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Previous
Puzzles of the Month + Solutions


April
 May 2010, Puzzle 124 
Back to PuzzleoftheMonth
page  Home 
Puzzle
# 124


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: ,
basic geometry knowledge.
Category:
dividingtheplane puzzle.
Keywords:
dissection, tiling, rectangles.
Related
puzzles:
 Squared
strip,
 Triangles
to square.
 Boia
che problema!
Aiutandoti con i quadratini di colore, ritaglia la
tavola qui sopra in 7 pezzi rettangolari in modo
che nessun pezzo possa contenerne un altro (ad esempio,
un rettangolo di 6x8 può contenerne uno di
4x7).
Parole
chiave: scomposizione geometrica, rettangoli.
Suggerisci un'altra
soluzione Chiudi
 Découpe à la
hache...
En utilisant les carrés comme guide, découpe
la table cidessus en 7 pièces rectangulaires
de sorte à ce qu’aucune pièce
ne puisse en contenir une autre (par exemple, un
rectangle de 6x8 carrés recouvre parfaitement
un rectangle de 4x7 carrés).
Mots
clés: découpage, recouvrement.
Propose une
autre solution Fermer

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



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 units^{2} rectangle, we will proceed empirically
as follows:
Rectangle 1: 1 x ... units^{2}
Rectangle 2: 2 x ... units^{2}
Rectangle 3: 3 x ... units^{2}
Rectangle 4: 4 x ... units^{2}
Rectangle 5: 5 x ... units^{2}
Rectangle 6: 6 x ... units^{2}
Rectangle 7: 7 x ... units^{2}
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)

Aside from rotations and reflections this tiling of
rectangles is unique
The
5 Winners of the Puzzle of the Month are:
Herbert Jones, USA  Charles
F. Espenlaub, USA  Cesco
Reale, Italy  Muhammad
Afifi, Egypt  Martin
Rick, South Africa
Congratulations!



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.archimedeslab.org
You can reuse 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! 




