Moscow Math Olympiad 1962 2 reflecting equilateral triangle

Question

I have difficulty to prove this problem from Moscow Mathematical Olympiad 1962.

We reflect an equilateral triangle with one marked side through one of its sides. Then we similarly reflect the resulting triangle, etc., until at a certain step the triangle returns to its initial position. Prove that the marked side also returns to its initial position

and also another one follows: Prove that the number of reflections is even.

Answer 1

Here's just sketch of the solution, following which, I think, you can do the rest.

For the second part, consider the following diagram. Let the whole plane of the paper be coloured this way. Either the triangle began its journey from one of the blacks or from one of the uncoloured boxes. It's clear that it moves into the the contrasting colour at each move. So, to return to its initial position, that is, to the same colour, it must complete an even number of moves.

For the first part, consider a triangle and number its sides as $1,2,3$. Now, you've the whole set of possible paths in the diagram, so, you can mark each triangle in the box as $1,2$ or $3$ depending on the number of the side which lands into the side. Note that every side can have an unique number. This is a strong hint.

Answer 2

Consider a triangular grid with the following regular colouring of its edges.

Answer 3

Here is an intuition, or lazy idea for your consideration:

Since equilateral triangles have angles of $60^0$, we can put together a prove based on the cyclic group under multiplication formed by the $6$-th complex roots in the unit circle, and considering the reflections as rotations of the marked side on the powers of $z^2$ (in red):