1.jpg?height=466&width=412&top_left_y=968&top_left_x=843)

Figure 2

The half-aperture angle of the cone is equal to the angle formed by a generator and the axis, for example, the angle $FAB$. In the triangle $FAB$ (Figure 2), one side is exactly an edge of the tetrahedron $ABCD$, and the other two sides are the heights of two faces of the tetrahedron. Therefore, if the half-aperture angle is $\alpha$, then:

$$
\cos \alpha=\frac{\frac{a}{2}}{a \cdot \frac{\sqrt{3}}{2}}=\frac{\sqrt{3}}{3}, \quad \text { from which } \quad 2 \alpha=109.5^{\circ}
$$

Based on the work of Judit Nagy (Miskolc, Földes F. Gymnasium, 1st year)