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

Home > Just Math
Cover the rectangle (Posted on 2019-01-10) Difficulty: 3 of 5
You have N rectangles (N > 1). The 2N numbers used by the length of short and long edges of these rectangles are all different positive integers.

You can create a larger rectangle by using all of these rectangles. The large rectangle is fully covered without any overlap or overflow of the smaller rectangles. What is the minimum possible area of the large rectangle?

Note: The large rectangle cannot be a square.

No Solution Yet Submitted by Danish Ahmed Khan    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
a guess | Comment 1 of 2
I think it may be impossible. (I could well be wrong.) I can almost see a proof: Two rectangles of different heights, side by side, have a gap at either their top or bottom or both. Adding yet another rectangle to fill a gap must produce yet a new gap. This is true for the sides aligned or the bases and tops aligned. Since new gaps are always made: there can be no closure to make a uniform rectangle. Perhaps....   
------------------------------------------------------------------
A comment added later, after the solution by B. Smith:

Surely his is the right answer since the integers are all minimal. 

My logical error was not seeing that the continuing "gap" can circle and the last added rectangle fills the first and last gap. 

Kudos, B. Smith! 

Edited on January 11, 2019, 1:55 am
  Posted by Steven Lord on 2019-01-10 23:04:33

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 (12)
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