1. ## Can we colour...

Can we colour every point of a circle, which radius equals $1$, with $2$ different colours so that every two points which are distant from each other in the lenght of $1$, are different colours? And why?

2. ## Re: Can we colour...

for any point on the circle there will be 2 points equidistant from it

think of an equilateral triangle inscribed in the circle. All three vertices are equidistant from one another.

So two colors won't do it. You'd need three.

3. ## Re: Can we colour...

I think you can do it. If you consider points on the circle that are at 0, 60, 120, 180, 24, and 300 degrees each is distance 1 away from its two neighbors. Therefore you can color point at 0, 120 and 24 degrees red and the points at 60, 180, and 300 degrees blue. Now rotate and infinitesimal amount clockwise and repeat. What you end up with is the circle with alternating 60-degree wide arcs of red and blue.

4. ## Re: Can we colour...

Originally Posted by ChipB
I think you can do it. If you consider points on the circle that are at 0, 60, 120, 180, 24, and 300 degrees each is distance 1 away from its two neighbors. Therefore you can color point at 0, 120 and 24 degrees red and the points at 60, 180, and 300 degrees blue. Now rotate and infinitesimal amount clockwise and repeat. What you end up with is the circle with alternating 60-degree wide arcs of red and blue.
do you mean 240?

Yep, you're correct.

5. ## Re: Can we colour...

Thanks for all your replies. I think that this is possible, because we can put circumcircle which radius equals 1 on regular hexagon which edge equals also 1.
We can colour every vertex of that hexagon with two colours so that every neighbouring vertexes are different colours.
Now we can turn that hexagon by angle from closed interval $(0,\frac{ \pi }{3})$. But how to proof that?

6. ## Re: Can we colour...

Originally Posted by exe43
Can we colour every point of a circle, which radius equals $1$, with $2$ different colours so that every two points which are distant from each other in the lenght of $1$, are different colours? And why?
Originally Posted by romsek
for any point on the circle there will be 2 points equidistant from it
think of an equilateral triangle inscribed in the circle. All three vertices are equidistant from one another.
So two colors won't do it. You'd need three.
First I will say that this is a well known problem and romsek's approach is correct. But the question must be stated very carefully and this one is not. Here are two webpages that help seeing what I mean.
Equilateral Triangles and Circular Segements

Here is the problem with the problem statement. Suppose the circle circumscribes an equilateral triangle of side $a=1$(see first webpage) then the radius of the circle is $\mathscr{R}=\dfrac{2h}{3}=\dfrac{2}{3} \dfrac{\sqrt{3}}{2}a=\dfrac{a}{\sqrt3}$
Note that the circumradius is not one unit if side of the triangle is one unit.

But there is a careful wording that leads to romsek's argument.

7. ## Re: Can we colour...

Originally Posted by exe43
Thanks for all your replies. I think that this is possible, because we can put circumcircle which radius equals 1 on regular hexagon which edge equals also 1.
We can colour every vertex of that hexagon with two colours so that every neighbouring vertexes are different colours.
Now we can turn that hexagon by angle from closed interval $(0,\frac{ \pi }{3})$. But how to proof that?
See my reply above #6. the circle will not have radius 1.

8. ## Re: Can we colour...

Remember the radius of the circle is 1 and the distance between points that we are interested is also 1. For a circle of radius 1 the inscribed equilateral triangle does not have side lengths of 1.

What I tried to explain was an approach of dividing the circle into six sectors, like the attached figure, where any two points that are exactly a distance of 1 away from each other are different colors.

9. ## Re: Can we colour...

Hmmmm. I think I don't understand. In the exercise is written "can we..." so if we present any example which solves that conditions, exercise will be solved.
What is wrong with my example?
Thanks.

10. ## Re: Can we colour...

But centre of a circle doesn't belong to the circle. Circle is
It is the set of all points in a plane that are at a given distance from a given point, the centre;
So we don't take centre of a circle into account. We only look on points from which we can make a hexagon like I presented in a pic above.

11. ## Re: Can we colour...

Originally Posted by ChipB
Remember the radius of the circle is 1 and the distance between points that we are interested is also 1. For a circle of radius 1 the inscribed equilateral triangle does not have side lengths of 1.

What I tried to explain was an approach of dividing the circle into six sectors, like the attached figure, where any two points that are exactly a distance of 1 away from each other are different colors.

There does not exist a unit circle that contains three points that are pair-wise one unit from one another.

Any set of three points that are pair-wise one unit from one another form determines an equilateral triangle of sides of length one. It can be circumscribed by a circle with a radius of $\dfrac{\sqrt3}{3}$.

12. ## Re: Can we colour...

Originally Posted by Plato
There does not exist a unit circle that contains three points that are pair-wise one unit from one another.
That's not what the OP asked about. What he said was: "every two points which are distant from each other in the lenght [sic] of 1." He doesn't ask about three points all equidistant from each other.

13. ## Re: Can we colour...

Yeah, and they have to be on a circle.