## Task 25/64

Given is a regular three-sided pyramid with the side edge $s=10 \mathrm{~cm}$ and the angle $\alpha=75^{\circ}$ between the side and base edge.

How long is the shortest path on the mantle of the pyramid that starts from one vertex of the base and goes once around the pyramid back to the starting point?

![](https://cdn.mathpix.com/cropped/2024_06_06_1d2be4837fb839c2b31dg-098.jpg?height=542&width=606&top_left_y=1111&top_left_x=291)
Imagine the mantle cut along the side edge where the starting point lies, and unwrapped into the plane. The shortest path between two points in the plane is the line segment connecting these points (see figure).

The mantle consists of three congruent isosceles triangles. From the angle sum theorem for a plane triangle, it follows that each angle between two side edges is $30^{\circ}$. For the angle $A S A^{\prime}$, it follows that $\measuredangle A S A^{\prime}=90^{\circ}$. Therefore,

$$
a a^{\prime}=s \sqrt{2}=10 \sqrt{2} \approx 14.14 \mathrm{~cm}
$$