bonanova Posted October 12, 2013 Report Share Posted October 12, 2013 What is the area of the smallest square that holds five unit squares without overlap? Quote Link to comment Share on other sites More sharing options...

0 harey Posted October 13, 2013 Report Share Posted October 13, 2013 (edited) Place four unit squares into the corners of the resulting square and the 5th one into the middle after a 45 degree rotation. The diagonal of the new square: d=sqrt(2)+1+sqrt(2) giving the surface 4.5+2*sqrt(2) (about 7.33). Edited October 13, 2013 by harey Quote Link to comment Share on other sites More sharing options...

0 dgreening Posted October 13, 2013 Report Share Posted October 13, 2013 I think the smallest square would be a 3 x 3. If we start with the most compact area to contain 4 unit squares, it would be a 2 x 2 square. The addition of another unit square will force it to add at least one row or column - thus forcing you to a 2x3 configuration. To meet the criteria of the problem, we must expand that to a 3 x 3 There are multiple configurations [many are just reflections of others # # # # # - - - - or # # # # - - # - - or # - # - # - # - # or ... Quote Link to comment Share on other sites More sharing options...

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What is the area of the smallest square that holds five unit squares without overlap?

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