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Can you fully cover a square with 7 rectangles such that:

  1. Every rectangle has 2:1 ratio, ie., length double its width.
  2. No part of any rectangle is outside the square.
  3. No two rectangles overlap.

Note that rectangles can have different size. This puzzle is from a Numberphile video (see link in comments) and is possibly well known, but I haven't seen it here.

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2 Answers 2

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Is it possible?

Yes

Why?

6x6 is the smallest possible area that can be divided into 7 possibly distinct X by 2X rectangles (36 = 18 + 8 + 5*2). And the following layout works:

 AAAAAA
 AAAAAA
 AAAAAA
 BBBBCC
 BBBBDD
 EEFFGG
 

Perhaps the intent is that the rectangles must all be different sizes, which would make it less trivial to solve.

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  • $\begingroup$ Correct and well done! Ok can you do it when all rectangles are different size? $\endgroup$ Jan 7 at 22:03
  • $\begingroup$ Here is the cool Numberphile video about this puzzle: youtube.com/watch?v=VZ25tZ9z6uI $\endgroup$ Jan 8 at 0:51
  • $\begingroup$ @DmitryKamenetsky: The video you link to explicitly allows matching sizes. It starts with dividing a square into two $1 \times 2$ rectangles of the same size. $\endgroup$ Jan 8 at 5:46
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    $\begingroup$ @Ross yes the original question allowed duplicate sizes. Now we are making it harder for fun :) $\endgroup$ Jan 8 at 6:14
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By visual inspection of the diagram below it can be seen that a 3x3 square is divided into 7 rectangles whose sides are in 2:1 ratio.

rectangles within square

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