In the quadrilateral $A B C D$, the points $K$ and $L$ are the closer to $A$ trisection points of sides $A B$ and $A D$, respectively, and the points $N$ and $M$ are the closer to $C$ trisection points of sides $C B$ and $C D$, respectively. Determine the ratio of the areas of quadrilaterals $A B C D$ and $K L M N$.
We need to distinguish two cases based on whether the quadrilateral $A B C D$ is convex or concave.

1988-11-377-1.eps

1. Figure

If the quadrilateral $A B C D$ is convex, then the area of the quadrilateral $K L M N$ can be calculated by subtracting the areas of the triangles $A L K, D M L, C N M$, and $B K N$ from the area of the quadrilateral $A B C D$ (1. Figure). The triangles $B K N$ and $B A C$ are similar because the ratios of two pairs of their sides are equal $(B K: B A = B N: B C = 2: 3)$, and the angles between these sides are the same. The ratio of their areas is equal to the square of the similarity ratio, that is:

$$
\frac{T_{B K N}}{T_{B A C}}=\frac{4}{9}
$$

Similarly, it can be shown that

$$
\frac{T_{D M L}}{T_{D C A}}=\frac{4}{9}
$$

From (1) and (2), it follows - using the fact that $T_{B A C} + T_{D C A} = T_{A B C D}$ - that

$$
T_{B K N} + T_{D M L} = \frac{4}{9} T_{A B C D}
$$

Similarly, it can be shown that:

$$
T_{A L K} + T_{C N M} = \frac{1}{9} T_{A B C D}
$$

Thus, using (3) and (4):

$$
\begin{gathered}
T_{K L M N} = T_{A B C D} - \left[ \left( T_{B K N} + T_{D M L} \right) + \left( T_{A L K} + T_{C N M} \right) \right] = \\
= T_{A B C D} - \left[ \frac{4}{9} T_{A B C D} + \frac{1}{9} T_{A B C D} \right] = \frac{4}{9} T_{A B C D} .
\end{gathered}
$$

Therefore, the desired ratio in this case is $\frac{4}{9}$.

If the quadrilateral $A B C D$ is concave, then using the point that divides the diagonal starting from the vertex of the concave angle in a 1:2 ratio, the quadrilateral $K L M N$ can be dissected into a hexagon that lies entirely within the quadrilateral $A B C D$ (2. Figure). Then, using similar triangles, it can be shown in essentially the same way as in the first case that the desired ratio is also $\frac{4}{9}$.

1988-11-378-1.eps

2.a Figure

1988-11-378-2. eps

2.b Figure

Based on the work of Ágoston Kolos (Szeged, Zrínyi I. Secondary School, 8th grade)

Note. If the points $K, L, M, N$ divide the sides in the ratio $m: n$, then the ratio of the areas of the two quadrilaterals is

$$
\frac{2 m \cdot n}{(m+n)^{2}}
$$

János Csirik (Szeged, Ságvári E. Practical Gymnasium, 2nd grade)