Crayons

Originally posted by LemonJello
I think Anthem's latest suggestion with hexagons will still work. We just need a proviso analogous to the previous ones (such as, the terminal points on the 1 unit length diagonals you mention would take the same color as the point directly to their left).

EDIT: To be more clear, here I should have specified "region" (not "point" ) directly to the left.

I was browsing in the math section of my university's library today, and I found a 600 page book on none other than this problem (well, almost. The book was using an infinite plane, rather than a 5x5 square). I skipped to the end, wondering what the answer was... and...

The answer is either 4, 5, 6, or 7 (colors, not crayons). No one knows.

The proof that 3 does not work is definitely doable, though, so I won't post the answer in case someone wants to try to figure it out. (If you just want to see the proof, google "Moser Spindle" )

Originally posted by Anthem
I was browsing in the math section of my university's library today, and I found a 600 page book on none other than this problem (well, almost. The book was using an infinite plane, rather than a 5x5 square). I skipped to the end, wondering what the answer was... and...

The answer is either 4, 5, 6, or 7 (colors, not crayons). No one knows.

The proof t wants to try to figure it out. (If you just want to see the proof, google "Moser Spindle" )

Hmmm...okay. I did not know so much study had already been devoted to this problem. I could not figure it out, either. But I did not really want to admit it. 🙂

EDIT: Indeed, it seems you are correct. I guess this is known as the Hadwiger-Nelson problem.