Problem 6.5. An equilateral triangle is divided into 36 identical equilateral triangles. For which $k$ can it be cut along the grid lines into $k$ identical polygons?
Answer: $1,3,4,9,12,36$.

Solution. First, note that $k$ must be a divisor of the number 36. In Fig. ??, the divisions into $1,3,4,9,12,36$ equal parts are shown. It remains to prove that the triangle cannot be divided into 2, 
![](https://cdn.mathpix.com/cropped/2024_05_06_3ea49ea8943c27893726g-3.jpg?height=508&width=868&top_left_y=1762&top_left_x=593)

Fig. 3: to the solution of problem $? ?$

6, and 18 parts.

We will color the triangle in white and black using a "chessboard" coloring, such that adjacent triangles are of different colors and the corner triangles are black (Fig. ??). Then we will have 15 white triangles and 21 black ones.

If we assume that we managed to divide the triangle into two equal parts, and one of these parts contains $n$ white and $m$ black triangles, then the other part will either also contain $n$ white and $m$ black, or vice versa. In the first case, the total number of black triangles in the entire triangle would be $2m$, but it is odd; in the second case, the total numbers of black and white triangles would both be $n+m$, but they are different. In either case, we arrive at a contradiction.

The division into 18 equal parts is impossible because each such part must consist of two adjacent triangles, i.e., one white and one black; summing all parts would again lead to a contradiction with the fact that the number of black and white triangles is not equal.

Finally, we will show that it is impossible to divide the triangle into 6 identical figures. Suppose such a division exists, then each figure consists of 6 small triangles. Let one of the figures contain $n$ black and $m$ white triangles. The other figures in the division consist either of $n$ black and $m$ white, or vice versa, of $m$ black and $n$ white. Here, $n$ and $m$ have the same parity, since $n+m=6$. Then, if we sum the number of black triangles in all six figures, we get the sum of six numbers of the same parity, i.e., an even value; however, the total number of black triangles is odd.

Remark 1. The proof that a division into two equal parts does not exist can also be done in another way: in exactly one of the parts of such a division, two vertices of the original triangle will be contained, which means that one of the parts will contain a segment equal in length to the side of the triangle, while the other will not.

Remark 2. The proof that a division into 18 "rhombuses" of two triangles each does not exist can also be done in another way. Consider the six triangles adjacent to one of the sides. The rhombus containing the outermost one is uniquely determined; the rhombus containing the next triangle is also uniquely constructed, and so on, but it is impossible to construct a rhombus with the sixth triangle.

## Criteria

If the case of dividing into 1 part is not considered in the answer or solution, points are not deducted. The fact that $k$ is a divisor of 36 is considered obvious; points are not deducted for using this fact without justification; points are not added for proving it.

A correct solution to the problem is worth 7 points. In the absence of such a solution, the following criteria are summed:

2 6. Correct examples of divisions into $k$ parts for all $k$ from $3,4,9$, $12,36$ are provided or described.

If there are examples for not all such $k$, but at least for two different ones from $3,4,9$, 12, 36, then instead of 2 points for this criterion, 1 is given.

2 6. It is proven that division into 2 parts is impossible.

2 6. It is proven that division into 6 parts is impossible.

Olympiad for Schoolchildren "Kurchatov" in Mathematics - 2019

Final Stage.

1 6. It is proven that division into 18 parts is impossible.

The following criterion is applied only in the absence of points for other advances:

16 . The correct answer is given.