Help me with this geometry question!

Question

First of all excuse my English it is not my first language and I’m not very familiar with the math terms. So this is the original question which I found the answer is 7 but now I have other questions. In the equilateral triangle ABC we send a light beam through The vertex A so that it hits D on side BC so that: BD/BC=1/5. How many times does the light beam have to reflect so that it goes back through vertex A? So the answer for this is 7. And I’ve tried this other way that if D is on BC so that BD/BC=1/4 which the answer is it needs to reflect 4 times. Now I found that if BD/BC=irrational number, there is no way for it to go through A again. But if BD/BC=rational number then it will always go back through A. I want to know why this is? The picture is the answer to the original question.

Answer 1

Tile the plane with translates of $ABC$ and its inverse, so the vertices form a lattice. (You might start by tiling the plane with regular hexagons and dividing each hexagon into 6 equilateral triangles, or you could draw three families of equi-spaced parallel lines tilted at (say) $0^\circ, 120^\circ, 240^\circ$.) Any straight line segment from a vertex to another (except for those that contain an edge of one of the tiles) corresponds to a closed reflection path of the sort you are interested in, and the number of triangles such a line segment visits is equal to the number of reflections minus plus.

The rationality claim can be seen as follows. The line from $A$ cuts the base at $P$ in the line segment $BC$, the claim is that the ratio $r_1=|BP|/|BC|$ is rational if and only if the line passes through another vertex in the lattice of vertices in the tessellation. Look at the successive intersections with edges parallel to $BC$, and compute the successive ratios $r_n = |B_nP_n|/|B_n C_n|$ where $B_n$ and $C_n$ are the endpoints of the newly cut parallel edge. The sequence of numbers $r_n$ obeys $r_n = n r_1 \bmod 1$, that is to say, there exists a sequence of integers $k_n$ such that $r_n = n r_1 - k_n$. For your line to pass through another vertex of the tessellation you need $r_n=0$ for some $n\ne 1$; this corresponds to $r_1= k_n/n$.

Judging by your diagram I think this is basically the idea you had, as hinted by the regular spacing of the dots you plotted on the $BC$ edge. But I think it is a bit easier to understand if you plot your diagram in the "looking glass" world by lifting your trajectory to the covering space.