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Previous Puzzles of the Month + Solutions

 
April - May 2010, Puzzle 124
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Send to a Friend Puzzle # 124

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story
 A chopping problem
  
Using the checkered pattern as a guide, cut out the board below into 7 rectangular pieces so that no piece can contain another one (for instance, a 6x8 rectangle completely covers a 4x7 rectangle).

Difficulty level: bulbbulb, basic geometry knowledge.
Category: dividing-the-plane puzzle.
Keywords: dissection, tiling, rectangles.
Related puzzles:
- Squared strip,
- Triangles to square.


Source of the puzzle:
© G. Sarcone.
You cannot reproduce any part of this page without prior written permission.
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Solution

So, each of the 7 rectangles should have one side WIDER and one side SHORTER to each other rectangle, so that neither of the rectangles can be placed inside the other one in such a way that corresponding sides are parallel.
Since we have to find 7 rectangles that tile a 22 x 13 units2 rectangle, we will proceed empirically as follows:
Rectangle 1: 1 x ... units2
Rectangle 2: 2 x ... units2
Rectangle 3: 3 x ... units2
Rectangle 4: 4 x ... units2
Rectangle 5: 5 x ... units2
Rectangle 6: 6 x ... units2
Rectangle 7: 7 x ... units2

Rectangle 1 should have the widest surface since it has the shortest height; however, the possible dimensions 1 x 22 and 1 x 21 should be discarted [any rectangle with sides n ≤ 13 x (22 - 21) would fit inside the latter one]... Then, proceeding by trial and error we obtain the following tiling:

1 x 18
2 x 16
3 x 13
4 x 11
5 x 10
6 x 9
7 x 7
(see image below)

solution 1

Aside from rotations and reflections this tiling of rectangles is unique


cup winnerThe 5 Winners of the Puzzle of the Month are:
Herbert Jones, USA USA flag - Charles F. Espenlaub, USA USA flag - Cesco Reale, Italy Italian flag - Muhammad Afifi, Egypt Egyptian flag - Martin Rick, South Africa South African flag

Congratulations!

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Math fact behind the puzzle
Incomparable Rectangles
Two rectangles, neither of which will fit inside the other, are said to be 'incomparable' (this is equivalent to one rectangle being both longer and narrower). The minimum possible number of incomparable rectangles needed to tile a larger rectangle is 7
.

 

© 2006 G. Sarcone, www.archimedes-lab.org
You can re-use content from Archimedes’ Lab on the ONLY condition that you provide credit to the authors (© G. Sarcone and/or M.-J. Waeber) and a link back to our site. You CANNOT reproduce the content of this page for commercial purposes.

You're encouraged to expand and/or improve this article. Send your comments, feedback or suggestions to Gianni A. Sarcone. Thanks!
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Previous puzzles of the month...
contents+solutions
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Solved Puzzles

Oct-Nov 09:
The Mark of Zorro
July-Sept 09:
radiolarian's shell
May-June 09:
circle vs square

Jan-Feb 09
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geometric mouse

Sept-Oct 08
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perpendicular or not...

July-Aug 08:
ratio of triangles

May-June 08:
geometry of the bees

Febr-March 08
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parrot sequence...

Dec 07-Jan 08
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Oct-Nov 2007
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Aug-Sept 07:
indecisive triangle

June-July 07:
Achtung Minen!

April-May 07:
soccer balls

Febr-March 07:
prof Gibbus' angle

Jan 07:
triangles to square

Aug-Sept 2006
:
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June-July 06
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squared strip

Apr-May 06:
intriguing probabilities

Febr-March 06:
cows & chickens
Dec 05-Jan 06:
red monad

Sept-Oct 2005
:
magic star

July-Aug 05:
cheese!

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