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
Sort by:
Top NewestThis 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 · 3 years, 4 months ago
Log in to reply
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 · 3 years, 4 months ago
Log in to reply
I have a one-line solution, given a 'well known fact'.
Hint: Take the obvious geometric interpretation. – Calvin Lin Staff · 3 years, 4 months ago
Log in to reply
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 · 3 years, 4 months ago
Log in to reply
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 · 3 years, 4 months ago
Log in to reply
Did you mean "proof question"? – Tan Li Xuan · 3 years, 4 months ago
Log in to reply
Yes, edited. Thanks. – Calvin Lin Staff · 3 years, 4 months ago
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 · 3 years, 4 months ago
Log in to reply
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 · 3 years, 4 months ago
Log in to reply
Okay. :D – Finn Hulse · 3 years, 4 months ago
Log in to reply