# Strange Sequences

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
6 years, 4 months ago

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

Hint: Take the obvious geometric interpretation.

Staff - 6 years, 4 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.

- 6 years, 4 months ago

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.

Staff - 6 years, 4 months ago

Did you mean "proof question"?

- 6 years, 4 months ago

Yes, edited. Thanks.

Staff - 6 years, 4 months 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.

- 6 years, 4 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.

- 6 years, 4 months ago

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

- 6 years, 4 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.

- 6 years, 4 months ago

Okay. :D

- 6 years, 4 months ago