## Task 2.

An equilateral triangle is given, with 9 points marked on each side, dividing each side into 10 congruent segments. These points are connected by a total of 27 lines parallel to the sides of the triangle. In this way, the triangle is divided into 100 small equilateral triangles. The area between two adjacent parallel lines is called a strip.

How many of the small triangles can we select so that no two selected triangles are within the same strip?
## Solution.

Let $A B C$ be the initial triangle and let $D, E$ and $F$ be the midpoints of sides $\overline{B C}, \overline{A C}$ and $\overline{A B}$, respectively. Then the triangles $A E F, D B F, D E C$ and $D E F$ are congruent. Let $a, b, c, d$ be the number of small triangles chosen in each of them, respectively.

![](https://cdn.mathpix.com/cropped/2024_05_30_fcf49bda6fab77cc12a3g-15.jpg?height=456&width=571&top_left_y=483&top_left_x=754)

Notice that the trapezoid $A B D E$ consists of five horizontal strips, so it can contain at most five chosen small triangles, i.e.,

$$
a+b+d \leqslant 5
$$

Similarly, we get

$$
a+c+d \leqslant 5, \quad b+c+d \leqslant 5
$$

Adding these three inequalities, we obtain

$$
2(a+b+c+d)+d \leqslant 15
$$

from which, due to $d \geqslant 0$, we conclude that $a+b+c+d \leqslant 7.5$, or $a+b+c+d \leqslant 7$.

The following example shows that it is possible to choose exactly 7 small triangles such that no two are within the same strip. Therefore, the answer is 7.