Anička got a cake in the shape of a rectangle for her birthday. She cut it using two straight cuts. One cut went through the longer sides of the rectangle, intersecting them at one-third of their length, and the other cut went through the shorter sides, intersecting them at one-fifth of their length. Neither cut was parallel to the sides of the rectangle, and at each vertex of the rectangle, either two shorter segments or two longer segments of the divided sides always met.

![](https://cdn.mathpix.com/cropped/2024_04_17_24585cf941083de07254g-3.jpg?height=402&width=552&top_left_y=553&top_left_x=752)

Anička ate the gray-marked piece of the cake. Determine what fraction of the cake this was.

(A. Bohiniková)
A rectangle is centrally symmetric with respect to the center, which is the intersection of the diagonals. According to this center, any two points lying on opposite sides of the rectangle and in the same ratio to the corresponding vertices are also symmetric. Such pairs of points also determine the cuts mentioned in the problem. These cuts and the diagonals of the rectangle thus intersect at a common point. One of these diagonals divides the gray quadrilateral into two triangles, as shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_04_17_24585cf941083de07254g-3.jpg?height=397&width=556&top_left_y=1475&top_left_x=750)

One side of the left triangle is exactly one-third of the longer side of the rectangle, and the height to this side is half of the shorter side of the rectangle. This triangle thus occupies

$$
\frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{2}=\frac{1}{12}
$$

of the area of the entire rectangle. One side of the right triangle is exactly one-fifth of the shorter side of the rectangle, and the height to this side is half of the longer side of the rectangle. This triangle thus occupies

$$
\frac{1}{2} \cdot \frac{1}{5} \cdot \frac{1}{2}=\frac{1}{20}
$$

of the area of the entire rectangle. The area of the gray quadrilateral is equal to

$$
\frac{1}{12}+\frac{1}{20}=\frac{5+3}{60}=\frac{2}{15}
$$

of the area of the entire rectangle. The eaten part thus constitutes $\frac{2}{15}$ of the entire cake.

Note. If we denote the length of the longer side of the rectangle by $a$ and the length of the shorter side by $b$, then we can express the previous considerations as follows:

$$
\frac{1}{2} \cdot \frac{a}{3} \cdot \frac{b}{2}+\frac{1}{2} \cdot \frac{b}{5} \cdot \frac{a}{2}=\ldots=\frac{2 a b}{15}
$$

The rectangle has an area of $a b$, so the eaten part constitutes $\frac{2}{15}$ of the entire cake.

Another solution. At the beginning, we will again show that the cuts intersect at the center of the rectangle. This point is also where the axes of the rectangle pass through, dividing it into four congruent rectangles. One of these rectangles is divided by the cuts into three parts consisting of the gray quadrilateral and two white right triangles.

![](https://cdn.mathpix.com/cropped/2024_04_17_24585cf941083de07254g-4.jpg?height=403&width=560&top_left_y=1089&top_left_x=748)

The area of the gray part can be expressed by subtracting the areas of the two white triangles from the area of the quarter rectangle. Given the previous notation, the areas of these triangles are:

$$
\frac{1}{2} \cdot\left(\frac{a}{2}-\frac{a}{3}\right) \cdot \frac{b}{2}=\frac{a b}{24} \quad \text { and } \quad \frac{1}{2} \cdot\left(\frac{b}{2}-\frac{b}{5}\right) \cdot \frac{a}{2}=\frac{3 a b}{40}
$$

Thus, the area of the gray region is:

$$
\frac{a b}{4}-\frac{a b}{24}-\frac{3 a b}{40}=\frac{(30-5-9) a b}{120}=\frac{2 a b}{15} .
$$

Again, we conclude that the eaten part constitutes $\frac{2}{15}$ of the cake.

Grading suggestion. 1 point for justifying that the cuts intersect at the center of the rectangle (this can be proven using central symmetry, similarity, or other specific considerations); 2 points for the areas of the auxiliary triangles; 2 points for the result; 1 point for the quality and completeness of the commentary.