Jeck and Lisa are playing a game on table dimensions $m \times n$ , where $m , n >2$. Lisa starts so that she puts knight figurine on arbitrary square of table.After that Jeck and Lisa put new figurine on table by the following rules: 1. Jeck puts queen figurine on any empty square of a table which is two squares vertically and one square horizontally distant, or one square vertically and two squares horizontally distant from last knight figurine which Lisa put on the table 2. Lisa puts knight figurine on any empty square of a table which is in the same row, column or diagonal as last queen figurine Jeck put on the table. Player which cannot put his figurine loses. For which pairs of $(m,n)$ Lisa has winning strategy? Proposed by Stijn Cambie
Problem
Source: European Mathematical Cup 2014, Senior Division, P2
Tags: geometry, rectangle, combinatorics unsolved, combinatorics
15.12.2014 10:32
Note that if we can somehow pair up the squares of the table such that every pair is a knight's move apart, then Jeck would win, as for every move Lisa makes, Jeck can just move to the other square in the pair. Now let's see which pairs (m,n) we can pair the squares in such a manner. Every letter represents a different pair. We can tile larger rectangles with these smaller rectangles. Case 1: For (m,n) = (2x,4y), we can tile it with 2x4 rectangles. Case 2: For (m,n) = (4x+2,4y+2) (x,y $\ge$ 1), we can start with a 6x6 square, tile a 6 by 4(y-1) rectangle as in Case 1, then tile the remaining 4(x-1) by 4y+2 rectangle as in Case 1. Case 3: For (m,n) = (2x+1,4y) (x $\ge$ 1), we can tile it with a row of 3x4 rectangles, then 2x4 rectangles. Case 4: For (m,n) = (4x-1,4y+2), we can start with a 3x6 rectangle, tile the 3 by 4(y-1) rectangle with 3x4 rectangles, then tile the remaining 4(x-1) by 4y+2 rectangle as in Case 1. Case 5: For (m,n) = (4x+1,4y+2) (x,y $\ge$ 1), we can start with a 5x6 rectangle, tile the 5 by 4(y-1) rectangle as in Case 3, then tile the remaining 4(x-1) by 4y+2 rectangle as in Case 1. Since all these cases are tileable as described above, Jeck wins for all these cases. Now the remaining cases are when m,n are both odd. When (m,n) are both odd (say (2x+1,2y+1)), Lisa wins, as she can, after placing her first move anywhere, pair up the (2x) squares along the same column as that first move, then after that, along each row, the remaining (2y) squares. Regardless of what Jeck moves, Lisa can just place her piece into the other square in the pair that Jeck placed his piece into. So Jeck usually wins, unlike IMO 2011.
16.12.2014 21:25
Someone has the junior version of test problems for this same competition?