2. Given a parallelogram $A B C D$ and points $X$ and $Y$ on its boundary, different from point $A$, such that lines $A X$ and $A Y$ divide this parallelogram into three parts of equal area. Determine the ratio of the area of triangle $A X Y$ to the area of parallelogram $A B C D$.

(David Hruška)
SOLUTION. For the lines $A X$ and $A Y$ to divide the parallelogram $A B C D$ into three parts, the points $X$ and $Y$ must be distinct and neither can lie on side $A B$ or side $A D$. Each of them must therefore lie on side $B C$ or side $C D$. In the following paragraph, we will explain that one of the points $X, Y$ must lie inside side $B C$ and the other inside side $C D$, given that all three parts of the divided parallelogram have one-third of its area.

Since the diagonal $A C$ bisects the area of the parallelogram $A B C D$, none of the three parts with one-third area can contain the entire triangle $A B C$ or the entire triangle $A C D$. However, if both points lie on side $B C$ as shown in the figure, one of the three parts would contain the entire triangle $A C D$. Similarly, the case where both points $X, Y$ lie on side $C D$ is also ruled out.

![](https://cdn.mathpix.com/cropped/2024_04_17_f65fa79cddf5b6ea0ee5g-3.jpg?height=322&width=642&top_left_y=1064&top_left_x=707)

Since the labels $X$ and $Y$ can be interchanged, we will assume further that point $X$ lies inside side $B C$ and point $Y$ inside side $C D$. The parallelogram $A B C D$ is then divided into two triangles $A B X, A Y D$ and quadrilateral $A X C Y$.

![](https://cdn.mathpix.com/cropped/2024_04_17_f65fa79cddf5b6ea0ee5g-3.jpg?height=320&width=636&top_left_y=1639&top_left_x=710)

Let $S$ be the area of the entire parallelogram $A B C D$. According to the problem, the areas of triangles $A B X$ and $A Y D$ are $S_{A B X}=S_{A Y D}=S / 3$. Since triangles $A B X$ and $A B C$ share the same height from vertex $A$, the lengths of the opposite sides are in the ratio

$$
|B X|:|B C|=S_{A B X}: S_{A B C}=(S / 3):(S / 2)=2: 3,
$$

from which $|B X|=\frac{2}{3}|B C|$, so $|C X|=\frac{1}{3}|B C|$. Similarly, by comparing triangles $A Y D$ and $A C D$, we get $|C Y|=\frac{1}{3}|C D|$. Together, we find that by the SAS similarity criterion, triangles $C X Y$ and $C B D$ are similar in the ratio $1: 3$. Therefore, the area of the first is

$$
S_{C X Y}=\left(\frac{1}{3}\right)^{2} \cdot S_{C B D}=\left(\frac{1}{3}\right)^{2} \cdot \frac{S}{2}=\frac{S}{18}
$$

The quadrilateral $A X C Y$ also has an area of $S / 3$ and is composed of triangles $A X Y$ and $C X Y$. We already know the area of the latter, so the area of the former is

$$
S_{A X Y}=\frac{S}{3}-S_{C X Y}=\frac{S}{3}-\frac{S}{18}=\frac{5 S}{18} .
$$

Conclusion. The desired ratio of areas is 5 : 18.

For a complete solution, award 6 points. In the case of partial solutions, award:

- 1 point for excluding the case where both points $X$ and $Y$ lie on one of the sides $B C$ or $C D$.
- 2 points for determining at least one of the ratios in which points $X, Y$ divide the respective sides $B C$ or $C D$.
- 2 points for calculating the ratio of the area of triangle $C X Y$ to the area of parallelogram $A B C D$.

Do not penalize for the absence of the exclusion of the positions of points $X$ and $Y$ on sides $A B$ and $A D$. Do not penalize for the omission of cases where $X$ or $Y$ coincide with one of the vertices $B, C, D$. However, deduct one point if there is no explanation for why both points $X$ and $Y$ cannot lie on the same side $B C$ or $C D$.