10. (9th grade) On the side $D E$ of a regular hexagon $A B C D E F$, a point $K$ is chosen such that the line $A K$ divides the area of the hexagon in the ratio $3: 1$. In what ratio does the point $K$ divide the side $D E$? Answer: $3: 1$.
Solution. Without loss of generality, we can assume that $S(A B C D E F)=6$. Note that then the areas of triangles $A B C$ and $A E F$ are 1, and the areas of triangles $A C D$ and $A D E$ are 2. Denoting $x=D K: D E$, we get that $S(\triangle A D K)=2 x, S(\triangle A E K)=2(1-x)$, from which, $3+2 x=3(1+2(1-x))$. Solving, we get $x=3 / 4$. Therefore, $D K: K E=3 / 4: 1 / 4=3: 1$.