tag:blogger.com,1999:blog-9124539381685751273.post3880502187908657409..comments2023-06-19T04:35:06.263-07:00Comments on Skeptic's Play: A painted plane IImillerhttp://www.blogger.com/profile/05990852054891771988noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-9124539381685751273.post-59148469281646780812008-08-23T11:23:00.000-07:002008-08-23T11:23:00.000-07:00Finally got the time to figure out the second part...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.<BR/><BR/>If there's an easier way, let us know! (eventually ...)Yoohttps://www.blogger.com/profile/08964658430856685576noreply@blogger.comtag:blogger.com,1999:blog-9124539381685751273.post-37377029138738862582008-08-19T15:26:00.000-07:002008-08-19T15:26:00.000-07:00Yeah, I got that far (the first one is simple enou...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. :(Yoohttps://www.blogger.com/profile/08964658430856685576noreply@blogger.comtag:blogger.com,1999:blog-9124539381685751273.post-61319470783829877042008-08-19T14:56:00.000-07:002008-08-19T14:56:00.000-07:00Hint!It is proof by contradiction, just like befor...Hint!<BR/><BR/>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.millerhttps://www.blogger.com/profile/05990852054891771988noreply@blogger.com