A cross-section of the pond's upper opening is a circle with a diameter of $100 \mathrm{~m}$, and plane sections through the center of this circle are equilateral triangles. Let $A$ and $B$ be two opposite points on the upper opening, $C$ the deepest point of the pond, and $H$ the midpoint of $B C$. What is the length of the shortest path connecting $A$ to $H$?
The task text does not mention it, but it is clear that the equilateral triangular cross-section must also pass through the apex of the cone. Furthermore, it is obvious that the shortest path must be found on the surface of the cone's lateral area.

![](https://cdn.mathpix.com/cropped/2024_05_02_77085136727424f7360eg-1.jpg?height=651&width=1526&top_left_y=257&top_left_x=294)

Therefore, we should unfold the lateral surface of the cone onto a plane. We know that the lateral surface is a sector of a circle whose radius is equal to the slant height of the cone, and the length of the arc is equal to the circumference of the base circle with a diameter of $100 \mathrm{~m}$. From the two ways of calculating the arc length,

$$
i=100 \pi=\frac{100 \cdot \pi \alpha}{180^{\circ}}
$$

it follows that the central angle of the sector is exactly $180^{\circ}$, meaning that the unfolded lateral surface is a semicircle with a radius of $100 \mathrm{~m}$.

Then, from the right triangle $A H C$, we can easily determine the length of the segment $A H$.

$$
A H=\sqrt{100^{2}+50^{2}}=50 \sqrt{5} \approx 111.8 \mathrm{~m}
$$

Note. Many interpreted the shortest path as the straight-line distance through the air. These received 0 points on their work. (R. Zs.)