Euclidian plane $\pi$ with all points either red, green or blue

Question

In the Euclidian plane $\pi$ all points are either red, green or blue. Prove that you can select three points $A$, $B$ and $C$ from plane $\pi$ so that the the triangle $ABC$ satisfies all the following conditions:

1. Points $A,B,C$ have the same color.
2. Circumscribed circle of triangle $ABC$ has diameter 1000.
3. One angle of the triangle $ABC$ is 1000 time bigger than one of the remaining two.

It's quite easy to construct a triangle satisfying (1) and (2) - just draw a heptagon inside a circle with diameter 1000. According to Dirichlet, such heptagon must have at least 3 vertexes of the same color so the triangle satisfying (1) and (2) clearly exists. But the trick is to satisy the third condition at the same time. Can we extrapolate this idea by usin regular poligon with 1000 or 2000 sides?

Answer 1

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:

$$W(3,1002) \leq 2^{2^{3^{2^{2^{1011}}}}}$$

Answer 2

Well, as you already say yourself, you can draw any heptagon inside that circle. What happens if six of the seven vertices come really close together?

Answer 3

Remark: The following works only if condition 3 is interpreted as

3. One angle of the triangle ABC is at least 1000 times bigger than one of the remaining two

instead of

3. 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.

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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 $$\angle CBA>\frac\pi2-\epsilon=1000\cdot\frac12\epsilon>1000\cdot\angle BAC$$