Baltic Way 1995 · Problem 14

The small triangles can be coloured in four colours as shown in Figure 2. Then each flea can only reach triangles of a single colour. Moreover, number the horizontal rows are numbered as in Figure 2, and note that with each move a flea jumps from a triangle in an even-numbered row to a triangle in an odd-numbered row, or vice versa. Hence, if all the fleas are to meet in one small triangle, then they must initially be located in triangles of the same colour and in rows of the same parity. On the other hand, if these conditions are met, then the fleas can end up all in some designated triangle (of the right colour and parity). When a flea reaches this triangle, it can jump back and forth between the designated triangle and one of its neighbours until the other fleas arrive.

It remains to find the values of $n$ for which the big triangle contains at least $n$ small triangles of one colour, in rows of the same parity. For any odd $n$ there are at least $1+2+\cdots+\frac{n+1}{2}=\frac{1}{8}\left(n^{2}+4 n+3\right) \geq n$ such triangles. For even $n \geq 6$ we also have at least $1+2+\cdots+\frac{n}{2}=\frac{1}{8}\left(n^{2}+2 n\right) \geq n$ triangles of the required kind. Finally, it is easy to check that for $n=2$ and $n=4$ the necessary set of small triangles cannot be found.

Hence it is possible for the fleas to meet in one small triangle for all $n$ except 2 and 4 .

Figure 2