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 5. Point $K$ is such that $\overrightarrow{L K}=\overrightarrow{F B}-3 \overrightarrow{A B}$. 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 $K F$.
Answer: $\frac{5 \sqrt{3}}{3}$.

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 $D E O$. Furthermore, since $\overrightarrow{F B}-3 \overrightarrow{A B}=\overrightarrow{F A}+2 \overrightarrow{B A}$, 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 $K F$ equals the radius of the circumscribed circle of the equilateral triangle with side length 5, from which the answer follows.

![](https://cdn.mathpix.com/cropped/2024_05_06_a567280d18f9fc82386fg-15.jpg?height=434&width=692&top_left_y=2347&top_left_x=1136)