Problem 4. Let $L$ be the intersection point of the diagonals $C E$ and $D F$ of a regular hexagon $A B C D E F$ with side length 4. Point $K$ is such that $\overrightarrow{L K}=3 \overrightarrow{F A}-\overrightarrow{F B}$. Determine whether point $K$ lies inside, on the boundary, or outside of $A B C D E F$, and also find the length of segment $K A$.
Answer: $\frac{4 \sqrt{3}}{3}$.

Solution. Let $O$ be the center of the hexagon. It is known that then $F E D O$ is a rhombus, from which $F D \perp E O$. Similarly, $E C \perp D O$. Therefore, point $L$ is the center of the equilateral triangle $D E O$.

Next, since $3 \overrightarrow{F A}-\overrightarrow{F B}=\overrightarrow{B A}+2 \overrightarrow{F A}$, point $K$ is also the center of an equilateral triangle, the position of which is shown in the figure. This means that point $K$ lies outside the hexagon and $K A$ equals the radius of the circumscribed circle of an equilateral triangle with side length 4, from which the answer follows.

![](https://cdn.mathpix.com/cropped/2024_05_06_a567280d18f9fc82386fg-07.jpg?height=557&width=562&top_left_y=201&top_left_x=1158)