Posted 25 Aug 2014 · Report post There are 10 blue balls on a circle, all have different mass. I put green balls between every 2 neighboring blue balls, which weighs the difference of those 2 blue balls. Prove that I can put those green balls in two groups that weigh same. 0 Share this post Link to post Share on other sites

0 Posted 25 Aug 2014 · Report post If Bi > Bi+1, put the green ball in pile 1 If Bi+1 > Bi, put the green ball in pile 2 At the end, both piles will weigh the same Effectively the weight of green ball = absolute (Bi - Bi+1) By sorting the piles as above, we take care of the absolute value of this subtraction. Since the weights are in a full cycle, the net (without absolute) values of green balls will be zero. 0 Share this post Link to post Share on other sites

0 Posted 26 Aug 2014 · Report post If Bi > Bi+1, put the green ball in pile 1 If Bi+1 > Bi, put the green ball in pile 2 At the end, both piles will weigh the same Effectively the weight of green ball = absolute (Bi - Bi+1) By sorting the piles as above, we take care of the absolute value of this subtraction. Since the weights are in a full cycle, the net (without absolute) values of green balls will be zero. But weight difference is always positive. 0 Share this post Link to post Share on other sites

0 Posted 26 Aug 2014 · Report post If Bi > Bi+1, put the green ball in pile 1 If Bi+1 > Bi, put the green ball in pile 2 At the end, both piles will weigh the same Effectively the weight of green ball = absolute (Bi - Bi+1) By sorting the piles as above, we take care of the absolute value of this subtraction. Since the weights are in a full cycle, the net (without absolute) values of green balls will be zero. But weight difference is always positive. Yes, that is why we sort them into two piles depending on whether the next ball is heavier or lighter. And the two piles would weigh the same 0 Share this post Link to post Share on other sites

0 Posted 27 Aug 2014 · Report post Call the ball with the lowest mass L. Call the ball with the highest mass H. D=mass(H)-mass(L) D=sum of masses of greens going from L to H clockwise D=sum of masses of greens going from L to H counterclockwise 0 Share this post Link to post Share on other sites

0 Posted 27 Aug 2014 · Report post If Bi > Bi+1, put the green ball in pile 1 If Bi+1 > Bi, put the green ball in pile 2 At the end, both piles will weigh the same Effectively the weight of green ball = absolute (Bi - Bi+1) By sorting the piles as above, we take care of the absolute value of this subtraction. Since the weights are in a full cycle, the net (without absolute) values of green balls will be zero. I don't understand this solution. What is Bi? 0 Share this post Link to post Share on other sites

0 Posted 27 Aug 2014 · Report post If Bi > Bi+1, put the green ball in pile 1 If Bi+1 > Bi, put the green ball in pile 2 At the end, both piles will weigh the same Effectively the weight of green ball = absolute (Bi - Bi+1) By sorting the piles as above, we take care of the absolute value of this subtraction. Since the weights are in a full cycle, the net (without absolute) values of green balls will be zero. I don't understand this solution. What is Bi? Bi is a blue ball at position i (i is from 1 to 10). For the tenth ball bi is blue ball 10 and Bi+1 is blue ball 1 0 Share this post Link to post Share on other sites

0 Posted 28 Aug 2014 · Report post If Bi > Bi+1, put the green ball in pile 1 If Bi+1 > Bi, put the green ball in pile 2 At the end, both piles will weigh the same Effectively the weight of green ball = absolute (Bi - Bi+1) By sorting the piles as above, we take care of the absolute value of this subtraction. Since the weights are in a full cycle, the net (without absolute) values of green balls will be zero. I don't understand this solution. What is Bi? Bi is a blue ball at position i (i is from 1 to 10). For the tenth ball bi is blue ball 10 and Bi+1 is blue ball 1 Good catch. I had done it very clumsy way. 0 Share this post Link to post Share on other sites

0 Posted 29 Aug 2014 · Report post Call the ten original balls a, b, c, ..., h, i, and j. The difference balls have individual weights of a-b, b-c, ...,h-i, i-j, j-a and a total weight of zero. The difference balls thus have individual weights of zero. (The original balls have the same weight.) Any two groups of zero-weight balls weigh the same. 0 Share this post Link to post Share on other sites

0 Posted 29 Aug 2014 · Report post Call the ten original balls a, b, c, ..., h, i, and j. The difference balls have individual weights of a-b, b-c, ...,h-i, i-j, j-a and a total weight of zero. The difference balls thus have individual weights of zero. (The original balls have the same weight.) Any two groups of zero-weight balls weigh the same. difference balls all have weight more than zero. If a>b<c, then first 2 difference balls weigh a-b and c-b etc... 0 Share this post Link to post Share on other sites

0 Posted 29 Aug 2014 · Report post I imagined 10 green balls. I misread the problem. 0 Share this post Link to post Share on other sites

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There are 10 blue balls on a circle, all have different mass. I put green balls between every 2 neighboring blue balls, which weighs the difference of those 2 blue balls. Prove that I can put those green balls in two groups that weigh same.

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