Estonian Math Competitions

Answer: 65%.

Let the area of the triangle $ABC$ be $S$ and the areas of triangles $AKM$, $BKN$, $MKN$ and $CMN$ be $S_1$, $S_2$, $S_3$ and $S_4$, respectively (Fig. 30). Then:

$S_2+S_3+S_4=\frac{2}{3}S$ since the l.h.s. is the area of the triangle $MBC$ while triangles $MBC$ and $ABC$ have equal altitudes and the ratio of the lengths of the corresponding bases is $\frac{2}{3}$; $S_1+S_3+S_4=\frac{3}{2}{S}_4$ since the l.h.s. is the area of the triangle $ANC$ while triangles $ANC$ and $MNC$ have equal altitudes and the ratio of the lengths of the corresponding bases is $\frac{3}{2}$; $S_2=\frac{1}{3}{S}_3$ because triangles $BNK$ and $MNK$ have equal altitudes and the ratio of the lengths of the corresponding bases is $\frac{1}{3}$; $S_1=\frac{1}{4}S$ because the ratio of altitudes of the triangles $AKM$ and $ABC$ is $\frac{3}{4}$ while the ratio of the lengths of the corresponding bases is $\frac{1}{3}$.



Substituting $S_1$ and $S_2$ from the latter two equations to the former two equations, we obtain the system of equations $$\{\begin{array}{l}\frac{1}{3}{S}_3+S_3+S_4=\frac{2}{3}S,\\ \frac{1}{4}S+S_3+S_4=\frac{3}{2}{S}_4.\end{array}$$ After adding $\frac{1}{3}{S}_4$ to both sides of the first equation and bringing the term with $S$ to the l.h.s., we obtain the equivalent system $$\{\begin{array}{l}\frac{4}{3}(S_3+S_4)-\frac{2}{3}S=\frac{1}{3}{S}_4,\\ (S_3+S_4)+\frac{1}{4}S=\frac{3}{2}{S}_4.\end{array}$$ Solving with respect to $S_3+S_4$ and $S_4$ yields $S_3+S_4=0.65S$. Hence the area of the quadrilateral $MKNC$ equals 65% of the area of the triangle $ABC$.



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Alternative solution.

Denote the area of a figure $\Pi$ by $S_{\Pi }$. From the conditions of the problem, we obtain the following:

$\frac{S_{ABM}}{S_{ABC}}=\frac{|AM|}{|AC|}=\frac{1}{3}$, whence $S_{ABM}=\frac{1}{3}{S}_{ABC}$ and $S_{CBM}=\frac{2}{3}{S}_{ABC}$; $\frac{S_{ABK}}{S_{ABM}}=\frac{|BK|}{|BM|}=\frac{1}{4}$, whence $S_{ABK}=\frac{1}{4}\cdot \frac{1}{3}{S}_{ABC}=\frac{1}{12}{S}_{ABC}$ and, consequently, $S_{AMK}=(\frac{1}{3}-\frac{1}{12})S_{ABC}=\frac{1}{4}{S}_{ABC}$; * $\frac{S_{BCK}}{S_{BCM}}=\frac{|BK|}{|BM|}=\frac{1}{4}$ (Fig. 31), whence $S_{BCK}=\frac{1}{4}\cdot \frac{2}{3}{S}_{ABC}=\frac{1}{6}{S}_{ABC}$ and $S_{MCK}=(\frac{2}{3}-\frac{1}{6})S_{ABC}=\frac{1}{2}{S}_{ABC}$.

Since $\frac{S_{ABN}}{S_{ACN}}=\frac{|BN|}{|CN|}=\frac{S_{KBN}}{S_{KCN}}$, we obtain $$\frac{S_{ABN}}{S_{ACN}}=\frac{S_{ABN}-S_{KBN}}{S_{ACN}-S_{KCN}}=\frac{S_{ABK}}{S_{AMK}+S_{MCK}}=\frac{\frac{1}{12}}{\frac{1}{4}+\frac{1}{2}}=\frac{\frac{1}{12}}{\frac{3}{4}}=\frac{1}{9}.$$

Hence $\frac{S_{ABN}}{S_{ABC}}=\frac{S_{ABN}}{S_{ABN}+S_{ACN}}=\frac{S_{ABN}}{S_{ABN}+9S_{ABN}}=\frac{1}{10}$, implying $S_{ABN}=\frac{1}{10}{S}_{ABC}$.

Altogether, we obtain $$\frac{S_{MKNC}}{S_{ABC}}=\frac{S_{ABC}-S_{ABN}-S_{AMK}}{S_{ABC}}=1-\frac{1}{10}-\frac{1}{4}=0.65,$$ meaning that the area of the quadrilateral $MKNC$ equals 65% of the area of the triangle $ABC$.