Jumping coins cannot build a bigger square

Question

Four coins lie on a table at the corners of a square of side 1. A legal move picks two coins, a "jumper" and a "pivot", and moves the jumper to the point on the far side of the pivot, exactly as far from the pivot as it was before, in a straight line (a reflection of the jumper through the pivot). The pivot does not move.

Using any number of such moves, can you arrange the four coins at the corners of a square of side 2? Prove your answer.

Accepted Answer

No, it is impossible. Coordinates. Place the starting square at (0, 0), (1, 0), (0, 1), (1, 1). A jump of coin J over pivot M sends J to the point 2M minus J (coordinate by coordinate). If all coins have whole-number coordinates, they still do afterwards, so every coin stays on the integer grid forever. The parity invariant. Look at the new position 2M minus J. Since 2M is even in each coordinate, 2M minus J has the same parity (odd or even) in each coordinate as J itself. So a coin's parity pattern never changes: a coin whose coordinates are (even, odd) will have coordinates (even, odd) afte…

Jumping coins cannot build a bigger square

Your answer

Discussion

Learn the concepts

Related questions