You should be able to solve this problem (standard for high school students in the Holy Roman Empire)

You should be able to solve this problem (standard for high school students in the Holy Roman Empire)

yes.

No, just think of three lines. Two of the three lines have the same color, and where they meet the other color is missing

easy

No.
Let there be 3 lines. There will be three intersection points (1 for each pair). Let two lines be red and one be blue (all three the same color obviously doesn't work). The intersection of the two red lines will not have a blue line through it.

>one blue line crossed by two non parallel red lines
>both colours present at any point where any two lines cross

not enough information to solve the problem

t. brainlet

What retard formulated this question?

hi OP

nah im the retard with the image

I honestly rage every time i see this bitch's face.
I want to punch her smug face so badly.

the amount of lines is finite or infinite?

Did you miss the first word of the fucking question?

english is not my first language

Do it you pussy.

neihter is mine fucking loo

We could add a blue line at that point, and a red line where the two blue lines intersect, and so on. You haven't proved it for all n>2 unless you prove that this process can never terminate regardless of how you arrange the lines.

Only if there are less than two two lines or all lines pass through the same.
The fact that you can do it for 1 and 2 lines is trivial.

If you have three lines they all either intersect at one point or three points, if they intersect at three points then there is one point where two lines are the same color.

If they all intersect at the same point any new line we draw will intersect all three lines, again ensuring that it will intersect with atleast one of the same colour.
QED

My bad, I thought the problem mentioned infinite lines. But still, it's not possible in the general case, so the answer is still no

let n be the total number of lines
>a single blue line of infinite length.
>met at the start by a red line, of irrelevant length, at 360/n degrees to the blue line
>followed by another red line at 2(360/n). the two red lines are placed an irrelevant (non zero) distance apart
>repeat n-2 times
suck it nerds

Yes.
Consider the case where no lines intersect. This is possible even if lines are not parallel.

The question was not "is it possible to arrange the lines in a way that..."

Yes.
Proof: for each point that only has 1 line color, just draw a different colored line that passes through it.

Lines are infinite length, those two first red lines will intersect at some point

If the set of all points that intersect is empty, then the conditions are automatically satisfied.

You didn't get the question. The question was not "does there exist a finite set of lines such that it is possible ...", but "for an arbitrary finite set of lines, is it possible ...", otherwise it would be trivially true if you just think of 0 lines of 1 line
Also, lines have no beginning and no end, so if two non-parallel lines are on the same plane, they have to meet somewhere

I don't need to show that it doesn't work for all n>2.
All I need to show is that there is some number, n, of lines arranged in some valid configuration that cannot be colored to meet the condition.
The question allows us to color the lines not add lines.
Learn to read the fucking question. jfc.

actually the question is more along the lines of "given a set of lines, is it possible..."

so OP is probably interested in an algorithm to decide the problem