KGS math club/solution 11 21

It is assumed that the initial square is the unit square with coordinates (0,0), (0,1), (1,0) and (1,1). Note that jumping always yields points with integer coordinates.

1) There can never be a larger square.

For a contradiction, assume there is a jumping sequence that enlarges the unit square to a square of side length 3, say. Then there is also a jumping sequence from this larger square back to the unit square (the sequence that reverses each step of the original sequence). This inverse sequence may be applied verbatim to the unit square itself, which would result in an even smaller square of side length 1/3. But this is not possible, since jumping always yields integer coordinates.

Note that this not only applies to squares: no configuration can ever be enlarged (or shrinked) by jumping.

2) Three points can never be collinear.

Consider a jump from A over B. Note that this is realized by adding 2*(B-A) to A. The difference vector 2*(B-A) has even coordinates. Hence, modulo 2, the points don't change: in each configuration ever achieved, modulo 2 the points are equal to the respective corners of the unit square. In particular, after each jump the four points are pairwise distinct modulo 2.

Claim: If three points are collinear, then two of them are equal modulo 2.

Let A, B, C three collinear points, B being the inner point. First, let AB = BC, then C = A+2*(B-A). Hence A and C are equal modulo 2, as claimed. Now, w.l.o.g., assume AB < BC. Then perform a jump from A over B, thus making the frog on A the new inner frog. Repeating this process finally yields the first case, where the inner point is equidistant to the outer points.

Summing up: The corners of the unit square are pairwise distinct modulo 2, hence the same is true in all possible configurations. As shown in the claim, in the case of three collinear points, two of them are equal modulo 2. Hence, such a configuration can never be achieved.

Greetings from "Georg" on KGS