8.3. In a convex 6-sided polygon $A B C D E F$, all angles are equal to $120^{\circ}$. The lengths of four of its sides are known: $B C=3, C D=5, D E=4$ and $A F=6$. Find the lengths of the two remaining sides of the hexagon. Justify your answer.

![](https://cdn.mathpix.com/cropped/2024_05_06_23fc7e256b8316916311g-5.jpg?height=542&width=605&top_left_y=1431&top_left_x=134)

To the solution of problem 8.3
Solution: Let the lines $A F$ and $B C$ intersect at point $G$, the lines $B C$ and $D E$ at point $H$, and the lines $D E$ and $A F$ at point $I$ - see the figure. Then $\angle A B G=\angle B A G=60^{\circ}$, so triangle $A B G$ is equilateral. The same is true for triangles $C D H$ and $E F I$. Let $A B=y, E F=x$. Then $G H=y+3+5=y+8, H I=5+4+x=9+x$, $I G=x+6+y$. But triangle $G H I$ is also equilateral (all its angles are $60^{\circ}$). Therefore,

$$
y+8=9+x=6+x+y
$$

From these equations, we find $x=2, y=3$.

Answer: 2 and 3.

Recommendations for checking:

| is in the work | points |
| :--- | :--- |
| Correct and justified answer | 7 points |
| An example of a hexagon satisfying the conditions of the problem is provided | 1 point |
| Incorrect solution or its absence | 0 points |