In the equilateral triangle $ABC$, $AB=2$ units, and the midpoint of side $AB$ is $C_{0}$. The regular hexagonal grid $KLMNOP$ with unit side length is placed on the side of $AB$ that does not contain the triangle, such that $K$ coincides with $C_{0}$, and $L$ coincides with $A$. The grid starts to roll around $K$ along the perimeter of the triangle and stops when the distance traveled by point $K$ is equal to the distance traveled by point $L$. In what position does the grid stop?
When the hexagon rotates, each vertex moves along an arc of a circle, except for the vertex around which the rotation occurs, which remains stationary and is the center of the circle. The radius of the circle is either a side of the hexagon or one of its diagonals. Their lengths are 1 and, for example, in the triangle $K M N$, using the Pythagorean theorem, $K N=2$, $K M=\sqrt{3}$. The angle of rotation is either $60^{\circ}$ or $180^{\circ}$.

The rotation starts around the vertex $K$ in the direction of vertex $B$. In the first rotation, $K$ remains stationary, so the distance it travels is 0. The vertex $L$ rotates around $K$ along a unit circle by $60^{\circ}$, which is one-sixth of a circle, and the hexagon's vertex $P$ moves to $B$, so one of its sides again coincides with a side of the equilateral triangle. The distance traveled by $L$ is $\frac{\pi}{3}$.

In the second rotation, the hexagon must be rotated $180^{\circ}$ around $P$ so that the side $P O$ aligns with $B C$. The distance traveled by $K$ is $\pi$, and the distance traveled by $L$ is $\sqrt{3} \pi$. And so on, with the distances traveled by the vertices $K$ and $L$ in each subsequent rotation as follows:

|  | 1. | 2. | 3. | 4. | 5. | 6. |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $L$ | $\frac{\pi}{3}$ | $\sqrt{3} \pi$ | $\frac{2 \pi}{3}$ | $\sqrt{3} \pi$ | $\frac{\pi}{3}$ | 0 |
| $K$ | 0 | $\pi$ | $\frac{\sqrt{3} \pi}{3}$ | $2 \pi$ | $\frac{\sqrt{3} \pi}{3}$ | $\pi$ |

After the second rotation, the distance traveled by $L$ is greater than the distance traveled by $K$, since $\sqrt{3} \pi > \pi$.

![](https://cdn.mathpix.com/cropped/2024_05_02_c4338ed73120fb5f8ce1g-1.jpg?height=987&width=1092&top_left_y=1073&top_left_x=511)

Let $L_{5}$ and $K_{5}$ denote the distances traveled after the 5th step, and compare these:

\[
\begin{aligned}
& L_{5}=\frac{\pi}{3}(1+3 \sqrt{3}+2+3 \sqrt{3}+1)=\frac{\pi}{3}(4+6 \sqrt{3}) \\
& K_{5}=\frac{\pi}{3}(3+\sqrt{3}+6+\sqrt{3}+3)=\frac{\pi}{3}(9+2 \sqrt{3})
\end{aligned}
\]

Form the difference $L_{5}-K_{5}$.

\[
\frac{\pi}{3}(4+6 \sqrt{3}-9-2 \sqrt{3})=\frac{\pi}{3}(4 \sqrt{3}-5)>0
\]

since $\sqrt{3}>1.73, 4 \sqrt{3}-5>4 \cdot 1.73-5=6.92-5>0$.

Thus, $L_{5}>K_{5}$ still holds. Similarly, we can prove that in the 3rd and 4th steps, $L$ traveled the greater distance and the two distances were never equal.

Now let's calculate the values of $L_{6}$ and $K_{6}$. $L$ remains stationary in the 6th step, so

\[
\begin{gathered}
L_{6}=L_{5}=\frac{\pi}{3}(4+6 \sqrt{3}) \\
K_{6}=\frac{\pi}{3}(12+2 \sqrt{3}) \quad \text { and clearly } K_{6}>L_{6}
\end{gathered}
\]

This means that at some point after the start of the 6th rotation, the distances traveled by $K$ and $L$ will be equal. We can precisely determine the position of $K$, knowing that it moves along a unit circle centered at the point $L=A$, and the distance it has traveled is the difference $L_{5}-K_{5}=\frac{\pi}{3}(4 \sqrt{3}-5)$. Let $\alpha$ be the angle of occurrence of $K$, then

\[
\begin{gathered}
\frac{\pi \alpha}{180^{\circ}}=\frac{\pi}{3}(4 \sqrt{3}-5), \quad \text { hence } \\
\alpha=60^{\circ}(4 \sqrt{3}-5) \approx 115^{\circ} 41^{\prime}
\end{gathered}
\]

Thus, the vertex $K$ almost reaches the center of the initial hexagon, which can also be interpreted as the hexagon rotating approximately $64^{\circ} 19^{\prime}$ in the opposite direction from its initial position.