3. In square $A B C D$, diagonals $A C$ and $B D$ intersect at point $O$. Point $K$ is the midpoint of side $A B$. Points $M$ and $N$ are chosen on sides $A D$ and $B C$ respectively such that rays $O K, O M$, and $O N$ divide the square into three parts of equal area. In what ratio does point $M$ divide side $A D$?
3. Answer. 5:1 Solution. Let the areas of ACO and BCO be $S$ (obviously, they are the same). Then, from the equality of the areas BLOK and AKOM, it follows that the areas of triangles BLO and AOM are equal, denote their areas by $\mathrm{S}_{1}$. Since the area of BCO is $2 \mathrm{~S}$, the area of OLC is $2 S-S_{1}$ and is equal to the area of OMD. Thus, the area of part OLCMD is $2 \mathrm{~S}-\mathrm{S}_{1}+2 \mathrm{~S}-\mathrm{S}_{1}+2 \mathrm{~S}=6 \mathrm{~S}-2 \mathrm{~S}_{1}$. From the equality of the areas of all three parts, we get the equation: $6 \mathrm{~S}-2 \mathrm{~S}_{1}=\mathrm{S}+\mathrm{S}_{1}$, hence, $\mathrm{S}=3 / 5 \mathrm{~S}_{1}$.

Consider triangles AOM and MOD, their areas are $\mathrm{S}_{1}$ and $2 \mathrm{~S}-\mathrm{S}_{1}=\mathrm{S}_{1} / 5$, respectively. These triangles have a common height, therefore, their bases AM and MD are in the ratio of their areas, i.e., $5: 1$.

Grading criteria. Correct solution - 7 points. Correctly expressed areas of AOM and OMD - $\mathbf{3}$ points. In other cases - $\mathbf{0}$ points.