Background:
This question arose purely recreationally and doesn't really fit into any context that I know of.
Let $A \sqcup B = \mathbb{R}$ be a partition of the real numbers into two sets
I have proved that for any positive $k$, either $A$ or $B$ contain $3$ elements $x < y < z$ which satisfy the relation:
$$ z - y = k(y - x) $$
The gist of the proof was a proof by contradiction by brute forcing several forced ratios and then showing the same element had to belong to both $A$ and $B$. My feeling was the proof came out too locally and that there was room for improvement and so I have a more general question:
Question:
Define a partition of the real numbers: $$\mathbb{R} = \bigsqcup_{j=1}^n A_j $$
Let $k_2, ..., k_{m-1}$ be a list of positive real numbers (possibly an infinite sequence). What conditions on $n$ and $m$ can I put so that I can guarantee there will be $m$ points $ x_1 < \dots < x_m$ in a single $A_j$ such that $$x_{i+1} - x_{i} = k_i (x_i - x_{i-1})$$ for all $1 < i < m$.
I am stuck for techniques because the original proof was quite crude and it wouldn't generalise. I also don't have any feel for what the answer will be because the sets you need to look at in this problem aren't at all nice or intuitive.
Thanks!
Update: It appears this has been solved for general $n$ in the case $m$ is finite by Rado in 1933 (as mentioned in the link Gerry posted: http://journals.cambridge.org/action/displayAbstract?fromPage=online&aid=190037) in the affirmative for a general sequence $x_1, \dots, x_m$ although not particularly for the case $x_1 < \dots < x_m$ although this may be a simple extension.
I believe the case for infinite matrices in the general partition regularity problem has not been solved and neither has this particular subproblem when we have an infinite ratio sequence.