Strange Sequences

Two real sequences a1,a2,a2014a_1,a_2,\ldots a_{2014} and b1,b2,b2014b_1,b_2,\ldots b_{2014} both satisfy that for any two distinct positive integers i,ji,j, there exists a positive integer kk distinct from i,ji,j such that

(aiaj)(bibk)=(aiak)(bibj)(a_i-a_j)(b_i-b_k)=(a_i-a_k)(b_i-b_j)

where 1i,j,k20141\le i,j,k \le 2014. Prove that

(aiaj)(bibk)=(aiak)(bibj)(a_i-a_j)(b_i-b_k)=(a_i-a_k)(b_i-b_j)

is true for all i,j,ki,j,k where 1i,j,k20141\le i,j,k\le 2014.

Source: me

#Algebra #Contradiction #Sequences #Proofs #CleverInterpretation

Note by Daniel Liu
7 years ago

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Comments

I have a one-line solution, given a 'well known fact'.

Hint: Take the obvious geometric interpretation.

Calvin Lin Staff - 7 years ago

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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.

Daniel Liu - 7 years ago

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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.

Calvin Lin Staff - 7 years ago

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@Calvin Lin Did you mean "proof question"?

Tan Li Xuan - 7 years ago

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@Tan Li Xuan Yes, edited. Thanks.

Calvin Lin Staff - 7 years ago

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.

Roger Lu - 7 years ago

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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.

Daniel Liu - 7 years ago

This is tough. I have the feeling PIE is a good place to start, but I'll work more on it later.

Finn Hulse - 7 years ago

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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.

Daniel Liu - 7 years ago

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Okay. :D

Finn Hulse - 7 years ago
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