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Angry Friends (Posted on 2004-08-17) |
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Six friends were playing a game of indoor soccer together, but got into a fight during their game. Upset with each other, they decided to position themselves in the (square) 300x300 foot gymnasium so as to maximize the distance between the closest pair. Where should they each stand?
solution
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Comment 15 of 15 |
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o--------+--------o The distance between the closest |A | B | should be 50*SQRT(13) feet, id | | | est, approximately 180.2775638 +--------o--------+ feet. The distances between | |C | friends would then be: | | | A&B:300 A&C:50*SQRT(13) o--------+--------o A&D:200 A&E:100*SQR(13) |D | E| A&F:150*SQRT(5) B&C:50*SQRT(13) | | | B&D:100*SQRT(13) B&E:200 +--------o--------+ B&F:150*SQRT(5) C&D:50*SQRT(13) F C&E:50*SQRT(13) C&F:200 D&E:300 D&F:50*SQRT(13) E&F:50*SQRT(13)
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Posted by Dej Mar
on 2010-08-06 10:20:53 |
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