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

Home > Shapes > Geometry
Quadrilateral fun (Posted on 2004-01-10) Difficulty: 4 of 5
Begin with an arbitrary, convex quadrilateral. Next, draw squares outwardly on the sides of the quadrilateral, and join the centers of opposite squares.

You might find the the two resulting lines are equal in length and intersect at precisely 90 degrees.

Prove (or disprove) the notion, that this is always true.

See The Solution Submitted by SilverKnight    
Rating: 4.0000 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: Proof | Comment 3 of 6 |
(In reply to Proof by Federico Kereki)

It's interesting that you used the work of a famous mathematician to solve this one. Just the other day, CNN announced that another noted mathematician, a Russian named Grigori Perelman, now has a definite answer for "Poincare's Conjecture", which ranks right up there with "Fermat's Last Theorem" among famous unsolved (until recently) mathematical problems. Perelman stands to get a cool million dollars for his achievement, although I'm sure he was motivated by a love of pure mathematics rather than a desire for monetary gain. Details of Perelman's coup can be found at:

http://www.theage.com.au/articles/2003/08/02/1059480600718.html
Edited on January 10, 2004, 12:45 pm
  Posted by Penny on 2004-01-10 12:42:02

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


Search:
Search body:
Forums (1)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (6)
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