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Topic: Colored Equilateral Triangles (Read 602 times) |
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Yournamehere
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An old but good one:
Suppose every point in the Euclidean plane is colored either red or blue. Prove there exists an equilateral triangle with all three corners the same color.
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| « Last Edit: Oct 23rd, 2003, 7:57pm by Icarus » |
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Pietro K.C.
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Re: New: Colored Equilateral Triangles
« Reply #1 on: Sep 21st, 2002, 8:39am » |
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Suppose an equilateral triangle with vertices A, B and C; there are 4 possible colourings of the vertices with red or blue (which we shall refer to as 0 and 1), not counting rotational symmetry. They are: (0,0,0), (1,1,1), (0,1,1), (1,0,0). Therefore, the probability that a triangle will not have all vertices the same color is 1/2. Since there are an infinite number of equilateral triangles, the probability that NONE of them have same-color vertices is 0. So there is at least 1 (and in fact an infinite number of) equilateral triangle with vertices of the same color. 
Unless someone went out of their way especially to prevent that from happening, and went through the trouble of choosing a color for EACH point in the plane. But that would just be mean.
Maybe later I'll post a different solution I got. 
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| « Last Edit: Sep 21st, 2002, 8:42am by Pietro K.C. » |
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Jamie
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Re: New: Colored Equilateral Triangles
« Reply #2 on: Sep 22nd, 2002, 4:31am » |
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This was quite a fun little puzzle. I solved it by contradiction: assume you can have such a layout and see what it looks like. Start by looking at any equilateral triangle. By considering rotational symmetry and the fact the problem is colour-symmetric we can say without loss of generality that it must look like this:
Now, the two blue points mean that the opposite point must be red:
Applying the same logic now to the two red points gives us another two blue points:
Finally, the same step again applied to the two outermost pairs of blue points gives us another four red points, and an red equilateral triangle (actually 2):
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Jamie
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Re: New: Colored Equilateral Triangles
« Reply #3 on: Jul 14th, 2003, 6:20am » |
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I've been thinking about variants on this problem. Does it work for squares as well? It seems intuitive that it should, but I haven't been able to prove it yet. What happens if we're allowed 3 different colours? What about n?
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tohuvabohu
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Re: New: Colored Equilateral Triangles
« Reply #4 on: Jul 16th, 2003, 2:36pm » |
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I think I proved the square variation, but it's too long to bother typing out. In short:
If there is no square, I can demonstrate that certain shapes must exist, such as 4 points of one color forming a T, and 6 points forming a line of 4 with an extra point below the second and above the third (or vice versa). Then it got tough. From that last shape I had to eliminate a number of possible extensions.
Here are a couple that I'll present as a simple puzzle. Find the square in each and see that none of the four points could be changed without creating a different square (Any .'s simply represent a point whose color I had not yet determined. In the last two puzzles, you have to find the . that can neither be colored 0 or 1).
100110
011101
010100
000001
101100
100111
11100
00111
01101
00000
10110
001...
011001
.01100
10101.
100001
.101..
........
..110...
..1011..
.000010.
.101....
..1.001.
.0.11...
........
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Jamie
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Re: New: Colored Equilateral Triangles
« Reply #5 on: Jul 16th, 2003, 3:14pm » |
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on Jul 16th, 2003, 2:36pm, tohuvabohu wrote:| I think I proved the square variation, but it's too long to bother typing out. |
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Ah. Proof by Fermat. My favourite
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James Fingas
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Re: New: Colored Equilateral Triangles
« Reply #6 on: Jul 17th, 2003, 12:15pm » |
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tohuvabohu,
I tried a different method, which didn't pan out. I'm not convinced either way. But here is a puzzle in response to your puzzles. Try and find a square in this one:
1111100010
0010101110
0100100100
0111001001
1100010011
1011010101
1110110001
1001101111
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tohuvabohu
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Re: New: Colored Equilateral Triangles
« Reply #7 on: Jul 17th, 2003, 1:13pm » |
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Your puzzle has many squares. I found 6 in just a few minutes.
11111D0010
001D101110
010010D100
01A1D01001
AB00E10F11
B0BA01E1F1
1ACEC10F01
100C1E1111
A shares one point with B and one with C, and E shares one point with F. I didn't bother looking for any larger ones or tilted at any other angles.
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James Fingas
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Re: New: Colored Equilateral Triangles
« Reply #8 on: Jul 17th, 2003, 1:46pm » |
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ah ... now I understand why we are getting slightly different answers ... I was only considering squares aligned with the grid. I guess the triangle question didn't get complicated enough that it would matter.
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