In an equilateral triangle $ABC$, $K$ is the midpoint of side $AB$, point $L$ lies one-third of the way along side $BC$ closer to point $C$, and point $M$ lies one-third of the way along side $AC$ closer to point $A$.

Determine what fraction of the area of triangle $ABC$ is occupied by triangle $KLM$.

(L. Růžičková)

Hint. The given points define additional triangles. Consider the areas of some of them.
The area of triangle $K L M$ is expressed as the difference between the area of triangle $A B C$ and the areas of triangles $A K M, K B L$, and $M L C$.

![](https://cdn.mathpix.com/cropped/2024_04_17_1ca83b28e9c956579120g-2.jpg?height=457&width=462&top_left_y=1288&top_left_x=800)

The ratio of the distances of points $M$ and $C$ from point $A$ is the same as the ratio of the distances of these points from the line $A B$. Therefore, the height of triangle $A K M$ from vertex $M$ is one-third of the height of triangle $A B C$ from vertex $C$. At the same time, the side $A K$ of the first triangle, opposite vertex $M$, is half the length of side $A B$ of the second triangle, opposite vertex $C$. Thus, triangle $A K M$ occupies $\frac{1}{3} \cdot \frac{1}{2}=\frac{1}{6}$ of the area of triangle $A B C$.

Similarly, the height of triangle $K B L$ from vertex $L$ is two-thirds of the height of triangle $A B C$ from vertex $C$, and the corresponding side $K B$ of the first triangle is half the length of side $A B$ of the second triangle. Thus, triangle $K B L$ occupies $\frac{2}{3} \cdot \frac{1}{2}=\frac{1}{3}$ of the area of triangle $A B C$.

Thirdly, the height of triangle $C M L$ from vertex $L$ is one-third of the height of triangle $A B C$ from vertex $B$, and the corresponding side $C M$ of the first triangle is two-thirds the length of side $A C$ of the second triangle. Thus, triangle $C M L$ occupies $\frac{1}{3} \cdot \frac{2}{3}=\frac{2}{9}$ of the area of triangle $A B C$.

In total, triangle $K L M$ occupies

$$
1-\frac{1}{6}-\frac{1}{3}-\frac{2}{9}=\frac{5}{18}
$$

of the area of triangle $A B C$.

Note. Observe that the given solution is valid for any triangle $A B C$.

With the assumption of equilateral triangles, the sides and heights of the individual triangles are indicated in the following figure (where $a$ denotes the side and $v$ the height of triangle $A B C$).

![](https://cdn.mathpix.com/cropped/2024_04_17_1ca83b28e9c956579120g-3.jpg?height=622&width=668&top_left_y=680&top_left_x=700)

With this notation, $S_{A B C}=\frac{1}{2} a v$ and $S_{K L M}=\frac{5}{36} a v$, thus $S_{K L M}=\frac{5}{18} S_{A B C}$.