In a school playground n children are standing so that the d

In a school playground, n children are standing so that the distances between each pair of children are all distinct. Each child is holding a ball. When their teacher blows a whistle, each child simultaneously throws his or her ball to the closest child. If n is odd and n is greater than 3, prove that there is now at least one child without a ball.

I understand that since the distances are distinct there must be one child who is standing the furthest distance away and therefore must be left without a ball. But mathematically how can this be represented?

Solution

Let there be n students numbered from 1,2,3,....n standing in a line. It can be clearly understood that each student will either be in possession of either[0 or 1 or 2] balls. It is clear that if atleast one of the student has 2 balls, then atleast one student will have 0 balls or is left out.

Step:1 On the blow of the whistle, we know that 1 has no option but to throw it to 2. Similarly n has to throw to n-1.

If 1-2 distance is lower than that of 2-3 then 1 gets the ball or else is left out. Similar is the case with (n-1)-(n-2) and (n-1)-n.

If either of \'1\' or \'n\' doesn\'t get the ball then end of the problem.

If both \'1\' and \'n\' are in possession of balls i.e.(1 gave to 2; 2 gave to 1 & n-1 gave to n and n gave to n-1), then \'1\' has 1 ball and \'n\' has 1 ball.

Now \'3\' and \'n-2\' can either have 0 or 1 balls. If it is 0 then end of the problem, else \'3\' and \'n-2\' will have to get the balls from \'4\' and \'n-3\' respectively(since \'2\' gave to \'1\' and \'n-1\' gave to \'n\')

You can see a pattern emerging. Now \'4\' and \'n-3\' can either have 1 or 2 balls(but not 0 ( since \'3\' has given to \'4\' and \'n-3\' has given to \'n-2\'.)).

You can see pairs exchanging. \'1\' to\'2\' and \'2\' to \'1\' ; \'2\' to \'3\' and \'3\' to \'4\'. similarly \'n\' to \'n-1\' and \'n-1\' to \'n\' ; \'n-2\' to \'n-1\' and \'n-1\' to \'n-2\'.

Now the middle is left out. He can\'t have the ball.

The problem can end at any stage. Ultimate step is the middle man not having the ball.