All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Shapes > Geometry
2 Colors 2 (Posted on 2003-08-25) Difficulty: 4 of 5
Suppose you have an infinite plane, and each point on the plane has been arbitrarily painted one of two colors.

Prove that there exists an equilateral triangle whose vertices are all the same color.

What is the fewest number of points needed to prove this?

See The Solution Submitted by DJ    
Rating: 4.3684 (19 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(4): 5-point proof | Comment 8 of 14 |
(In reply to re(3): 5-point proof by Kelsey)

The very first part of the proof is that you start with two points that are different in color. In your diagram, those two points have the same color.

The proof lies not only on the number of points or even solely by virtue of their arrangement, but it hinges on how you select the points.

You have to start with two differently-colored points, and similarly, the last two points chosen depend on whatever the color of the midpoint happens to be.

The problem doesn't say that all points have to be randomly selected, and indeed, they cannot be.
  Posted by DJ on 2003-08-26 21:14:49

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information