Two real sequences a1,a2,…a2014 and b1,b2,…b2014 both satisfy that for any two distinct positive integers i,j, there exists a positive integer k distinct from i,j such that
(ai−aj)(bi−bk)=(ai−ak)(bi−bj)
where 1≤i,j,k≤2014. Prove that
(ai−aj)(bi−bk)=(ai−ak)(bi−bj)
is true for all i,j,k where 1≤i,j,k≤2014.
Source: me
#Algebra
#Contradiction
#Sequences
#Proofs
#CleverInterpretation
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Comments
I have a one-line solution, given a 'well known fact'.
Hint: Take the obvious geometric interpretation.
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I think the solution you found is the intended one, although I wouldn't really call it a one-liner. Also, assume you need to prove that well known fact since that's basically most of the problem.
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Haha, my definition of one-liner is basically that you just need that 1 idea, and things fall naturally after that.
This is a nice proof question. I really like the fact that is referenced, in part because it was one of the first applications of external combinatorics that I came across in Artur Engel, which showed me how beautiful such arguments could be.
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This is basically Sylvester–Gallai theorem: given a finite number of points in the plane, either all the points are collinear or there is a line containing exactly 2 points. I think best proof is using extreme principle.
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Yea, that's basically the solution. I also know of a proof by contradiction, but I'll try to find a proof using the extremal principle.
This is tough. I have the feeling PIE is a good place to start, but I'll work more on it later.
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Good luck; this is a really tough problem to crack. I tried to turn it into a problem but I failed so this is the best I can do.
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Okay. :D