11.3. In a regular hexagonal pyramid SABCDEF (ABCDEF is the base), the lateral edge is equal to $a$, and the plane angle at vertex $S$ is $10^{\circ}$. An ant crawls along the surface of the pyramid from vertex $A$, aiming to visit all lateral edges (possibly at the vertices) and return to point $A$. What is the length of its shortest path?
Answer: $a$.

Solution. "Cut" the pyramid SABCDEF along the edge SA and make a net (see Fig. 11.3).

Then any path on the lateral surface of the pyramid, satisfying the condition, will be a broken line on the net, connecting points $A$ and $A_{1}$. The shortest path from $A$ to $A_{1}$ is the length of the segment $A A_{1}$.

Notice that in the isosceles triangle $A S A_{1}$, the angle at vertex $S$ is $60^{\circ}$. Therefore, this triangle is equilateral, so $A A_{1} = a$.

Note that it is not necessary to specify the trajectory of the ant's movement on the pyramid itself.

Grading criteria.

![](https://cdn.mathpix.com/cropped/2024_05_06_2e033c978783fd869408g-2.jpg?height=928&width=939&top_left_y=804&top_left_x=958)

"+" A complete and justified solution is provided

" $\pm$ ” A generally correct solution is provided, containing minor gaps or inaccuracies (for example, it is stated that the shortest path on the net is $A A_{1}$, but its length is not specified)

“-" Only the answer is provided

“-" An incorrect solution is provided or it is missing