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# Black and white plane

## Question

Each point of the plane is painted either red or blue.

Prove, that there exists equilateral triangle with all vertices of the same color.

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Find two points on the plane that are the same color (say blue), call them 1 and 2. Find a the points 3 and 4 that are equidistant from 1 and 2 as they are from each other, i.e. form equilateral triangles with 1 and 2, these must be the other color (red). Find point 5 which forms an equilateral triangle with points 3 and 4., this point must be the original color (blue). Find point 6 which forms an equilateral triangle with points 2 and 5, this must be the second color (red). Now find point 7 which form equilateral triangles with points 3 and 6 AND points 1 and 5, hence an equilateral triangle with vertices of the same color must exist.

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the white grids are points..try to get rid of equilateral triangle from forming by picking red or blue

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the white grids are points..try to get rid of equilateral triangle from forming by picking red or blue

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the white grids are points..try to get rid of equilateral triangle from forming by picking red or blue

i see white equilateral here ,you should atleast pick one blue..but you know were proving that there are 9 points in a plane in such a way that if you try to

paint them either of the two colors you shall have atleast 1 equilateral triangle with same vertices color. I dont know how many if there are more points

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the white grids are points..try to get rid of equilateral triangle from forming by picking red or blue

i see white equilateral here ,you should atleast pick one blue..but you know were proving that there are 9 points in a plane in such a way that if you try to

paint them either of the two colors you shall have atleast 1 equilateral triangle with same vertices color. I dont know how many if there are more points

I meant for the white points to be blue. I don't see an equilateral triangle if colored as such.

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I meant for the white points to be blue. I don't see an equilateral triangle if colored as such.

I see two. They are larger than the smallest equilateral triangle that could be formed from the white points in TSLF's post.
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i guess Bmad is correct.. i err because 1/2 tangent is not 60 deg.. but i meant them to be 60 deg as possible in my gridding..

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..and we can draw an equilateral triangle for both small and big ones if we don't consider just the center points of 3 squares.

Edited by TimeSpaceLightForce
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..if you pick the set of points, and the painter picks the color...

5

4 2 6

1 3 7

for example:

o o o

o o o

o o o

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..if you pick the set of points, and the painter picks the color...

5

4 2 6

1 3 7

for example:

o o o

o o o

o o o

No matter how you paint this set of 9 points, one-colored equilateral triangle will appear -

as Yoruichi-san proved using only 7 of these points.

Therefore I consider your former answer a sort of a partial proof.

Still I would give you the "best answer", if no real proof (as provided by Yoruichi-san) would appear.

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..if you pick the set of points, and the painter picks the color...

5

4 2 6

1 3 7

for example:

o o o

o o o

o o o

No matter how you paint this set of 9 points, one-colored equilateral triangle will appear -

as Yoruichi-san proved using only 7 of these points.

Therefore I consider your former answer a sort of a partial proof.

Still I would give you the "best answer", if no real proof (as provided by Yoruichi-san) would appear.

I think of this puzzle as game type and its interesting. Reminds me of "GO".

Y-san is refutable, I agree with her approach for such triangle. .. but you see we look for possibilities

to make a puzzle do its best. This is why i posted the pointer & painter problem that can also

shows the proof . But it is better suited for your puzzle if the points can switch to black or white like Go pieces.

On the other hand i guess I'll just wait for the minimum set of points to be posted.

The best answer mark should never an issue for the puzzlers here. I like giving them ..

Thanks to you and kudos!

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Lol...TSLF, you're not very good at paint by numbers, are you? (I'm just kidding )

I don't think you're following my proof...it's ordered in such a way for a reason...

The fact that you can paint the arrangement of 7 points a way in which there are no equilateral triangles does not refute my argument, since my argument was not that there is no way to paint such an arrangement as such. My argument was that if you start with the assumption/null hypothesis "there exists no equilateral triangles with points all the same color" then you would eventually reach a contradiction, which I was able to show in 7 points.

I.e. I did not start with a set of 7 points and try to paint them, I started with 2 points that were the same color (which it's pretty trivial to prove must exist), and found the corresponding points that I could make logical conclusions about based on the null hypothesis, and showed the null hypothesis was eventually cornered into a contradiction, hence must be false.

In your diagram, points 1 and 2 are different colors, which kind of misses the point (no pun intended ). Start with any two points that are the same color, say your 3 and 6, or your 1 and 3 (everything would just be rotated), or your 2 and 7 (everything would be rotated and scaled up), or any two points of the same color, and I'm confident you will see how each step of the argument is proven true by the previous steps and geometry .

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reputable means "honorable" different to "rufute your argument" , i did not disagree. did I?

Yet If i follow your method of find 2 points of same color and counter color the 3rd point of equilateral vertex..i was lost!

11 9

8 5

4 2

1 3 7 10

6

12

it can go on.. guess im really not that good on paint a point, but as i explain to witzar..my solution is for points that "can be" red or blue so however the points change colors (like on -off bulbs) there always exist an equilateral triangle..

Edited by TimeSpaceLightForce
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Lol...sorry, read the 'p' as an 'f'...damb phone screens

...you're just not trying hard enough ;P

Seriously though, I know you're smart enough to see the point that 2 and 5 form an equilateral triangle with .

The point is (again, no pun intended) that the minimum number of points needed to prove the null hypothesis false is 7, but you can find an infinite number of points that don't if you select the wrong points...it's like a game you're playing against the null hypothesis, you can win in 7 moves, but you can stall the game for as long as you like...

Have you ever played gomoku? (We call it wu zhi lian, or "connect five") I love the game, I'm very good at it, if I do say so myself *whistling*, and the winning strategy is usually a forcing strategy that makes your opponent put pieces at certain spots to prevent you from connecting five (or an open-ended four). So I guess I like proofs that are the same way *shrugs*.

You're free to have your opinion about which kind of proof is better, I don't really care about getting 'best answer' or w/e so if witzar wants to give it to you, he can be my guest, but in the end, it's his call what he likes better .

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Oh, and, btw, TSLF, when I said I was "confident that you'd see", I meant I was confident in your intelligence to be able to find 7-move win, lol...

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