BMAD Posted September 28, 2013 Report Share Posted September 28, 2013 What is the radius of the smallest circle that can enclose all 52 non-overlapping cards of an ordinary deck of playing cards? And what is the configuration of the cards? Assume the smaller dimension of the cards is 5 and the larger is 7. Quote Link to comment Share on other sites More sharing options...

0 superprismatic Posted September 28, 2013 Report Share Posted September 28, 2013 Make 9 rows of cards with each row centered on the ones vertically adjacent to it: Row 1: 3 cards abutted on their length 5 sides (row has length 21) Row 2: 5 cards abutted on their length 5 sides (row has length 35) Row 3: 6 cards abutted on their length 5 sides (row has length 42) Row 4: 7 cards abutted on their length 5 sides (row has length 49) Row 5: 10 cards abutted on their length 7 sides (row has length 50) Row 6: 7 cards abutted on their length 5 sides (row has length 49) Row 7: 6 cards abutted on their length 5 sides (row has length 42) Row 8: 5 cards abutted on their length 5 sides (row has length 35) Row 9: 3 cards abutted on their length 5 sides (row has length 21) The radius of the smallest circle containing the cards is 25.9326. Quote Link to comment Share on other sites More sharing options...

0 phil1882 Posted September 28, 2013 Report Share Posted September 28, 2013 the best i can do a 6x5 rectangle, with an additional 6 on the top and bottom, and 5 left and right. gives a circle of radius 31.9 Quote Link to comment Share on other sites More sharing options...

0 BMAD Posted September 28, 2013 Author Report Share Posted September 28, 2013 Make 9 rows of cards with each row centered on the ones vertically adjacent to it: Row 1: 3 cards abutted on their length 5 sides (row has length 21) Row 2: 5 cards abutted on their length 5 sides (row has length 35) Row 3: 6 cards abutted on their length 5 sides (row has length 42) Row 4: 7 cards abutted on their length 5 sides (row has length 49) Row 5: 10 cards abutted on their length 7 sides (row has length 50) Row 6: 7 cards abutted on their length 5 sides (row has length 49) Row 7: 6 cards abutted on their length 5 sides (row has length 42) Row 8: 5 cards abutted on their length 5 sides (row has length 35) Row 9: 3 cards abutted on their length 5 sides (row has length 21) The radius of the smallest circle containing the cards is 25.9326. I would like to see what you mean. Quote Link to comment Share on other sites More sharing options...

0 superprismatic Posted September 28, 2013 Report Share Posted September 28, 2013 Make 9 rows of cards with each row centered on the ones vertically adjacent to it: Row 1: 3 cards abutted on their length 5 sides (row has length 21) Row 2: 5 cards abutted on their length 5 sides (row has length 35) Row 3: 6 cards abutted on their length 5 sides (row has length 42) Row 4: 7 cards abutted on their length 5 sides (row has length 49) Row 5: 10 cards abutted on their length 7 sides (row has length 50) Row 6: 7 cards abutted on their length 5 sides (row has length 49) Row 7: 6 cards abutted on their length 5 sides (row has length 42) Row 8: 5 cards abutted on their length 5 sides (row has length 35) Row 9: 3 cards abutted on their length 5 sides (row has length 21) The radius of the smallest circle containing the cards is 25.9326. I would like to see what you mean. I wish I could make a picture of it, but I have never done anything like that. I've only used text. Perhaps a text picture will help. Each 5x7 card is represented by a 5x7 matrix containing 35 instances of a single digit. First and second rows: 111111122222223333333 111111122222223333333 111111122222223333333 111111122222223333333 111111122222223333333 44444445555555666666677777778888888 44444445555555666666677777778888888 44444445555555666666677777778888888 44444445555555666666677777778888888 44444445555555666666677777778888888 I can't do the third row because the next card would have half of its 7-long side having blank space above it, and the other half having half of card 4 above it. Quote Link to comment Share on other sites More sharing options...

0 phil1882 Posted September 28, 2013 Report Share Posted September 28, 2013 here's a picture of superprismatics solution. Quote Link to comment Share on other sites More sharing options...

0 superprismatic Posted September 28, 2013 Report Share Posted September 28, 2013 here's a picture of superprismatics solution. cardscircle.JPG Thats a nice picture, phil That's a nice picture, phil. It's missing the two six-long rows (3rd & 7th). Would you add them? I'm impressed that you could do such a good job on the two rows closest to the middle row, as they are each 49 units long whereas the middle row is 50 -- making it difficult to center them. Quote Link to comment Share on other sites More sharing options...

0 phil1882 Posted September 28, 2013 Report Share Posted September 28, 2013 sorry should have read closer Quote Link to comment Share on other sites More sharing options...

0 BMAD Posted September 28, 2013 Author Report Share Posted September 28, 2013 well done Phil. I was having a hard time visualizing what Superprismatic meant. Quote Link to comment Share on other sites More sharing options...

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What is the radius of the smallest circle that can enclose all 52 non-overlapping cards of an ordinary deck of playing cards? And what is the configuration of the cards? Assume the smaller dimension of the cards is 5 and the larger is 7.

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