Can you paint a plane using 2 colours, so that any 2 points which are separated by 10 cm will have different colours?
[Math] Painting plane with 2 colors (2 points separated by 10 cm different colors )
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Related Solutions
[Math] Can three distinct points in the plane always be separated into bounded regions by four lines
Okay, I think this works. By scaling and rotation, we can assume that two of the points are $(0,0)$ and $(0,1)$. Then the other point is $(x,y)$. Now the problem can be solved if the third point is $(1,0)$, with something like
Now if $x\ne 0$, the linear transformation $A=\pmatrix{x&0\\y&1}$ maps the point $(0,1)$ to $(x,y)$ and fixes the other two points, and also maps each green line to some new line, so $A$ applied to each line gives you four lines which enclose the points $(0,1), (0,0)$ and $(x,y)$.
If the third point is collinear with the other two points then it is easy to come up with the four lines that work.
Just make a cone that contains the two top points and another which contains the two bottom points. Then only the middle point will be in the intersection of the cones.
This is the subject well-known open problem called the Hadwiger-Nelson Problem. The problem asks for the exact minimum number of colors that we can color the plane with, so that no two points of distance one are the same color.
You have observed that we cannot do it with only 2 colors, and asked, can we do it with 3 colors? The answer to that is no as well: on the Wikipedia page, we see that this is proven with the Mosser Spindle. It is a collection of 7 points in the plane, with 11 edges of length 1 between them, such that these points cannot be colored with only 3 colors.
Therefore it is known that 2 or 3 colors are impossible. It is not known, however, if it can be done with 4 colors. It is known that it can be done in 7 colors, so the minimum number of colors is either 4, 5, 6, or 7 -- but we don't know which!
In fact, it is expected by some that the exact minimum depends on the infamous Axiom of Choice.
UPDATE: Remarkable news earlier this year in April 2018: amateur mathematician Aubrey de Grey showed that 4 colors is impossible, thus narrowing the minimum down to 5, 6, or 7. To do this, he constructed a set of about 1500 points in the plane and proved that it is impossible to color them with 4 colors. This is the first progress on the problem in 60 years. You can read about it in de Grey's paper (which is quite readable). You can also read about it in this blog post and this Quanta magazine article.
Best Answer
No you can't, because there are three points that are vertexes of an equilateral triangle with a side of $10\,\rm cm$ and you can't have all 3 vertex colored differently.