Euclidian plane $\pi$ with all points either red, green or blue
This actually solves the original problem:
One angle of the triangle ABC is exactly 1000 times bigger than one of the remaining two
It took me a few days to figure this out. I really don't pretend to look smart by answering my own question. But it got a solid number of upvotes and I got a hint from a guy familiar with Van der Waerden's theorem.
For any given positive integers $r$ and $k$, there is some number $N$ such that if the integers $\{1, 2, ..., N\}$ are colored, each with one of $r$ different colors, then there are at least k integers in arithmetic progression all of the same color.
The least such number is called Van der Waerden's number $W(r,k)$.
Draw a circle of diameter $1000$. Divide this circle in at least $N=W(3, 1002)$ equal segments (exact number is unknown but it definitely exists) and denote the dividing points with $M_i$, $(i=1,2,...,N)$. Dividing points can have any of the $3$ specified colors.
According to Van der Waerden's theorem it is guaranted that we'll have an arithmetic progression of $1002$ integers that represent indexes of points of the same color:
$$A=M_k, B=M_{k+d}, M_{k+2d}, M_{k+3d}, ...,C=M_{k+1001d}$$
Because all points are equidistant, arc $\stackrel\frown{BC}$ is exactly 1000 times longer than arc $\stackrel\frown{AB}$ which means that in monochromatic triangle $ABC$:
$$\angle A=1000\angle C$$
According to Wikipedia, the best upper bound for the minimal number of points in this case is:
$$W(3,1002) \leq 2^{2^{3^{2^{2^{1011}}}}}$$
Remark: The following works only if condition 3 is interpreted as
- One angle of the triangle ABC is at least 1000 times bigger than one of the remaining two
instead of
- One angle of the triangle ABC is exactly 1000 times bigger than one of the remaining two
I'll have to go back to the drawing board for the exactly-variant.
The problem can be solved with many more colours, as long as the number of colours is less than the cardinality of the continuum.
Let $\epsilon=\frac1{1002}\pi$.
Pick any circle $\mathcal C$ of diameter $1000$ around some point $O$. On $\mathcal C$ pick an arc $\stackrel\frown {UV}$ of arc length $<\epsilon$ (thanks to Henning Mankolm for suggesting this improvement). As $\mathcal C$ has continuum-many points and we have less than continuum-many colours, there exists a colour, say blue, such that $\stackrel\frown {UV}$ has at least three blue interior points $A,B,C$ (labelled so that $U,A,B,C,V$ is clockwise order). Then $\angle BAC=\frac12\angle BOC<\frac12\epsilon$ and similarly $\angle ACB<\frac12\epsilon$, hence $$\angle CBA>\frac\pi2-\epsilon=1000\cdot\frac12\epsilon>1000\cdot\angle BAC$$