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
Note by
Daniel Liu
4 years, 6 months ago
No vote yet
1 vote
Easy Math Editor
*italics*or_italics_**bold**or__bold__paragraph 1
paragraph 2
[example link](https://brilliant.org)> This is a quote# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"2 \times 32^{34}a_{i-1}\frac{2}{3}\sqrt{2}\sum_{i=1}^3\sin \theta\boxed{123}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.
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.
Log in to reply
I have a one-line solution, given a 'well known fact'.
Hint: Take the obvious geometric interpretation.
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.
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.
Log in to reply
Did you mean "proof question"?
Log in to reply
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.
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.
Log in to reply
Okay. :D
Log in to reply