# No Equilateral Triangles, Please

Ten dots are arranged in a triangular grid as in the applet below. Each dot may be one of two colors: orange or red. If the centers of three dots of the same color form an equilateral triangle, the triangle is shown in blue. By clicking on the dots you can switch between the two possible colors. The task is to find a coloring of the dots without monochromatic triangles, or prove that this is impossible.

What if applet does not run? |

|Activities| |Contact| |Front page| |Contents| |Geometry| |Store|

Copyright © 1996-2017 Alexander Bogomolny

The answer to the problem (call it Problem A) is that there is always going to be a monochromatic triangle. The configuration pops up in a solution to a problem (call it Problem B) from *Which Way Did the Bicycle Go?*: is it possible to divide the plane into two sets so that neither set contains the vertices of an equilateral triangle? The two colors indicate the two sets the plane is split into. A negative solution to Problem A implies a negative solution to Problem B.

I'll give two (negative) solutions to Problem A. The first comes from the book. The second, which seems to me a little simpler, is home made.

What if applet does not run? |

### Solution 1

Suppose that there is a configuration with no equilateral monochromatic triangle. Without loss of generality assume dot 5 is red. Since points 3, 4, 9 form an equilateral triangle, at least one of them must be red. Let it be 4. Then clearly 2 and 8 must be orange, since either would form an equilateral triangle with 4 and 5. Since 2, 8, and 6 form an equilateral triangle, 6 must be red, so 3 must be orange. But now if 1 is red then dots 1, 4, 6 for a monochromatic (red) triangles, and if it is orange, then triangle 1, 2, 3 is monochromatic (orange).

### Solution 2

Dots 2, 3, 6, 9, 8, 4 form a regular hexagon with center at dot 5. If all six vertices are of the same color we are done. Assume both colors are present. There are three pairs of diametrically opposite dots: 2-9, 3-8, and 4-6. There are two possibilities. Either all three pairs are monochromatic or at least one of them combines dots of different color. In the first case, we may assume without loss of generality we may that 2-9 and 3-8 are red, while 4-6 is orange. Then either triangle 1, 4, 6 is red or triangle 1, 2, 3 is orange. A contradiction. In the latter case, assume 2 is red while 9 is orange. Since triangle 3, 4, 9 is equilateral, both 3 and 4 can't be orange. Assume 3 is red. Similarly, both 6 and 8 can't be red. Thus assume 6 is orange. Then either triangle 2, 3, 5 is red, or triangle 5, 6, 9 is orange. Again, a contradiction.

Note that in both solutions, dots 7 and 10 do not make an appearance. Does it mean that they are not needed for the proofs?

### References

- J. Konhauser, D. Velleman, S. Wagon,
*Which Way Did the Bicycle Go?*, MAA, 1996, #20

- Ramsey's Theorem
- Party Acquaintances
- Ramsey Number R(3, 3, 3)
- Ramsey Number R(4, 3)
- Ramsey Number R(5, 3)
- Ramsey Number R(4, 4)
- Geometric Application of Ramsey's Theory
- Coloring Points in the Plane and Elsewhere
- Two Colors - Two Points
- Three Colors - Two Points
- Two Colors - All Distances
- Two Colors on a Straight Line
- Two Colors - Three Points
- Three Colors - Bichromatic Lines
- Chromatic Number of the Plane
- Monochromatic Rectangle in a 2-coloring of the Plane
- Two Colors - Three Points on Circle
- Coloring a Graph
- No Equilateral Triangles, Please

|Activities| |Contact| |Front page| |Contents| |Geometry| |Store|

Copyright © 1996-2017 Alexander Bogomolny

62324353 |