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Difference between revisions of "KGS math club/solution 6 1"

(→‎Solution for rational weigths)
(→‎Solution for rational weigths)
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== Solution for rational weigths ==
 
== Solution for rational weigths ==
  
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'''First, reduce the problem a bit'''
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'''First, reduce the problem a bit.'''
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We number the balls from 0 to 2n.
 
We number the balls from 0 to 2n.
  
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'''Then solve.'''
 
'''Then solve.'''
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Let <math>W</math> be the total weight of the balls. If we remove each ball in turn from the collection, we know that <math>\forall i: 2 | (W-w_i)</math>, that is, each remaining ball-mass is divisible by two.
 
Let <math>W</math> be the total weight of the balls. If we remove each ball in turn from the collection, we know that <math>\forall i: 2 | (W-w_i)</math>, that is, each remaining ball-mass is divisible by two.
  

Revision as of 05:21, 18 July 2009

Problem statement

I'm going to write a solution for general $n$ (number of the balls in a group, set to 5 if you like). The problem statement is thus following:

You have a collection of $2n+1$ balls with the property that if you remove any one of the balls, the other $2n$ can be split into two groups of $n$ so that each weighs the same. If you assume that all of the balls have rational weight, there is a cute proof that they all must weigh the same. Can you find a proof? Can you find a way to extend the result to the general case where the balls have real weights?

Solution for rational weigths

First, reduce the problem a bit.

We number the balls from 0 to 2n.

If we subtract the same mass from each ball, the property outlined in the problem statement shouldn't change. Each group of $n$ balls will lose $n*w_{subtracted}$ of their weight, preserving equality or inequality.

Since the weights are assumed to be rational, we can transform them all to integers for example by multiplying them by the product of all their divisors. Then we can subtract the smallest integer from all of them, leaving the smallest zero. If all balls weigh the same, we are done with the proof; if not, at least some of the integers must be positive. Now, we divide the whole bunch of integers by their greatest common divisor, leaving us with gcd = 1. So, the whole transformation was essentially $w_i = (p_i/q_i - p_j/q_j) * k$, where $p_i/q_i$ are the original rational weights, $w_i$ is the resulting integer, $j$ is the ball weighing the least and $k$ a certain rational number, namely $k = Q / gcd_i((p_i/q_i - p_j/q_j) * Q)$, where $Q = \prod_i q_i$

TL;DR: unless all weights are the same, there surely must exist integers $w_i$ such that $w_j=0$ for some $j$ and $gcd_{w_i>0}(w_i) = 1$ and that the integers $w_i$, each associated with the ball number $i$, can be grouped according to the desired property exactly the same way the original balls can. That is, $\sum_{i\in G}w_i=\sum_{i\in H}w_i$ exactly when the balls in G weigh the same in total as the balls in H.

Then solve.

Let $W$ be the total weight of the balls. If we remove each ball in turn from the collection, we know that $\forall i: 2 | (W-w_i)$, that is, each remaining ball-mass is divisible by two.

Combining these, we get $\forall i,j: 2 | (w_i-w_j)$ by subtracting the case i from the case j.

This would mean that all $w_i$ are of form $a + 2b$ $(a\geq 0,b\geq 0)$, that is, their difference is divisible by 2 as shown. But since the smallest $w_j=0$, we have $a=0$. Thus we can drop the $a$ and write $w_i = 2b$ $(b>0)$. But that would mean that $gcd(w_i) = 2$, a contradiction!

Solution for real weights

Take a basis of $\mathbb{R}$ as $\mathbb{Q}$ vector space. If the weights of the balls have the property of the problem also each of the coordinates of the weights on this base share the same property (This is because taking coordinates is a linear map). Then (using the result for rationals weights) follows that each coordinate is constant and hence all the weights are constant

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