In a plane, there is an infinite triangular grid consists of equilateral triangles whose lengths of the sides are equal to $ 1$, call the vertices of the triangles the lattice points, call two lattice points are adjacent if the distance between the two points is equal to $ 1;$ A jump game is played by two frogs $ A,B,$ "A jump" is called if the frogs jump from the point which it is lying on to its adjacent point, " A round jump of $ A,B$" is called if first $ A$ jumps and then $ B$ by the following rules: Rule (1): $ A$ jumps once arbitrarily, then $ B$ jumps once in the same direction, or twice in the opposite direction; Rule (2): when $ A,B$ sits on adjacent lattice points, they carry out Rule (1) finishing a round jump, or $ A$ jumps twice continually, keep adjacent with $ B$ every time, and $ B$ rests on previous position; If the original positions of $ A,B$ are adjacent lattice points, determine whether for $ A$ and $ B$,such that the one can exactly land on the original position of the other after a finite round jumps.
Problem
Source: Chinese TST
Tags: calculus, integration, rotation, combinatorics proposed, combinatorics
13.04.2008 15:13
Frogs can exchange positions only if they are separated by a distance of integral multiple of 3 and it is allowed for one frog to sit on other. If they are adjacent originally like in the picture (wlog) using rule 1 does not help but to increase the distance between the frogs. By using the rule 2 , red frog can move around blue frog on the green hexagon wlog( we can have blue frog moving around red frog but it is still same as this case). Red frog can takes positions 1,3,5. While red frog sits at any of these positions, by rule 1 (second aspect) red frog moves to center to the hexagon and blue frog moves to diagonally opposite vertex that happens to be a even vertex. From here blue frog cannot reach using rule 2 as it is sitting in a even vertex. If frogs were initially not adjacent, we have two cases: 1. If frogs are not separated by a distance of multiple of 3 we can bring them adjacent to each other on the straight line connecting them by rule 1. By a cross over(second aspect of rule 1) we create a difference of 3 between the frogs and hence frogs cannot exchange position. So one of the frog has to stay static and the other has to rotate on the hexagon to take the diagonally opposite position. This we showed in the previous case as not possible. 2. If frogs are separated by a distance of multiple of 3 we can bring them at a distance of 0 and then the cross over, we can make them exchange positions.
Attachments:
