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 Angry Friends (Posted on 2004-08-17)
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?

 No Solution Yet Submitted by ThoughtProvoker Rating: 3.6667 (6 votes)

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 New Solution | Comment 5 of 15 |

I don't know if this is THE maximized distance, but I think it is larger than what has been found so far.

I placed two players on two opposite corners of the gym.  Then I placed the other 4 players such that one was on each side of the gym.

To maximize the distances, I placed them such that they were all equidistant to each other.  So I am basically dealing with a hexagon with equal sides, but NOT equal angles.  Almost like chopped off two opposite corners of the square.  By symmetry, I am chopping off a 45 degree angle.

My basic equation to solve is X (the distance a long a side of the square between a corner and side player) + Y (the leg of a 45 degree right triangle with hypotenuse = X) = 300 ft = X + X*cos(45) = X * (1 + 1/sqrt(2))

So X = 300ft / (1 + 1/sqrt(2)) = 175.736 ft.

 Posted by nikki on 2004-08-17 09:57:31

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