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# n-colored straight lines on painted plane

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(1a) Paint the plane using 3 colors in such way that no 3-colored straight line exists.
(1b) Prove that if plane is painted with 4 colors, then a 3-colored straight line exists.
(2a) Paint the plane using as many colors as you can in such way that no 4-colored straight line exists.
(2b) Let m be the number of colors you used in (2a). Prove that if plane is painted with (m+1) colors, then a 4-colored straight line exists.
(3) What about 5-colored straight lines?
(4) What about n-colored straight lines?

Note: I don't have solutions to all of the problems above. Partial solutions are welcome.

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Posted (edited) · Report post

I think this works (basically paint the whole plane one color...then paint 2 intersecting lines another color...and then paint the intersection point of those two lines a third color):

Edited by Pickett
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Posted (edited) · Report post

I think this works (basically paint the whole plane one color...then paint 2 intersecting lines another color...and then paint the intersection point of those two lines a third color):

(1a) solved

You could use as many blue lines as you want (if all of them are intersecting in black point and if some red points remain). One blue line seems most elegant to me. Just wandering, why you chose to use two blue lines.

Edited by witzar
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Posted (edited) · Report post

I think this works (basically paint the whole plane one color...then paint 2 intersecting lines another color...and then paint the intersection point of those two lines a third color):

(1a) solved

You could use as many blue lines as you want (if all of them are intersecting in black point and if some red points remain). One blue line seems most elegant to me. Just wandering, why you chose to use two blue lines.

good question...I guess I just immediately went to quadrants when talking about "the plane"...

Edited by Pickett
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Posted (edited) · Report post

(1a) solved

You could use as many blue lines as you want (if all of them are intersecting in black point and if some red points remain). One blue line seems most elegant to me. Just wandering, why you chose to use two blue lines.

There's something about the dimensionality that suggests why it's impossible to paint with four colors. In the simple (elegant) solution, color 1 is painted 0-dimensionally, color 2 is painted with 1 dimension, color 3 has 2 dimensions. If you paint with a 4th color, it would require another dimension to keep from violating the 3-colored line rule. But since you're limited to a 2-D plane instead of 3-D space, you can't do it.

Edited by HoustonHokie
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Posted · Report post

You could add points or another intersecting line and preclude being able to draw a 4-colored line. Here are some examples:

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Posted · Report post

(1a) solved

You could use as many blue lines as you want (if all of them are intersecting in black point and if some red points remain). One blue line seems most elegant to me. Just wandering, why you chose to use two blue lines.

There's something about the dimensionality that suggests why it's impossible to paint with four colors. In the simple (elegant) solution, color 1 is painted 0-dimensionally, color 2 is painted with 1 dimension, color 3 has 2 dimensions. If you paint with a 4th color, it would require another dimension to keep from violating the 3-colored line rule. But since you're limited to a 2-D plane instead of 3-D space, you can't do it.

While that's a good observation, it's not strictly a requirement.

the whole plane is painted with one color except for one straight line that is painted using 2 other colors. The colors used to paint the line are equal in the sense that neither needs to be 0-dimensional. For example, half the line is one color and the other half is another.

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Posted · Report post

A more convincing proof of 1B.

If there are 4 colors then take 4 points such that you have one of every color, the first 3 form a triangle and so the fourth could be inside or outside the triangle (if it was on one of the edges then that already makes a line with 3 colors)

The two cases look like this:

In the left case take the center point that is on the intersection, if it was colored 1 or 4 then the line between 2 and 3 would have three different colors, if it was colored 2 or 3 then the line between 1 and 4 would have three colors.

In the right case it's almost exactly like above just instead take the point that is the intersection between the 3-4 line and the continuation of the 1-2 line.

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Posted (edited) · Report post

I think you can do infinite colors, take the unit circle and paint it's circumference with as many colors as you like, the rest of the plane paint blue, now any straight line could intersect with the circle in exactly 0 1 or 2 points, so the maximum number of different colors is 3.

Edited by Anza Power
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Posted · Report post

A more convincing proof of 1B.

If there are 4 colors then take 4 points such that you have one of every color, the first 3 form a triangle and so the fourth could be inside or outside the triangle (if it was on one of the edges then that already makes a line with 3 colors)

The two cases look like this:

In the left case take the center point that is on the intersection, if it was colored 1 or 4 then the line between 2 and 3 would have three different colors, if it was colored 2 or 3 then the line between 1 and 4 would have three colors.

In the right case it's almost exactly like above just instead take the point that is the intersection between the 3-4 line and the continuation of the 1-2 line.

(1b) solved. Well done.

You can treat both cases as one. Provided we have 4 points of different color: either some 3 of them are collinear (thus forming 3-colored line) or they are vertices of a (not necessary convex) quadrilateral. No matter how intersection of quadrilateral's diagonals is colored, one of the diagonals is a 3-colored line.

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Posted · Report post

I think you can do infinite colors, take the unit circle and paint it's circumference with as many colors as you like, the rest of the plane paint blue, now any straight line could intersect with the circle in exactly 0 1 or 2 points, so the maximum number of different colors is 3.

Well done!

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