Friday, August 15, 2008

A painted plane II

Your fears have been realized: I am presenting more painted plane problems!

Let's say I've painted each point on an infinite, continuous plane. Each point is either painted red or blue. Prove that there must exist three points of the same color which form the corners of an equilateral triangle.

I've painted the plane again, this time with three colors. Each point is either red, blue, or green. Prove that there must either exist three points of the same color, or of three different colors, such that the points form the corners of an equilateral triangle.

The solution has now been posted! see here

Oh, and yes, there will be even more painted plane puzzles in the future.


miller said...


It is proof by contradiction, just like before. Start with any two points of the same color, and assume there are no monochromatic triangles. Remember that the triangles can be any size and orientation.

Yoo said...

Yeah, I got that far (the first one is simple enough to do in my head), but I haven't been able to devote the time to do the other half yet. :(

Yoo said...

Finally got the time to figure out the second part. I did it the hard way, starting out with a equilateral triangle with one red corner and two green corners. With an exhaustive case analysis (fortunately, there weren't as many cases as I had feared) and plotting points extending the original triangle to form other triangles, I found that sooner or later you always end up with a triangle that violates the assumption.

If there's an easier way, let us know! (eventually ...)