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.
Let ,
Then, for any given , we have for all
part (a):
In order to find out the conditions for which is divisible by two 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 three 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 .
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 where and , thus :
subcase 1 : :
and
since , then for this subcase.
subcase 2 : :
and
since , then for this subcase.
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 |