1996 IMO Problems/Problem 1
Problem
We are given a positive integer and a rectangular board with dimensions , . The rectangle is divided into a grid of unit squares. The following moves are permitted on the board: one can move from one square to another only if the distance between the centers of the two squares is . The task is to find a sequence of moves leading from the square with as a vertex to the square with as a vertex.
(a) Show that the task cannot be done if is divisible by or .
(b) Prove that the task is possible when .
(c) Can the task be done when ?
Solution
First we define the rectangular board in the cartesian plane with centers of the unit squares as integer coordinates and the following coordinates for the squares at the corners of , , , , as follows: , , ,
Let be the coordinates of the piece after move with the initial position of the piece.
, and also .
Let ,
Then, for any given , we have for all
Part (a):
In order to find out the conditions for which is divisible by 2 we are going to look at the following three cases:
(1) When both and are divisible by .
(2) When both and are odd.
(3) When one of and is even and the other one is odd.
Case (1): Since and ,
then .
Thus, for , this case is a valid one.
Case (2): Since and ,
then .
Thus, for , this case is a valid one.
Case (3): Since and , or and ,
then .
Thus, for , this case is NOT a valid one.
Having proved that Case (1) and Case (2) are the only valid cases for we are going to see what happens for both cases when we start with a square where both coordinates are odd:
if ,
then for case (1):
and for case (2):
and
This means that when is divisible by two, when starting at no matter how many moves you make all will either be or .
Since the ending coordinate is and , then the task cannot be done if is divisible by .
Now we look at the conditions for which is divisible by 3 by looking at the following three cases:
(4) When both and are divisible by .
(5) When one of them is not divisible by and the other one is.
(6) When neither nor is divisible by .
Case (4): Since and ,
then .
Thus, for , this case is a valid one.
Case (5): Since either or and the other ,
then .
Thus, for , this case is NOT a valid one.
Case (6): Since and ,
then .
Thus, for , this case is NOT a valid one.
Having proved that Case (4) is the only valid case for we are going to see what happens when we start with a square that is
if , then .
Therefore starting at any no matter how many moves the piece will always land at another square . Since we start at , then the piece will always land at squares that are .
Since final square is , then the task cannot be done if is divisible by .
In summary the task cannot be done if is divisible by or is divisible by .
Part (b):
When we have
Therefore the moves can have the following values:
Since we start at and we want to end at we can write the following valid moves:
Having shown that all moves comply with the possible values for and that all are inside the rectangular grid, then the task is possible when .
Part (c):
When we have
Therefore the moves can have the following values:
Let be a set of the squares of the rectangle with the condition for . That is, the reminder of when divided by 2 is the same as the reminder of when divided by 2 where is the smallest integer that is not less than . If that condition is true, then
Now let's look at a case (c1) where and , thus :
c1 subcase 1 : :
and
since , then for this subcase.
c1 subcase 2 : :
This move is not allowed because the only values of such that are . Adding or subtracting 9 from any of these numbers will either make or and will put the piece outside of the board.
In summary, for case (c1),
Now let's look at a case (c2) where and , thus :
c2 subcase 1 : :
and
since , then for this subcase.
c2 subcase 2 : :
Notice that the only values of such that are . Adding 9 to the values of 1,2, or 3 or subtracting 9 from the values of 10, 11, or 12, will result in . However, subtracting 9 from the values of 1,2, or 3, or adding 9 for the values of 10,11, or 12 will put the piece outside the board and not a valid move. Furthermore, for the values of 4 and 9, adding or subtracting 9 from them will result in the piece being outside of the board and not a valid move. Therefore for all valid moves in this subcase which are 9 added to 1, 2, or 3 or 9 subtracted from 10, 11, or 12, the result will be
since , then for this subcase.
In summary, for case (c2),
Therefore if then
Now we look at the starting point: :
Since and , then and all subsequent squares after that
Now let's look at the end point: :
Since and , then and
Since , then there are no valid moves starting from that will end in
Therefore the task cannot be done when .
~Tomas Diaz. orders@tomasdiaz.com
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1996 IMO (Problems) • Resources | ||
Preceded by First Problem |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |