Two real sequences \(a_1,a_2,\ldots a_{2014}\) and \(b_1,b_2,\ldots b_{2014}\) both satisfy that for any two distinct positive integers \(i,j\), there exists a positive integer \(k\) distinct from \(i,j\) such that

\[(a_i-a_j)(b_i-b_k)=(a_i-a_k)(b_i-b_j)\]

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

\[(a_i-a_j)(b_i-b_k)=(a_i-a_k)(b_i-b_j)\]

is true for all \(i,j,k\) where \(1\le i,j,k\le 2014\).

*Source: me*

## Comments

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TopNewestThis 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 · 2 years, 10 months ago

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– Daniel Liu · 2 years, 10 months ago

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.Log in to reply

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

Hint:Take the obvious geometric interpretation. – Calvin Lin Staff · 2 years, 10 months agoLog in to reply

– Daniel Liu · 2 years, 10 months ago

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.Log in to reply

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 · 2 years, 10 months ago

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– Tan Li Xuan · 2 years, 10 months ago

Did you mean "proof question"?Log in to reply

– Calvin Lin Staff · 2 years, 10 months ago

Yes, edited. Thanks.Log in to reply

This is tough. I have the feeling PIE is a good place to start, but I'll work more on it later. – Finn Hulse · 2 years, 10 months ago

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– Daniel Liu · 2 years, 10 months ago

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.Log in to reply

– Finn Hulse · 2 years, 10 months ago

Okay. :DLog in to reply